Free math worksheets

- Look for the point where the two lines cross. - Read both coordinates of that intersection from the grid.

1. A solution of a system is a point that lies on both lines. 2. The two lines intersect at \((2, 3)\). 3. Therefore, the system's solution is \((2, 3)\).

\((2, 3)\)

- One coordinate is already given directly by the first equation. - Substitute that value into the other equation to find the remaining coordinate.

1. The first equation gives \(x = 6\). 2. Substitute \(6\) for \(x\) in the second equation: \(6 + y = 10\), so \(y = 4\). 3. The solution is \((6, 4)\).

\((6, 4)\)

- Look for a pair of variable terms that cancel when the equations are added. - After one variable is eliminated, solve the resulting one-variable equation. - Substitute the value you find into either original equation.

1. Add the equations. The \(y\)-terms cancel, giving \(2x=8\). 2. Divide by \(2\): \(x=4\). 3. Substitute into \(x+y=7\): \(4+y=7\), so \(y=3\). 4. The solution is \((4, 3)\).

Adding the equations gives \(2x=8\). Therefore, the solution is \((4, 3)\).

- At an intersection, the two equations have the same x-value and y-value. - After solving for x, substitute into either equation to find y. - A system solution must satisfy both original equations.

1. Set the equations equal: \(2x-1=-x+5\). Then \(3x=6\), so \(x=2\). Substituting gives \(y=3\), so \(S=(2,3)\). 2. For \(g_1\), \(2(2)-1=3\). For \(g_2\), \(-2+5=3\). Thus \((2,3)\) satisfies both equations.

a) \(S=(2, 3)\) b) Yes. Substitution gives \(y=3\) in both equations.

- What equation can you write when two functions have the same output at the same x-value? - After finding x, substitute it into one equation to find y. - A solution to a system must satisfy both original equations.

1. At the intersection, \(h(x)=k(x)\). Solve \(0.8x+2=-1.2x+6\), giving \(2x=4\) and \(x=2\). 2. Substitute \(x=2\): \(h(2)=0.8\cdot2+2=3.6\). Thus the intersection is \((2,3.6)\). 3. Check the second equation: \(k(2)=-1.2\cdot2+6=3.6\). The point satisfies both equations.

\((2, 3.6)\)

- What must be true about the two function values at an intersection? - Move all x-terms to one side before solving. - Multiplying by a common denominator may make the fraction equation easier.

1. Set the function values equal: \(\frac{2}{5}x + 3 = -\frac{1}{2}x - 6\). 2. Combine the x-terms: \(\frac{2}{5}x + \frac{1}{2}x = -9\), so \(\frac{9}{10}x = -9\). 3. Solving gives \(x = -10\). 4. Substitute into \(f\): \(y = \frac{2}{5} \cdot (-10) + 3 = -1\). 5. Therefore, \(S = (-10, -1)\).

\(S = (-10, -1)\)

- What must be true about the two function values at an intersection? - What y-coordinate does every point on the x-axis have? - Use your result from part a) to justify part b).

1. Set the function values equal: \(3x-9=-x+3\). Then \(4x=12\), so \(x=3\). Substituting gives \(k(3)=0\), so the intersection point is \((3,0)\). 2. A point lies on the x-axis exactly when its y-coordinate is \(0\). Since the intersection has y-coordinate \(0\), it lies on the x-axis.

a) \((3, 0)\) b) The intersection lies on the x-axis because its y-coordinate is \(0\).

- At an intersection, what must be true about the two y-values for the same x-value? - After solving for x, substitute into either equation to find y. - Check the ordered pair in the other equation.

1. Set the equations equal: \(-2x+6=0.5x-4\). Then \(10=2.5x\), so \(x=4\). 2. Substitute \(x=4\): \(y=-2(4)+6=-2\). The intersection is \((4,-2)\). 3. Check \(h(4)=0.5(4)-4=-2\), so the point satisfies both equations.

\((4, -2)\)

- At an intersection, what must be true about the two function values? - After finding the common x-value, substitute it into either function. - Which coordinate of the intersection gives the solution of \(f(x)=g(x)\)?

1. Solve \(2x-3=-x+6\). Then \(3x=9\), so \(x=3\). Substituting gives \(y=3\), so the intersection is \((3,3)\). 2. At the intersection, both functions have the same input and output. Here both equal \(3\) when \(x=3\), so \(x=3\) is the solution of the equation \(f(x)=g(x)\).

a) \(x=3\), and the intersection point is \((3, 3)\). b) The intersection shows the input where the two function values are equal.

- What must be true about the two function values at an intersection? - After finding the common x-value, how can you find the y-coordinate? - Recall the signs of coordinates in each quadrant.

1. Set the function values equal: \(1.5x - 2 = -x + 3\). 2. Then \(2.5x = 5\), so \(x = 2\). 3. Substitute into \(g\): \(g(2) = -2 + 3 = 1\). Thus \(S = (2, 1)\). 4. Both coordinates are positive, so \(S\) lies in Quadrant I.

\(S = (2, 1)\), in Quadrant I.

- Use one equation for the total number of tickets and another for the total cost. - Solve the simpler equation for one variable. - Substitute that expression into the cost equation.

1. Let \(x\) be the number of youth tickets and \(y\) the number of adult tickets. The system is \(x+y=25\) and \(12x+18y=360\). 2. From \(x+y=25\), write \(x=25-y\). 3. Substitute: \(12(25-y)+18y=360\). Then \(300+6y=360\), so \(y=10\). 4. Substitute into \(x+y=25\): \(x=15\).

\(15\) youth tickets and \(10\) adult tickets

- Look for the point shared by both lines. - Read the x-coordinate and y-coordinate from the grid. - Write the intersection as an ordered pair.

1. A solution of a graphed system is a point that lies on both lines. 2. The two lines intersect at \((2, 1)\). 3. Therefore, the solution of the system is \((2, 1)\).

\((2, 1)\)

- Find the point where both lines pass through the same grid location. - Read both coordinates carefully, including the sign of the y-coordinate. - How many intersection points do the two distinct nonparallel lines have?

1. Read the common point of the two lines from the coordinate grid: \((2, -1)\). 2. The lines cross only once in the plane. 3. Therefore, the system has the unique solution \((2, -1)\).

The solution is \((2, -1)\). It is unique because the two lines intersect at exactly one point.

- The system's solution is the common point of the two lines. - Use the grid to read the x-coordinate first and then the y-coordinate. - Write a one-point solution set using the ordered pair.

1. The solution must lie on both graphed lines. 2. The lines intersect at \((2, 3)\). 3. Therefore, the solution set is \(\{(2, 3)\}\).

The solution set is \(\{(2, 3)\}\).

- The two base angles of an isosceles triangle are congruent. - The three interior angles of a triangle sum to \(180^\circ\). - Express the vertex angle using one base-angle variable.

1. Let \(b\) be the measure of each base angle. 2. The vertex angle is \(b+15^\circ\). 3. The interior angles of a triangle sum to \(180^\circ\), so \(2b+(b+15^\circ)=180^\circ\). 4. Simplify: \(3b+15^\circ=180^\circ\), so \(b=55^\circ\). 5. The vertex angle is \(55^\circ+15^\circ=70^\circ\).

The two base angles are \(55^\circ\) each, and the vertex angle is \(70^\circ\).

- The two acute angles in a right triangle are complementary. - Express one acute angle in terms of the other. - Substitute that expression into the \(90^\circ\) angle sum.

1. The acute angles of a right triangle sum to \(90^\circ\). 2. Since \(\alpha=3\beta+12^\circ\), write \((3\beta+12^\circ)+\beta=90^\circ\). 3. Simplify: \(4\beta+12^\circ=90^\circ\), so \(\beta=19.5^\circ\). 4. Then \(\alpha=90^\circ-19.5^\circ=70.5^\circ\).

\(\alpha=70.5^\circ\) and \(\beta=19.5^\circ\).

- A graphical solution is the point common to both lines. - Read the x-coordinate from the horizontal axis and the y-coordinate from the vertical axis. - Check that the intersection lies exactly on grid lines.

1. Locate the point where the two graphed lines intersect. 2. The intersection is at \((1, 2)\). 3. Therefore, the system's solution is \((1, 2)\).

\((1, 2)\)

- Look for the crossing point of the two lines. - Use the tick labels to read its coordinates. - Write the x-coordinate first.

1. Find the point that lies on both graphed lines. 2. The lines meet at \((2, 2)\). 3. Therefore, the solution is \((2, 2)\).

\((2, 2)\)

- Find where each line crosses the vertical axis to identify its \(y\)-intercept. - Use two clearly readable points on each line to determine its slope. - The intersection of the two lines is the ordered pair that satisfies both equations.

1. Line \(g\) has \(y\)-intercept \(1\). Using the points \((0, 1)\) and \((1, 2)\), its slope is \(1\), so its equation is \(y = x + 1\). 2. Line \(h\) has \(y\)-intercept \(4\). Using the points \((0, 4)\) and \((2, 3)\), its slope is \(-0.5\), so its equation is \(y = -0.5x + 4\). 3. The lines intersect at \((2, 3)\). 4. Therefore, the solution set is \(\{(2, 3)\}\).

(I) \(y = x + 1\); (II) \(y = -0.5x + 4\); solution set: \(\{(2, 3)\}\)

- The y-intercepts give the initial heights. - The magnitude of each negative slope is the burn rate. - The intersection represents equal heights at the same time.

1. Candle A begins at \(6\) inches and decreases at \(0.75\,\text{in./h}\), so \(A(x)=-0.75x+6\). 2. Candle B begins at \(9\) inches and decreases at \(1.5\,\text{in./h}\), so \(B(x)=-1.5x+9\). 3. Set the functions equal: \(-0.75x+6=-1.5x+9\). Then \(0.75x=3\), so \(x=4\). 4. Substitute \(x=4\) into either function to get \(y=3\). The candles are both \(3\) inches tall after \(4\) hours.

a) Candle A: initial height \(6\,\text{in.}\), burn rate \(0.75\,\text{in./h}\); Candle B: initial height \(9\,\text{in.}\), burn rate \(1.5\,\text{in./h}\) b) After \(4\) hours, both are \(3\) inches tall.

- The y-intercepts are the starting balances. - The slopes are the weekly saving rates. - The intersection represents equal balances.

1. Sarah starts with \(\$40\), while Luke starts with \(\$10\), so Sarah has more initially. 2. Luke's balance increases by \(\$5\) per week, while Sarah's increases by \(\$2\) per week, so Luke saves more each week. 3. Their equations are \(L(x)=5x+10\) and \(S(x)=2x+40\). 4. Set them equal: \(5x+10=2x+40\), so \(3x=30\) and \(x=10\). 5. The common balance is \(5\cdot10+10=60\), or \(\$60\).

a) Sarah starts with more; Luke saves more each week, \(\$5\) per week versus \(\$2\) per week. b) After \(10\) weeks; \(\$60\)

- A point must lie on every line to solve the entire system. - Find the one location shared by all three lines. - Read the coordinates from the grid.

1. A solution of the three-equation system must lie on all three graphed lines at once. 2. The three lines share the point \((2, 2)\). 3. Therefore, the solution is \((2, 2)\).

\((2, 2)\)

- Choose one variable and compare its coefficients in the two equations. - What single multiplication would make those coefficients opposites? - After adding the equations, solve the resulting one-variable equation. - Substitute back into an original equation to find the other coordinate.

1. Multiply the second equation by \(-2\): \(-2x-4y=-20\). 2. Add this equation to \(2x+y=11\). The \(x\)-terms cancel, giving \(-3y=-9\). 3. Divide by \(-3\): \(y=3\). 4. Substitute into \(x+2y=10\): \(x+6=10\), so \(x=4\). 5. The solution is \((4, 3)\).

One valid setup is to multiply the second equation by \(-2\), giving \(-2x-4y=-20\). Adding it to the first equation gives \(-3y=-9\). Therefore, the solution is \((4, 3)\).

- Translate each purchase into an equation. - Use a system-solving method that eliminates one variable. - Scale one equation so one pair of variable coefficients matches.

1. Let \(x\) be the price of one watermelon and \(y\) the price of one mango, in dollars. 2. Write the system \(4x + 6y = 52\) and \(5x + 3y = 47\). 3. Multiply the second equation by \(2\): \(10x + 6y = 94\). 4. Subtract the first equation: \(6x = 42\), so \(x = 7\).

d) \(\$7\)

- For each runner, identify the starting position and the distance added each hour. - A linear position model has the form starting position plus rate times time. - At the catch-up time, what must be true about the two position values?

1. Alex starts at \(0\) and gains \(8\) miles each hour, so \(s_A(t)=8t\). 2. Blake starts at mile marker \(2\) and gains \(6\) miles each hour, so \(s_B(t)=2+6t\). 3. At the catch-up time the positions are equal: \(8t=2+6t\). Thus \(2t=2\), so \(t=1\) hour. 4. Substitute \(t=1\): \(s_A(1)=8\) and \(s_B(1)=8\). Alex catches Blake at mile marker \(8\).

\(s_A(t)=8t\) and \(s_B(t)=2+6t\). Alex catches Blake after \(1\) hour at mile marker \(8\).

- Represent each plan with a cost equation using the same variables. - At the intersection, the same ordered pair satisfies both equations. - Once you find the break-even point, compare the plans' per-minute rates to decide which is cheaper afterward.

1. At \(30\) minutes, QuickRide costs \(30\cdot\$0.30=\$9.00\) and CityCar costs \(\$10.00+30\cdot\$0.10=\$13.00\). At \(60\) minutes, QuickRide costs \(60\cdot\$0.30=\$18.00\) and CityCar costs \(\$10.00+60\cdot\$0.10=\$16.00\). 2. Let \(x\) be monthly driving minutes and \(y\) be monthly cost. The system is \(y=0.30x\) and \(y=10+0.10x\). Equating the expressions gives \(0.30x=10+0.10x\), so \(0.20x=10\) and \(x=50\). Then \(y=0.30\cdot50=15\). 3. The lines intersect at \((50,15)\). Because CityCar has the smaller per-minute rate, it is less expensive for more than \(50\) minutes of monthly driving. 4. The solution means both plans cost \(\$15.00\) at \(50\) minutes of monthly driving.

a) At \(30\) minutes: QuickRide \(\$9.00\), CityCar \(\$13.00\). At \(60\) minutes: QuickRide \(\$18.00\), CityCar \(\$16.00\). b) \(y=0.30x\) and \(y=10+0.10x\); the intersection is \((50,15)\). CityCar is less expensive for more than \(50\) minutes. c) At \(50\) minutes of monthly driving, both plans cost \(\$15.00\).

- Express the mother’s current age in terms of the daughter’s current age. - Write a second equation for their ages 6 years from now. - Both ages increase by the same amount. - Substitute one age relationship into the other equation.

1. Let \(L\) be Leah’s current age and \(M\) be Ms. Miller’s current age. 2. The current-age relationship is \(M = 4L\). 3. The relationship 6 years from now is \(M + 6 = 3(L + 6)\). 4. Substitute \(M = 4L\) into the second equation: \(4L + 6 = 3(L + 6)\). 5. Distribute: \(4L + 6 = 3L + 18\). 6. Subtract \(3L\) and then subtract \(6\): \(L = 12\). 7. Find Ms. Miller’s age: \(M = 4 \cdot 12 = 48\).

Leah is 12 years old, and Ms. Miller is 48 years old.

- At the break-even point, what must be true about the two total-cost expressions? - After finding the common mileage, substitute it into either cost equation. - Compare the per-mile rates to decide which plan becomes cheaper after the intersection.

1. At the break-even point, both equations have the same \(x\) and \(y\), so set \(0.30x+45=0.50x+25\). 2. Subtract \(0.30x\) and \(25\): \(20=0.20x\), so \(x=100\). 3. Substitute \(x=100\) into either equation: \(y=0.30\cdot100+45=75\). The break-even point is \((100,75)\). 4. Plan A has the higher starting fee but smaller per-mile rate, so Plan B is less expensive for trips shorter than \(100\) miles, and Plan A is less expensive for trips longer than \(100\) miles.

The break-even point is \((100,75)\). Plan B is less expensive for trips under \(100\) miles; Plan A is less expensive for trips over \(100\) miles.

- At an intersection point, both functions have the same input and output. - Compare the two function values for each listed x-value. - Solve the equation for \(x\), then substitute to find \(y\). - Check the point in both equations.

1. The values are: <table><tr><th>\(x\)</th><th>0</th><th>1</th><th>2</th><th>3</th></tr><tr><td>\(g(x)\)</td><td>4</td><td>3</td><td>2</td><td>1</td></tr><tr><td>\(h(x)\)</td><td>-2</td><td>0</td><td>2</td><td>4</td></tr></table> Both functions have value \(2\) when \(x=2\), so the intersection is \((2,2)\). 2. Verify algebraically: \(-x+4=2x-2\). Adding \(x\) and then \(2\) gives \(6=3x\), so \(x=2\). 3. Substitute into either function: \(g(2)=-2+4=2\). Thus, the intersection point is \((2,2)\).

The intersection point is \((2, 2)\).

- How do you find the slope between two points? - Which given point reveals the y-intercept directly? - What condition do the function values satisfy at an intersection?

1. The slope of \(g\) is \(m=\frac{5-1}{2-0}=2\). Point \(P(0,1)\) gives the y-intercept \(1\), so \(g\) has equation \(y=2x+1\). 2. Set the equations equal: \(2x+1=-x+7\). Then \(3x=6\), so \(x=2\). Substituting gives \(y=5\), so \(S=(2,5)\).

a) \(y=2x+1\) b) \(S=(2, 5)\)

- If you know the x-coordinate of the intersection, how can you find its y-coordinate? - What must be true about that point for the second line? - Substitute the point into the equation containing the unknown slope.

1. Find the y-value of the intersection using \(f\): \(f(-2) = \frac{1}{2} \cdot (-2) + 1 = 0\). 2. Thus the intersection point is \((-2, 0)\). 3. Substitute this point into \(g(x) = mx + 4\): \(0 = -2m + 4\). 4. Then \(2m = 4\), so \(m = 2\).

\(m = 2\)

- Start by finding where the first two lines have equal function values. - How can you check whether a given point lies on another line? - Substitute the x-coordinate of the intersection into \(h\) and compare the y-value.

1. Set \(f(x) = g(x)\): \(3x - 4 = -x + 8\). 2. Then \(4x = 12\), so \(x = 3\). 3. Find the y-value: \(f(3) = 3 \cdot 3 - 4 = 5\). Thus \(S = (3, 5)\). 4. Check \(h\) at \(x = 3\): \(h(3) = 2 \cdot 3 - 1 = 5\). 5. Since \(h(3) = 5\), the graph of \(h\) also passes through \(S\).

The intersection of \(f\) and \(g\) is \(S = (3, 5)\). The graph of \(h\) also passes through \(S\).

- Use the fully known function to find the y-value at \(x = 4\). - What must be true about both functions at their intersection? - Use the y-value you found to solve for the missing parameter.

1. Use \(g\) to find the y-value of the intersection: \(g(4) = -1.2 \cdot 4 + 10 = 5.2\). 2. Since the point is on both lines, \(f(4) = 5.2\). 3. Substitute into \(f\): \(0.8 \cdot 4 + b = 5.2\). 4. Then \(3.2 + b = 5.2\), so \(b = 2\). 5. The intersection point is \((4, 5.2)\).

\(b = 2\) and the intersection point is \((4, 5.2)\).

- How do you find a line's slope from two points? - How can you write a line equation from a point and a slope? - Once you have both equations, how do you find their common point?

1. For \(h_1\), the slope is \(m_1 = \frac{9 - 5}{4 - 2} = 2\). Using \(A(2, 5)\), \(5 = 2 \cdot 2 + b_1\), so \(b_1 = 1\). Thus \(h_1(x) = 2x + 1\). 2. For \(h_2\), use \(C(6, 1)\): \(1 = -0.5 \cdot 6 + b_2\), so \(b_2 = 4\). Thus \(h_2(x) = -0.5x + 4\). 3. Set the equations equal: \(2x + 1 = -0.5x + 4\). 4. Then \(2.5x = 3\), so \(x = 1.2\). 5. Substitute into \(h_1\): \(y = 2 \cdot 1.2 + 1 = 3.4\). 6. Therefore, the intersection point is \((1.2, 3.4)\).

The intersection point is \((1.2, 3.4)\).

- What must be true about the coordinates of a point where two lines intersect? - After finding the common x-coordinate, substitute it into either equation to find y. - For the third line, what information is fixed by a known slope and one point?

1. Set the equations equal: \(1.5x+2=-0.5x+6\). Then \(2x=4\), so \(x=2\). Substituting gives \(y=5\), so \(S=(2,5)\). 2. Write \(k\) as \(y=-2x+b\). Using \(S(2,5)\), \(5=-2\cdot2+b\), so \(b=9\). Therefore, \(k\) has equation \(y=-2x+9\).

1. \(S=(2, 5)\) 2. \(y=-2x+9\)

- First write an equation for each diagonal from its two endpoints. - How do you find the slope of a line through two points? - Once you have both equations, set them equal to find their common point.

1. For diagonal \(AC\), the slope is \(m_{AC} = \frac{3 - 0}{4 - (-2)} = \frac{1}{2}\). Using point \(A(-2, 0)\), its equation is \(y = 0.5x + 1\). 2. For diagonal \(BD\), the slope is \(m_{BD} = \frac{0 - 4}{4 - 0} = -1\). Since \(B(0, 4)\) is the y-intercept, its equation is \(y = -x + 4\). 3. Set the equations equal: \(0.5x + 1 = -x + 4\). 4. Then \(1.5x = 3\), so \(x = 2\). 5. Substitute into either diagonal: \(y = -2 + 4 = 2\). Therefore, \(S = (2, 2)\).

\(S = (2, 2)\)

- First write an equation for the path determined by the two given points. - At the crossing point, what must be true about the two y-values? - Check the point you find in both equations.

1. For Car 1, the slope is \(m = \frac{1 - 4}{5 - (-1)} = -\frac{1}{2}\). 2. Use point \(Q(5, 1)\) to find the y-intercept: \(1 = -0.5 \cdot 5 + b\), so \(b = 3.5\). Thus the first path is \(y = -0.5x + 3.5\). 3. Set the two path equations equal: \(-0.5x + 3.5 = 2x - 4\). 4. Then \(7.5 = 2.5x\), so \(x = 3\). 5. Substitute into the second equation: \(y = 2 \cdot 3 - 4 = 2\). Therefore, \(S = (3, 2)\).

\(S = (3, 2)\)

- Use the perimeter formula for a rectangle. - Express the two adjusted side lengths. - What must be true about the sides of a square? - Add or subtract the resulting equations.

1. Let \(l\) be the longer side and \(w\) the shorter side, in centimeters. 2. The perimeter equation is \(2(l + w) = 30\), so \(l + w = 15\). 3. The adjusted sides of the square are equal, so \(l - 3 = w + 2\), or \(l - w = 5\). 4. Add the equations: \((l + w) + (l - w) = 15 + 5\). 5. Then \(2l = 20\), so \(l = 10\). 6. Substitute into \(l + w = 15\): \(10 + w = 15\), so \(w = 5\).

The original rectangle has side lengths \(10\,\text{cm}\) and \(5\,\text{cm}\).

- Write and simplify the perimeter equation. - Write expressions for the original and new areas. - Be careful with signs when subtracting the new area. - After simplifying, solve the resulting linear system.

1. Let \(l\) be the original length and \(w\) the original width, in feet. 2. The perimeter equation is \(2(l + w) = 40\), so \(l + w = 20\). 3. The original area is \(lw\), and the new area is \((l + 2)(w - 2)\). 4. Because the area decreases by \(12\,\text{ft}^2\), \(lw - (l + 2)(w - 2) = 12\). 5. Expand and simplify: \(lw - (lw - 2l + 2w - 4) = 12\), so \(2l - 2w + 4 = 12\), or \(l - w = 4\). 6. Add \(l + w = 20\) and \(l - w = 4\): \(2l = 24\), so \(l = 12\). 7. Then \(w = 20 - 12 = 8\).

The garden bed was originally \(12\,\text{ft}\) long and \(8\,\text{ft}\) wide.

- Choose variables for the base fee and the price per gigabyte. - Compare how the total cost changes when the data use changes. - Dividing the difference in cost by the difference in data use gives the per-gigabyte price. - Subtracting the equations eliminates the base fee.

1. Let \(G\) be the monthly base fee and \(p\) the price per gigabyte, in dollars. 2. Write the system \(G + 5p = 17.50\) and \(G + 12p = 31.50\). 3. Subtract the first equation from the second: \(7p = 14\), so \(p = 2\). 4. Substitute into the first equation: \(G + 5 \cdot 2 = 17.50\). 5. Solve: \(G + 10 = 17.50\), so \(G = 7.50\).

The monthly base fee is \(\$7.50\), and the price is \(\$2.00\) per gigabyte.

- Use the perimeter formula for a rectangle. - Translate “\(3\) less than twice” into an algebraic equation. - Substitute the expression for the longer side into the perimeter equation.

1. Let \(l\) be the longer side and \(w\) the shorter side, in centimeters. 2. The perimeter equation is \(2(l + w) = 54\), so \(l + w = 27\). 3. The side-length relationship is \(l = 2w - 3\). 4. Substitute into the perimeter equation: \((2w - 3) + w = 27\). 5. Simplify: \(3w - 3 = 27\), so \(3w = 30\) and \(w = 10\). 6. Then \(l = 2 \cdot 10 - 3 = 17\).

The side lengths are \(17\,\text{cm}\) and \(10\,\text{cm}\).

- Express who has saved more and by how much. - Represent doubling and tripling with your variables. - One equation already gives one variable in terms of the other, so consider substitution.

1. Let \(x\) be the amount Jordan has saved and \(y\) the amount Casey has saved. 2. Write the system: \(x = y + 20\) and \(2x + 3y = 240\). 3. Substitute \(y + 20\) for \(x\): \(2(y + 20) + 3y = 240\). 4. Solve: \(2y + 40 + 3y = 240\), so \(5y = 200\) and \(y = 40\). 5. Then \(x = 40 + 20 = 60\).

Jordan has saved \(\$60\), and Casey has saved \(\$40\).

- Convert both travel times to hours before finding speeds. - A tailwind adds to the plane's still-air speed, while a headwind subtracts from it. - Which operation on the two equations will eliminate the wind speed?

1. Convert the travel times to hours: \(1\) hour \(30\) minutes is \(1.5\) hours, and \(1\) hour \(48\) minutes is \(1.8\) hours. 2. With the tailwind, the ground speed is \(450\div1.5=300\,\text{mph}\). With the headwind, the ground speed is \(450\div1.8=250\,\text{mph}\). 3. Let \(v_p\) be the plane's speed in still air and \(v_w\) the wind speed. Then \(v_p+v_w=300\) and \(v_p-v_w=250\). 4. Add the equations: \(2v_p=550\), so \(v_p=275\). Then \(275+v_w=300\), so \(v_w=25\).

The plane's speed in still air is \(275\,\text{mph}\), and the wind speed is \(25\,\text{mph}\).

- Multiply one equation so both equations have the same notebook coefficient. - After finding the individual prices, calculate the regular price of the bundle quantities. - Compare the regular total with the sale price.

1. Let \(h\) be the price of one notebook and \(s\) the price of one pen. The system is \(4h+3s=11\) and \(2h+5s=9\). Multiply the second equation by \(2\): \(4h+10s=18\). Subtract the first equation to get \(7s=7\), so \(s=1\). Then \(2h+5=9\), so \(h=2\). 2. At regular prices, \(6\) notebooks and \(8\) pens cost \(6\cdot2+8\cdot1=20\) dollars. Since \(18.50<20\), the sale bundle saves \(\$1.50\).

a) One notebook costs \(\$2\), and one pen costs \(\$1\). b) Yes. The regular price is \(\$20.00\), so the bundle saves \(\$1.50\).

- Write a perimeter equation for the original rectangle. - Express the adjusted length and width. - Write and simplify a second perimeter equation. - Subtract the equations to eliminate one variable.

1. Let \(x\) be the original length and \(y\) the original width, in centimeters. 2. The original perimeter gives \(2x + 2y = 50\), or \(x + y = 25\). 3. The new perimeter gives \(2(x + 5) + 2(2y) = 80\). 4. Simplify: \(2x + 10 + 4y = 80\), so \(x + 2y = 35\). 5. Subtract \(x + y = 25\) from \(x + 2y = 35\): \(y = 10\). 6. Then \(x + 10 = 25\), so \(x = 15\).

The original length is \(15\,\text{cm}\), and the original width is \(10\,\text{cm}\).

- Express a \(20\%\) increase and a \(20\%\) decrease using decimal multipliers. - Write one equation for the current total and one for the projected total. - Solve the system using substitution. - Check that both student counts are whole numbers.

1. Let \(a\) be the current number of students in Homeroom A and \(b\) the current number in Homeroom B. 2. The current total gives \(a + b = 60\). 3. Apply the projected percent changes: \(1.2a + 0.8b = 62\). 4. Solve the first equation for \(b\): \(b = 60 - a\). 5. Substitute: \(1.2a + 0.8(60 - a) = 62\). 6. Simplify: \(1.2a + 48 - 0.8a = 62\), so \(0.4a = 14\) and \(a = 35\). 7. Then \(b = 60 - 35 = 25\).

Homeroom A currently has \(35\) students, and Homeroom B currently has \(25\) students.

- Write one equation for each condition. - One equation expresses one variable directly in terms of the other, which makes substitution useful. - Substitute the first value you find into either original equation to determine the other value.

1. The first condition gives \(x = 1.5y\), and the second gives \(x + y = 35\). 2. Substitute \(1.5y\) for \(x\) in the second equation: \(1.5y + y = 35\). 3. Combine like terms: \(2.5y = 35\), so \(y = 14\). 4. Substitute into the first equation: \(x = 1.5 \cdot 14 = 21\). 5. Check: \(21 + 14 = 35\) and \(21 = 1.5 \cdot 14\).

The equations are \(x = 1.5y\) and \(x + y = 35\). The ordered pair is \((21, 14)\).

- Write the relationship between the original length and width. - Write formulas for the original and new perimeters. - Subtract the original perimeter from the new perimeter. - Use the side-length relationship after finding one dimension.

1. The relationship between the original sides is \(l = w + 4\). 2. The original perimeter is \(2l + 2w\). 3. After the length is doubled, the new perimeter is \(2(2l) + 2w = 4l + 2w\). 4. The increase in perimeter gives \((4l + 2w) - (2l + 2w) = 20\). 5. Simplify: \(2l = 20\), so \(l = 10\). 6. Substitute into \(l = w + 4\): \(10 = w + 4\), so \(w = 6\).

The original length is \(10\,\text{cm}\), and the original width is \(6\,\text{cm}\).

- Track how each person’s amount changes when money is transferred. - Translate “twice as much” and “the same amount” into equations. - Write one equation for each statement. - Simplify both equations before choosing a solution method.

1. Let \(a\) be Anna’s savings and \(b\) Ben’s savings. 2. If Ben gives Anna \(\$15\), the relationship is \(a + 15 = 2(b - 15)\). 3. If Anna gives Ben \(\$5\), the relationship is \(a - 5 = b + 5\). 4. Rewrite the equations as \(a - 2b = -45\) and \(a - b = 10\). 5. From the second equation, \(a = b + 10\). Substitute into the first: \((b + 10) - 2b = -45\). 6. Solve: \(-b = -55\), so \(b = 55\). 7. Then \(a = 55 + 10 = 65\).

Anna has saved \(\$65\), and Ben has saved \(\$55\).

- How many intersections do you need to find the three vertices of a triangle formed by three lines? - Which pairs of lines should you intersect? - On the line \(x = 2\), the x-coordinate of every point is already known. How can you use that?

1. Find the intersection of \(g_1\) and \(g_2\): \(2x - 4 = -0.5x + 6\). Then \(2.5x = 10\), so \(x = 4\). Substituting into \(g_1\) gives \(y = 2 \cdot 4 - 4 = 4\), so one vertex is \((4, 4)\). 2. For the intersection of \(g_1\) and \(g_3\), use \(x = 2\). Then \(y = 2 \cdot 2 - 4 = 0\), so a second vertex is \((2, 0)\). 3. For the intersection of \(g_2\) and \(g_3\), use \(x = 2\). Then \(y = -0.5 \cdot 2 + 6 = 5\), so the third vertex is \((2, 5)\).

The vertices are \((4, 4)\), \((2, 0)\), and \((2, 5)\).

- Write an equation for the rectangle’s perimeter. - Relate the square’s side length to the rectangle’s longer side. - Use the difference between the two perimeters. - Substitute the longer side into the rectangle equation.

1. Let \(l\) be the length and \(w\) the width of the rectangular field, in yards. 2. The rectangle’s perimeter gives \(2(l + w) = 300\), so \(l + w = 150\). 3. The square’s side length is \(l\), and its perimeter is \(300 + 160 = 460\). Therefore, \(4l = 460\). 4. Solve: \(l = 115\). 5. Substitute into \(l + w = 150\): \(115 + w = 150\), so \(w = 35\).

The rectangular field is \(115\,\text{yd}\) long and \(35\,\text{yd}\) wide.

- Write an equation for the perimeter. - Set the adjusted area equal to the original area. - Expand the product \((x + 3)(y - 2)\). - Use substitution after simplifying both equations.

1. Let \(x\) be the side that is increased and \(y\) the side that is decreased, in centimeters. 2. The perimeter gives \(2(x + y) = 40\), so \(x + y = 20\). 3. The unchanged area gives \((x + 3)(y - 2) = xy\). 4. Expand and simplify: \(xy - 2x + 3y - 6 = xy\), so \(-2x + 3y = 6\). 5. From \(x + y = 20\), write \(x = 20 - y\). 6. Substitute: \(-2(20 - y) + 3y = 6\). 7. Simplify: \(-40 + 5y = 6\), so \(5y = 46\) and \(y = 9.2\). 8. Then \(x = 20 - 9.2 = 10.8\).

The original side lengths are \(10.8\,\text{cm}\) and \(9.2\,\text{cm}\).

- Choose a variable for each type of animal. - Write one equation for the total number of heads and another for the total number of legs. - For part b), consider whether a sum of even numbers can be odd.

1. Let \(x\) be the number of chickens and \(y\) the number of sheep. The system is \(x+y=25\) and \(2x+4y=72\). From \(x=25-y\), substitute to get \(2(25-y)+4y=72\). Then \(2y=22\), so \(y=11\) and \(x=14\). 2. Every animal contributes an even number of legs. A sum of even numbers must be even, but \(75\) is odd, so the visitor's claim is impossible.

a) There are \(14\) chickens and \(11\) sheep. b) The claim is impossible because every animal contributes an even number of legs, so the total cannot be odd.

- Choose variables for the amounts of the two juices. - Write one equation for the total volume. - Write another equation for the total cost using each price per quart.

1. Let \(x\) be the amount of Juice A and \(y\) the amount of Juice B, in quarts. 2. Write the system \(x + y = 20\) and \(1.20x + 2.00y = 32.80\). 3. Solve the first equation for \(y\): \(y = 20 - x\). 4. Substitute: \(1.2x + 2(20 - x) = 32.8\). 5. Simplify: \(1.2x + 40 - 2x = 32.8\), so \(-0.8x = -7.2\) and \(x = 9\). 6. Then \(y = 20 - 9 = 11\).

The class should use \(9\,\text{qt}\) of Juice A and \(11\,\text{qt}\) of Juice B.

- Recall the sum of the interior angles of a triangle. - Express both \(\beta\) and \(\gamma\) in terms of \(\alpha\). - Substitute those expressions into the angle-sum equation. - Check that the three results total \(180^\circ\).

1. The angle sum is \(\alpha+\beta+\gamma=180^\circ\). 2. The given relationships are \(\beta=2\alpha\) and \(\gamma=\beta-30^\circ=2\alpha-30^\circ\). 3. Substitute into the angle-sum equation: \(\alpha+2\alpha+(2\alpha-30^\circ)=180^\circ\). 4. Simplify: \(5\alpha-30^\circ=180^\circ\), so \(5\alpha=210^\circ\) and \(\alpha=42^\circ\). 5. Then \(\beta=84^\circ\) and \(\gamma=54^\circ\).

\(\alpha=42^\circ\), \(\beta=84^\circ\), and \(\gamma=54^\circ\)

- Use the fact that the base angles of an isosceles triangle are congruent. - Express the vertex angle in terms of one base angle. - Use the sum of the interior angles of a triangle.

1. Because the triangle is isosceles, \(\alpha=\beta\). 2. The angle sum is \(2\alpha+\gamma=180^\circ\). 3. The given relationship is \(\gamma=1.5(\alpha+\beta)=1.5(2\alpha)=3\alpha\). 4. Substitute into the angle-sum equation: \(2\alpha+3\alpha=180^\circ\), so \(5\alpha=180^\circ\) and \(\alpha=36^\circ\). 5. Therefore, \(\beta=36^\circ\) and \(\gamma=108^\circ\).

\(\alpha=36^\circ\), \(\beta=36^\circ\), and \(\gamma=108^\circ\)

- Represent a two-digit number using its tens digit and ones digit. - Translate the digit sum into an equation. - Write an expression for the number with its digits reversed. - Use the stated difference to form a second equation.

1. Let \(x\) be the tens digit and \(y\) the ones digit. The original number is \(10x + y\). 2. The digit sum gives \(x + y = 12\). 3. The reversed number is \(10y + x\), so \(10y + x = 10x + y + 18\). 4. Simplify the second equation: \(-x + y = 2\). 5. Add \(x + y = 12\) and \(-x + y = 2\): \(2y = 14\), so \(y = 7\). 6. Then \(x = 12 - 7 = 5\), so the original number is \(10 \cdot 5 + 7 = 57\).

The original number is \(57\).

- Write the number as \(10t + u\) and its digit sum as \(t + u\). - Translate “\(6\) less than seven times the digit sum” carefully. - Use the relationship between the two digits to substitute for one variable. - Check both conditions after finding the number.

1. Let \(t\) be the tens digit and \(u\) be the ones digit. The number is \(10t + u\). 2. The digit relationship is \(t = u + 2\). 3. The value relationship is \(10t + u = 7(t + u) - 6\). 4. Substitute \(t = u + 2\): \(10(u + 2) + u = 7(u + 2 + u) - 6\). 5. Distribute and combine like terms: \(11u + 20 = 14u + 8\). 6. Subtract \(11u\) and then subtract \(8\): \(12 = 3u\), so \(u = 4\). 7. Find the tens digit: \(t = 4 + 2 = 6\). The number is \(64\). 8. Check: the digit sum is \(6 + 4 = 10\), and \(7 \cdot 10 - 6 = 64\). Also, \(6 = 4 + 2\).

The number is \(64\).

- Use the first scale to find the weight of one wooden block. - Carry that block weight into the equation for the second scale. - Write one equation for each balanced scale. - On the second scale, find what must be added to the two blocks to total \(20\,\text{oz}\).

1. Let \(x\) be the weight of one wooden block in ounces. The first scale gives \(3x + 8 = 32\). 2. Subtract \(8\): \(3x = 24\). Divide by \(3\): \(x = 8\). 3. Use the second scale: \(2x + y = 20\). 4. Substitute \(x = 8\): \(2 \cdot 8 + y = 20\). 5. Simplify and solve: \(16 + y = 20\), so \(y = 4\).

The unknown weight is \(4\,\text{oz}\).

- Identify the two speeds that combine to produce the ground speed. - Add the speeds when Julia walks with the walkway. - Subtract the walkway speed when she walks against it. - Add the equations to eliminate one variable.

1. Let \(v_J\) be Julia’s walking speed and \(v_W\) the walkway’s speed, in meters per second. 2. Walking with the walkway gives \(v_J + v_W = 2.5\). 3. Walking against the walkway gives \(v_J - v_W = 0.7\). 4. Add the equations: \(2v_J = 3.2\), so \(v_J = 1.6\). 5. Substitute into the first equation: \(1.6 + v_W = 2.5\), so \(v_W = 0.9\).

Julia’s walking speed is \(1.6\,\text{m/s}\), and the walkway’s speed is \(0.9\,\text{m/s}\).

- Choose variables for the two amounts being mixed. - Write one equation for the total volume. - Write another equation for the amount of pure fruit juice. - Solve the system using substitution or elimination.

1. Let \(x\) be the number of gallons of Drink A and \(y\) the number of gallons of Drink B. 2. The total volume gives \(x + y = 60\). 3. The amount of fruit juice gives \(0.40x + 0.70y = 0.50 \cdot 60\), or \(0.4x + 0.7y = 30\). 4. Solve the first equation for \(y\): \(y = 60 - x\). 5. Substitute: \(0.4x + 0.7(60 - x) = 30\). 6. Simplify: \(0.4x + 42 - 0.7x = 30\), so \(-0.3x = -12\) and \(x = 40\). 7. Then \(y = 60 - 40 = 20\).

The company should mix \(40\,\text{gal}\) of Drink A and \(20\,\text{gal}\) of Drink B.

- Use one equation for the total number of tickets. - Write a second equation for total revenue. - Express one ticket count in terms of the other. - For equal ticket counts, divide the fixed total equally between the two types.

1. Let \(x\) be the number of adult tickets and \(y\) the number of student tickets. The system is \(x+y=240\) and \(8x+5y=1440\). From \(y=240-x\), substitute to get \(8x+5(240-x)=1440\). Then \(3x=240\), so \(x=80\) and \(y=160\). 2. Equal numbers would mean \(120\) adult tickets and \(120\) student tickets. The revenue would be \(120\cdot8+120\cdot5=1560\) dollars.

a) The festival sold \(80\) adult tickets and \(160\) student tickets. b) The revenue would have been \(\$1560\).

- Identify the two parts that make up the total weight. - After three-fourths of the apples are sold, what fraction remains? - Write one equation for the original weight and one for the later weight. - Consider subtracting the equations to eliminate the crate’s weight.

1. Let \(x\) be the weight of the empty crate and \(y\) the original weight of the apples, in pounds. 2. Write the system \(x + y = 31\) and \(x + \frac{1}{4}y = 10\). 3. Subtract the second equation from the first: \(\frac{3}{4}y = 21\), so \(y = 28\). 4. Substitute into \(x + y = 31\): \(x + 28 = 31\), so \(x = 3\).

The empty crate weighs \(3\,\text{lb}\).

- Choose variables for the two unknown weights. - Write an equation for each combination of items. - Decide which system-solving method is most efficient. - Check the weights in both original combinations.

1. Let \(r\) be the weight of one backpack and \(s\) the weight of one sleeping bag, in pounds. 2. Write the system \(r + 2s = 11\) and \(2r + s = 10\). 3. Solve the second equation for \(s\): \(s = 10 - 2r\). 4. Substitute into the first equation: \(r + 2(10 - 2r) = 11\). 5. Simplify: \(r + 20 - 4r = 11\), so \(-3r = -9\) and \(r = 3\). 6. Then \(s = 10 - 2 \cdot 3 = 4\).

One backpack weighs \(3\,\text{lb}\), and one sleeping bag weighs \(4\,\text{lb}\).

- Write an expression for the amount remaining in each tank after a given time. - Translate “equal amounts” into an equation. - Write a system with the initial total and the equal-amount condition. - First determine how much water drains from each tank in \(50\) minutes.

1. Let \(x\) be the initial amount in Tank 1 and \(y\) the initial amount in Tank 2, in gallons. 2. The initial total gives \(x + y = 900\). 3. After \(50\) minutes, Tank 1 has lost \(3 \cdot 50 = 150\) gallons and Tank 2 has lost \(5 \cdot 50 = 250\) gallons. 4. Equal remaining amounts give \(x - 150 = y - 250\), or \(x = y - 100\). 5. Substitute into the total equation: \((y - 100) + y = 900\), so \(2y = 1000\) and \(y = 500\). 6. Then \(x = 500 - 100 = 400\).

Tank 1 originally contained \(400\,\text{gal}\), and Tank 2 originally contained \(500\,\text{gal}\).

- For the first equation, choose pairs whose coordinates add to \(10\). - Substitute the expression for \(y\) from the second equation into the first. - A common solution must satisfy both equations.

1. For equation (I), two examples are \((0,10)\) and \((5,5)\). For equation (II), two examples are \((0,2)\) and \((2,4)\). 2. Substitute \(y=x+2\) into equation (I): \(x+x+2=10\). Then \(2x=8\), so \(x=4\) and \(y=6\). The common solution is \((4,6)\).

a) For (I): \((0, 10)\), \((5, 5)\). For (II): \((0, 2)\), \((2, 4)\). b) \((4, 6)\)

- Choose variables for the two ticket prices. - Write one revenue equation for each night. - Simplify the equations before solving. - Find the regular price for the group and compare it with the package price.

1. Let \(s\) be the student ticket price and \(a\) the adult ticket price. The system is \(60s+20a=440\) and \(45s+30a=480\). Simplify to \(3s+a=22\) and \(3s+2a=32\). Subtract to get \(a=10\), then \(3s+10=22\), so \(s=4\). 2. At regular prices, the group would pay \(5\cdot4+2\cdot10=40\) dollars. Since \(35<40\), the package saves \(\$5\).

a) A student ticket costs \(\$4\), and an adult ticket costs \(\$10\). b) Yes. The package costs \(\$5\) less than the \(\$40\) regular price.

- Use distance equals rate times time for each observation. - Write one equation for the combined distances. - Carefully translate which train travels \(140\) miles farther in the second comparison. - Substitute a repeated distance expression into the first equation.

1. Let \(v_A\) and \(v_B\) be the speeds of Trains A and B, in miles per hour. 2. The first condition gives \(4v_A + 3v_B = 620\). 3. The second condition gives \(2v_A = 3v_B - 140\), or \(3v_B = 2v_A + 140\). 4. Substitute into the first equation: \(4v_A + (2v_A + 140) = 620\). 5. Simplify: \(6v_A + 140 = 620\), so \(6v_A = 480\) and \(v_A = 80\). 6. Then \(3v_B = 2 \cdot 80 + 140 = 300\), so \(v_B = 100\).

Train A travels at \(80\,\text{mph}\), and Train B travels at \(100\,\text{mph}\).

- Identify what remains unchanged in both packing arrangements. - Choose variables for the number of crates and the number of apples. - When two crates are empty, how many crates are actually used? - Write an expression for the total number of apples in each arrangement.

1. Let \(x\) be the number of crates and \(y\) the number of apples. 2. The first packing arrangement gives \(y = 12x + 15\). 3. In the second arrangement, only \(x - 2\) crates are used, so \(y = 15(x - 2)\). 4. Set the expressions equal: \(12x + 15 = 15(x - 2)\). 5. Simplify: \(12x + 15 = 15x - 30\), so \(3x = 45\) and \(x = 15\). 6. Then \(y = 12 \cdot 15 + 15 = 195\).

There are \(195\) apples and \(15\) crates.

- Convert the early and late times from minutes to hours. - Write a distance expression for each speed scenario. - The distance is the same in both scenarios. - Solve for the scheduled time, then calculate the distance.

1. Let \(t\) be the scheduled travel time in hours and \(d\) the distance in miles. 2. Convert the time differences: \(10\) minutes is \(\frac{1}{6}\) hour, and \(5\) minutes is \(\frac{1}{12}\) hour. 3. The two scenarios give \(d = 60\left(t + \frac{1}{6}\right)\) and \(d = 80\left(t - \frac{1}{12}\right)\). 4. Set the distance expressions equal: \(60t + 10 = 80t - \frac{20}{3}\). 5. Solve: \(20t = 10 + \frac{20}{3} = \frac{50}{3}\), so \(t = \frac{5}{6}\) hour, or \(50\) minutes. 6. Find the distance: \(d = 60\left(\frac{5}{6} + \frac{1}{6}\right) = 60\).

The destination is \(60\,\text{mi}\) away, and the scheduled travel time is \(50\) minutes.

- Write an expression for the gift price in each situation. - Combine the students' contributions with the amount short or left over. - Set the two expressions for the same gift price equal. - Divide the total cost equally for part b).

1. Let \(x\) be the number of students and \(y\) the price of the gift. The two situations give \(y=4.50x+12\) and \(y=5.50x-12\). Setting them equal gives \(4.50x+12=5.50x-12\), so \(x=24\). Then \(y=4.50\cdot24+12=120\), so the gift costs \(\$120\). 2. The exact contribution is \(120\div24=5\), so each student should contribute \(\$5.00\).

a) There are \(24\) students, and the gift costs \(\$120\). b) Each student should contribute \(\$5.00\).

- Account for the additional people in the second situation. - Express the rental cost using the contributions and the shortage or surplus. - Use parentheses for the size of the larger group. - Set the two expressions for the rental cost equal.

1. Let \(x\) be the original number of people and \(M\) the rental cost, in dollars. 2. The first situation gives \(M = 20x + 40\). 3. The second situation gives \(M = 18(x + 5) - 20\). 4. Set the expressions equal: \(20x + 40 = 18(x + 5) - 20\). 5. Simplify: \(20x + 40 = 18x + 70\), so \(2x = 30\) and \(x = 15\). 6. The rent is \(M = 20 \cdot 15 + 40 = 340\) dollars. 7. Check: \(18 \cdot (15 + 5) - 20 = 340\).

The original group had \(15\) people, and the cabin rental cost was \(\$340\).

- Use distance equals rate times time. - In each scenario, the cyclists’ distances add to \(60\) miles. - Carefully determine how long each cyclist rides in the second scenario. - Write one equation for each scenario.

1. Let \(v_A\) be Anton’s speed and \(v_B\) Beatrice’s speed, in miles per hour. 2. In the first scenario, their distances after \(2\) hours total \(60\) miles: \(2v_A + 2v_B = 60\). 3. In the second scenario, Anton rides for \(1\) hour and Beatrice rides for \(4\) hours, so \(v_A + 4v_B = 60\). 4. Simplify the first equation: \(v_A + v_B = 30\), so \(v_A = 30 - v_B\). 5. Substitute into the second equation: \((30 - v_B) + 4v_B = 60\). 6. Simplify: \(30 + 3v_B = 60\), so \(v_B = 10\). 7. Then \(v_A = 30 - 10 = 20\).

Anton rides at \(20\,\text{mph}\), and Beatrice rides at \(10\,\text{mph}\).

- Choose variables for the two admission prices. - Write one cost equation for each group. - Simplify the equations before solving the system. - For part b), find the individual-ticket total and compare it with the pass price.

1. Let \(x\) be the adult admission price and \(y\) the child admission price, in dollars. The system is \(5x+10y=90\) and \(3x+12y=84\). Simplify to \(x+2y=18\) and \(x+4y=28\). Subtract to get \(2y=10\), so \(y=5\). Then \(x+2(5)=18\), so \(x=8\). 2. Individual tickets for \(2\) adults and \(3\) children cost \(2\cdot8+3\cdot5=31\) dollars. Since \(31<35\), the family pass costs \(\$4\) more and does not save money.

a) Adult admission is \(\$8\), and child admission is \(\$5\). b) No. Individual tickets cost \(\$31\), which is \(\$4\) less than the family pass.

- Write one equation for total distance. - Use time equals distance divided by speed for each part of the trip. - Add the two travel times to get the total time. - Clear the fractions before solving the system.

1. Let \(x\) be the distance traveled in Eco mode and \(y\) the distance traveled in Sport mode, in miles. 2. The total distance gives \(x + y = 54\). 3. The total time gives \(\frac{x}{18} + \frac{y}{27} = 2.5\). 4. Multiply the time equation by \(54\): \(3x + 2y = 135\). 5. From the distance equation, \(y = 54 - x\). 6. Substitute: \(3x + 2(54 - x) = 135\). 7. Simplify: \(x + 108 = 135\), so \(x = 27\). 8. Then \(y = 54 - 27 = 27\).

Alex travels \(27\,\text{mi}\) in Eco mode and \(27\,\text{mi}\) in Sport mode.

- Write one equation for the amount of nitrogen in each mixture. - Multiply each volume by its unknown concentration. - Express one concentration in terms of the other and substitute. - Convert the decimal solutions back to percentages.

1. Let \(x\) and \(y\) be the nitrogen concentrations of Fertilizers A and B, written as decimals. 2. The first \(5\)-gallon mixture gives \(3x + 2y = 5 \cdot 0.14 = 0.70\). 3. The second \(5\)-gallon mixture gives \(x + 4y = 5 \cdot 0.10 = 0.50\). 4. Solve the second equation for \(x\): \(x = 0.50 - 4y\). 5. Substitute into the first equation: \(3(0.50 - 4y) + 2y = 0.70\). 6. Simplify: \(1.50 - 12y + 2y = 0.70\), so \(-10y = -0.80\) and \(y = 0.08\). 7. Then \(x = 0.50 - 4 \cdot 0.08 = 0.18\). 8. Therefore, Fertilizer A is \(18\%\) nitrogen and Fertilizer B is \(8\%\) nitrogen.

Fertilizer A is \(18\%\) nitrogen, and Fertilizer B is \(8\%\) nitrogen.

- Find the total value of each \(10\)-pound blend. - Write a system using the two pure-bean prices. - Multiply one equation so a variable can be eliminated. - Compare the pure-bean prices and the proportion of each bean in the blends.

1. Let \(a\) be the price per pound of Arabica and \(r\) the price per pound of Robusta, in dollars. Each blend contains \(10\,\text{lb}\), so the total-value equations are \(4a+6r=10\cdot12.80=128\) and \(7a+3r=10\cdot14.30=143\). Multiply the second equation by \(2\) to get \(14a+6r=286\). Subtract the first equation: \(10a=158\), so \(a=15.80\). Then \(4\cdot15.80+6r=128\), so \(r=10.80\). 2. Arabica is more expensive than Robusta. Blend 2 contains \(70\%\) Arabica, while Blend 1 contains only \(40\%\), so Blend 2 has the higher average price.

a) Arabica costs \(\$15.80\) per pound, and Robusta costs \(\$10.80\) per pound. b) Blend 2 costs more because it contains a larger proportion of the more expensive Arabica beans.

- Translate each person’s statement into an equation. - Choose variables for the two amounts of money. - One equation can be solved for a variable and substituted into the other. - Represent “two-fifths” as a fraction or decimal.

1. Let \(L\) be Lucas’s amount and \(M\) Mia’s amount, in dollars. 2. Translate the statements into the system \(L + \frac{2}{5}M = 30\) and \(M + \frac{1}{2}L = 30\). 3. Solve the second equation for \(M\): \(M = 30 - \frac{1}{2}L\). 4. Substitute into the first equation: \(L + \frac{2}{5}\left(30 - \frac{1}{2}L\right) = 30\). 5. Simplify: \(L + 12 - \frac{1}{5}L = 30\), so \(\frac{4}{5}L = 18\) and \(L = 22.5\). 6. Then \(M = 30 - \frac{1}{2} \cdot 22.5 = 18.75\).

Lucas has \(\$22.50\), and Mia has \(\$18.75\).

- Write one equation for each separate pouring situation. - Include only the amount added to the receiving cup in each equation. - Check that your solution satisfies both conditions. - Add the two original amounts to find the total.

1. Let \(x\) be the original amount in Cup A and \(y\) the original amount in Cup B, in fluid ounces. The system is \(x+\frac{1}{3}y=10\) and \(y+\frac{1}{2}x=12.5\). Multiply the first equation by \(3\): \(3x+y=30\), so \(y=30-3x\). Substitute into the second equation: \(30-3x+\frac{1}{2}x=12.5\). Then \(-\frac{5}{2}x=-17.5\), so \(x=7\), and \(y=9\). 2. The total amount is \(7+9=16\,\text{fl oz}\).

a) Cup A originally contained \(7\,\text{fl oz}\), and Cup B contained \(9\,\text{fl oz}\). b) The cups contained \(16\,\text{fl oz}\) altogether.

- Choose variables for the two bike prices. - Translate each comparison into an equation that includes the helmet price. - Simplify one equation and use substitution. - Check both original comparisons with your results.

1. Let \(x\) be the price of Jordan’s mountain bike and \(y\) the price of Leo’s mountain bike, in dollars. 2. The first condition gives \(x + 120 = 3(y + 40)\). 3. The second condition gives \(y + 120 = \frac{1}{2}(x + 40)\). 4. Simplify the first equation: \(x + 120 = 3y + 120\), so \(x = 3y\). 5. Substitute into the second equation: \(y + 120 = \frac{1}{2}(3y + 40)\). 6. Multiply by \(2\): \(2y + 240 = 3y + 40\), so \(y = 200\). 7. Then \(x = 3 \cdot 200 = 600\).

Jordan’s mountain bike costs \(\$600.00\), and Leo’s mountain bike costs \(\$200.00\).

- Write one equation for each comparison. - Use parentheses so each extension is added before the lengths are compared. - Solve one equation for a variable and substitute. - Clear the fraction by multiplying the equation by \(4\).

1. Let \(a\) be the original length of Rod A and \(b\) the original length of Rod B, in centimeters. 2. The first condition gives \(a + 50 = 2(b + 10)\). 3. The second condition gives \(b + 50 = \frac{3}{4}(a + 10)\). 4. Solve the first equation for \(a\): \(a = 2b - 30\). 5. Substitute into the second equation: \(b + 50 = \frac{3}{4}(2b - 20)\). 6. Multiply by \(4\): \(4b + 200 = 3(2b - 20)\). 7. Simplify: \(4b + 200 = 6b - 60\), so \(260 = 2b\) and \(b = 130\). 8. Then \(a = 2 \cdot 130 - 30 = 230\).

Rod A is \(230\,\text{cm}\) long, and Rod B is \(130\,\text{cm}\) long.

- Express the touring-bike price using the basic-bike price. - Write two expressions for the racing-bike price. - Set those expressions equal and solve.

1. Let \(b\) be the price of the basic bike and \(r\) the price of the racing bike, in dollars. The touring bike costs \(4b\). 2. The two comparisons for the racing bike give the system \(r = 4b + 1200\) and \(r = b + 4800\). 3. Set the two expressions for \(r\) equal: \(4b + 1200 = b + 4800\). 4. Solve: \(3b = 3600\), so \(b = 1200\). 5. The touring bike costs \(4 \cdot 1200 = 4800\). 6. The racing bike costs \(1200 + 4800 = 6000\).

The basic bike costs \(\$1200.00\), the touring bike costs \(\$4800.00\), and the racing bike costs \(\$6000.00\).

- Calculate the boat’s ground speed in each direction. - Upstream speed is the still-water speed minus the current speed. - Downstream speed is the still-water speed plus the current speed. - A raft with no motor travels at the speed of the current.

1. The upstream ground speed is \(\frac{24}{2} = 12\,\text{mph}\), and the downstream ground speed is \(\frac{24}{1.5} = 16\,\text{mph}\). 2. Let \(v_b\) be the boat’s speed in still water and \(v_c\) the speed of the current. Then \(v_b - v_c = 12\) and \(v_b + v_c = 16\). 3. Add the equations: \(2v_b = 28\), so \(v_b = 14\). Then \(v_c = 2\). 4. A raft with no motor moves at the current’s speed, so its travel time is \(\frac{24}{2} = 12\) hours.

a) The boat’s speed in still water is \(14\,\text{mph}\), and the current’s speed is \(2\,\text{mph}\). b) The raft would take \(12\) hours.

- Use one equation for the total number of bottles and one for the total volume. - Include each bottle size in the volume equation. - For part b, compare the amount of juice held by a small bottle with the amount held by a large bottle.

1. Let \(x\) be the number of \(16\)-fluid-ounce bottles and \(y\) the number of \(24\)-fluid-ounce bottles. The system is \(x + y = 200\) and \(16x + 24y = 3840\). 2. From \(x = 200-y\), substitute into the volume equation: \(16(200-y)+24y=3840\). 3. Simplify: \(3200+8y=3840\), so \(y=80\) and \(x=120\). 4. Larger bottles hold more juice per bottle. Increasing their proportion increases the average amount of juice per bottle, so fewer bottles are needed for the same total volume.

a) The company fills \(120\) bottles of the \(16\)-fluid-ounce size and \(80\) bottles of the \(24\)-fluid-ounce size. b) The total number of bottles would decrease because more juice would be packaged in each bottle on average.

- How can you read the slope and \(y\)-intercept of each line from a graph? - What does the intersection of two lines represent for the solution set of a system? - What does it mean for the solution set when two distinct lines are parallel? - How can you check an equation read from a graph by substituting points on the line?

1. For graph a), line \(g_1\) crosses the \(y\)-axis at \((0, -3)\) and has slope \(2\), so its equation is \(y = 2x - 3\). Line \(g_2\) crosses the \(y\)-axis at \((0, 0)\) and has slope \(-1\), so its equation is \(y = -x\). For graph b), line \(h_1\) crosses the \(y\)-axis at \((0, 2)\) and has slope \(-0.5\), so its equation is \(y = -0.5x + 2\). Line \(h_2\) crosses the \(y\)-axis at \((0, -1)\) and has slope \(-0.5\), so its equation is \(y = -0.5x - 1\). 2. In graph a), the lines intersect at \((1, -1)\), so the solution set is \(\{(1, -1)\}\). In graph b), the lines have the same slope but different \(y\)-intercepts, so they are parallel and the solution set is \(\emptyset\).

1. Graph a): (I) \(y = 2x - 3\); (II) \(y = -x\). Graph b): (I) \(y = -0.5x + 2\); (II) \(y = -0.5x - 1\). 2. Graph a): \(\{(1, -1)\}\). Graph b): \(\emptyset\).

- What does the intersection of two lines represent for the corresponding system of equations? - Pay attention to the scale on each axis. Is the intersection exactly at a grid point? - Which algebraic methods can you use to solve a system of linear equations? - It may help to rewrite both equations in the form \(y = mx + b\). - How can you check your calculated solution?

1. From the graph, the intersection appears to be near \((1.5, 1.5)\). An exact reading is difficult because the intersection is not at a grid point. 2. Rewrite equation (I) as \(y = -x + 3\) and equation (II) as \(y = 1.5x - 1\). 3. Set the expressions equal: \(-x + 3 = 1.5x - 1\). Then \(4 = 2.5x\), so \(x = 1.6\). 4. Substitute into equation (I): \(y = 3 - 1.6 = 1.4\). The exact intersection is \((1.6, 1.4)\). 5. The exact point \((1.6, 1.4)\) is close to the graph-based estimate \((1.5, 1.5)\).

a) A reasonable estimate is about \((1.5, 1.5)\). The intersection is not at a grid point, so the exact coordinates are difficult to read. b) The exact intersection is \((1.6, 1.4)\). c) The algebraic result is close to the estimate from the graph.

- Identify each rider’s starting distance and rate of change. - Use the form \(s = mt + b\) for each equation. - At the intersection, both riders have the same time and distance. - Set the two distance expressions equal to calculate the intersection.

1. Jordan starts at \(0\) miles and travels \(15\) miles each hour, so \(s_J(t) = 15t\). Maya starts \(10\) miles from Jordan’s home and travels \(10\) miles each hour, so \(s_M(t) = 10t + 10\). 2. The graph’s lines intersect at \((2, 30)\). This means that after \(2\) hours, both riders are \(30\) miles from Jordan’s home, so Jordan catches Maya there. 3. Set the distance expressions equal: \(15t = 10t + 10\). Subtract \(10t\) to get \(5t = 10\), so \(t = 2\). Substitute into either equation: \(s = 15 \cdot 2 = 30\).

1. Jordan: \(s_J(t) = 15t\); Maya: \(s_M(t) = 10t + 10\) 2. The intersection is \((2, 30)\). After \(2\) hours, Jordan catches Maya \(30\,\text{mi}\) from Jordan’s home. 3. Solving \(15t = 10t + 10\) gives \(t = 2\), and substitution gives \(s = 30\).

- Read each line's y-intercept and use two grid points to find its slope. - At an intersection point, what must be true about the two function values? - Solve the equation formed by setting the two expressions equal. - Check the ordered pair in both equations.

1. Line \(f\) crosses the y-axis at \((0, 1)\). Using \((0, 1)\) and \((2, 2)\), its slope is \(\frac{1}{2}\), so \(f(x)=\frac{1}{2}x+1\). 2. Line \(g\) crosses the y-axis at \((0, 5)\). Using \((0, 5)\) and \((1, 4)\), its slope is \(-1\), so \(g(x)=-x+5\). 3. At the intersection, \(\frac{1}{2}x+1=-x+5\). Then \(\frac{3}{2}x=4\), so \(x=\frac{8}{3}\). 4. Substitute \(x=\frac{8}{3}\) into either equation to get \(y=\frac{7}{3}\). Therefore \(S\left(\frac{8}{3}, \frac{7}{3}\right)\).

a) \(f(x)=\frac{1}{2}x+1\) and \(g(x)=-x+5\) b) \(S\left(\frac{8}{3}, \frac{7}{3}\right)\)

- Read the y-intercept of each line from the graph. - Use two points on each line to determine its slope. - Recall the slope-intercept form \(y = mx + b\). - At an intersection point, the two functions have the same output for the same input. - After finding the x-coordinate, substitute it into either equation to find the y-coordinate.

1. For \(f\), the y-intercept is \(2\). Using the points \((0, 2)\) and \((2, 3)\), the slope is \(\frac{3 - 2}{2 - 0} = \frac{1}{2}\). Therefore, \(f(x) = \frac{1}{2}x + 2\). 2. For \(g\), the y-intercept is \(-1\). Using the points \((0, -1)\) and \((1, -2)\), the slope is \(\frac{-2 - (-1)}{1 - 0} = -1\). Therefore, \(g(x) = -x - 1\). 3. Set the expressions equal: \(\frac{1}{2}x + 2 = -x - 1\). 4. Then \(\frac{3}{2}x = -3\), so \(x = -2\). 5. Substituting into \(g\) gives \(y = -(-2) - 1 = 1\). Thus, \(S = (-2, 1)\).

a) \(f(x) = \frac{1}{2}x + 2\) and \(g(x) = -x - 1\). b) \(S = (-2, 1)\).

- Check the axis scales before estimating each intersection. - Use the marked points to find the slope and y-intercept of each line. - Recall the slope-intercept form \(y = mx + b\). How can two points be used to find \(m\)? - To find an exact intersection, set the two function expressions equal and solve for \(x\). - Substitute the x-coordinate into either equation to find the corresponding y-coordinate.

1. In graph a), the intersection is about \((1.4, 1.6)\). In graph b), the intersection is about \((-0.8, 2.6)\). 2. For graph a), \(f\) passes through \((0, 3)\) and \((3, 0)\), so its slope is \(-1\) and \(f(x) = -x + 3\). Line \(g\) passes through \((1, 1)\) and \((3, 4)\), so its slope is \(\frac{4 - 1}{3 - 1} = 1.5\). Using \((1, 1)\) gives a y-intercept of \(-0.5\), so \(g(x) = 1.5x - 0.5\). 3. For graph a), set the equations equal: \(-x + 3 = 1.5x - 0.5\). Then \(3.5 = 2.5x\), so \(x = 1.4\). Substitution gives \(y = 1.6\). Thus, \(S = (1.4, 1.6)\). 4. For graph b), \(f\) passes through \((0, 1)\) and \((1, -1)\), so \(f(x) = -2x + 1\). Line \(g\) passes through \((0, 3)\) and \((2, 4)\), so its slope is \(0.5\) and \(g(x) = 0.5x + 3\). 5. For graph b), set the equations equal: \(-2x + 1 = 0.5x + 3\). Then \(-2 = 2.5x\), so \(x = -0.8\). Substitution gives \(y = 2.6\). Thus, \(S = (-0.8, 2.6)\).

1) a) About \((1.4, 1.6)\); b) about \((-0.8, 2.6)\). 2) a) \(f(x) = -x + 3\) and \(g(x) = 1.5x - 0.5\); b) \(f(x) = -2x + 1\) and \(g(x) = 0.5x + 3\). 3) a) \(S = (1.4, 1.6)\); b) \(S = (-0.8, 2.6)\).

- Represent the decrease and increase as slopes. - Identify the starting amount for each linear equation. - At the intersection of the two graphs, the functions have the same value.

1. Write a linear equation for each tank’s amount after \(t\) hours: \(A(t) = -50t + 500\) and \(B(t) = 150t + 100\). 2. Set the expressions equal: \(-50t + 500 = 150t + 100\). 3. Solve: \(400 = 200t\), so \(t = 2\). 4. Substitute \(t = 2\): \(B(2) = 150 \cdot 2 + 100 = 400\). 5. Both tanks contain \(400\,\text{gal}\) after \(2\) hours.

After \(2\) hours, both tanks contain \(400\,\text{gal}\) of water.

- Use where each graph crosses the vertical axis to find its \(y\)-intercept. - Use the rise and run between two clear points to determine the coefficient of \(x\). - To calculate the intersection, set the two function expressions equal.

1. The graph of \(f\) crosses the y-axis at \(1\). Using \((0, 1)\) and \((1, 2)\), the slope is \(1\), so \(f(x) = x + 1\). 2. The graph of \(g\) crosses the y-axis at \(5\). Using \((0, 5)\) and \((2, 1)\), the slope is \(-2\), so \(g(x) = -2x + 5\). 3. Set the expressions equal: \(x + 1 = -2x + 5\). Then \(3x = 4\), so \(x = \frac{4}{3}\). 4. Substitute into either equation to get \(y = \frac{7}{3}\). Thus the exact intersection is \(\left(\frac{4}{3}, \frac{7}{3}\right)\), or approximately \((1.33, 2.33)\).

a) \(f(x) = x + 1\) and \(g(x) = -2x + 5\) b) The intersection is \(\left(\frac{4}{3}, \frac{7}{3}\right)\), or approximately \((1.33, 2.33)\).

- Use two clear points on each line to determine its slope. - When you set the function expressions equal, collect the x-terms on one side. - Check whether your calculated point is consistent with your estimate from the graph.

1. For \(f\), the graph shows a y-intercept of \(3\) and points such as \((0, 3)\) and \((3, 0)\), so the slope is \(-1\). Thus, \(f(x) = -x + 3\). 2. For \(g\), the graph shows a y-intercept of \(1\) and points such as \((0, 1)\) and \((4, 2)\), so the slope is \(\frac{2 - 1}{4 - 0} = 0.25\). Thus, \(g(x) = 0.25x + 1\). 3. Set the functions equal: \(-x + 3 = 0.25x + 1\). 4. Then \(2 = 1.25x\), so \(x = 1.6\). 5. Substitute to get \(y = -1.6 + 3 = 1.4\). To two decimal places, the intersection is \(S = (1.60, 1.40)\).

Estimate: about \(S = (1.5, 1.5)\). To two decimal places: \(S = (1.60, 1.40)\).

- First determine each line's equation in slope-intercept form \(y = mx + b\). - Read the y-intercept where each line crosses the y-axis. - Use two clear points to calculate each slope. - At the intersection, both functions have the same output for the same input.

1. For \(f\), the marked points \((0, 1)\) and \((4, 3)\) give slope \(\frac{3 - 1}{4 - 0} = 0.5\), and the y-intercept is \(1\). Therefore, \(f(x) = 0.5x + 1\). 2. For \(g\), the marked points \((1, 3)\) and \((3, 1)\) give slope \(\frac{1 - 3}{3 - 1} = -1\). Using \((3, 1)\), the y-intercept is \(4\), so \(g(x) = -x + 4\). 3. Set the functions equal: \(0.5x + 1 = -x + 4\). 4. Then \(1.5x = 3\), so \(x = 2\). 5. Substituting gives \(y = -2 + 4 = 2\). The calculated intersection is \((2, 2)\), which matches the graph.

a) The graph gives an intersection of about \((2, 2)\). b) The equations are \(f(x) = 0.5x + 1\) and \(g(x) = -x + 4\), and the calculated intersection is \((2, 2)\), matching the graph.

- The y-intercept represents a setup fee. - The intersection represents equal cost for the same quantity. - For part c), find which graph lies lower after the intersection. - Evaluate both costs at \(x=20\).

1. Line \(p\) has a y-intercept of \(60\), so it represents Vendor A. Line \(q\) passes through the origin, so it represents Vendor B. 2. Set the cost equations equal: \(4x+60=10x\). Then \(60=6x\), so \(x=10\), and the common cost is \(\$100\). 3. For whole-number quantities greater than \(10\), Vendor A's graph is lower, so Vendor A is less expensive starting at \(11\) shirts. 4. For \(20\) shirts, Vendor B charges \(10\cdot20=\$200\) and Vendor A charges \(4\cdot20+60=\$140\). The savings are \(\$60\).

a) Line \(p\): Vendor A; line \(q\): Vendor B b) \((10, 100)\); both charge \(\$100\) for \(10\) shirts c) \(11\) shirts d) \(\$60\)

- The y-intercept is the initial volume. - The slope is the inflow rate. - Use the displayed time interval for each domain. - Set the equations equal to find when the volumes match.

1. Tank 1 starts with \(25\) gallons and reaches \(50\) gallons after \(5\) minutes. Its slope is \(m=\frac{50-25}{5}=5\), so \(V_1(x)=5x+25\), with \(0\le x\le8\). 2. Tank 2 starts at \(0\) and reaches \(50\) gallons after \(5\) minutes. Its slope is \(10\), so \(V_2(x)=10x\), with \(0\le x\le8\). 3. Set the equations equal: \(10x=5x+25\). Then \(5x=25\), so \(x=5\). 4. Substitute \(x=5\) into either equation to get \(y=50\) gallons.

a) \(V_1(x)=5x+25\), with \(0\le x\le8\); \(V_2(x)=10x\), with \(0\le x\le8\) b) After \(5\) minutes; \(50\) gallons

- Rewrite each equation in the form \(y = mx + b\). - Compare the slopes and y-intercepts with the diagrams. - Check the intersection in both original equations.

1. Rewrite each equation in slope-intercept form. Equation (I) becomes \(y = -x + 4\), with slope \(-1\) and y-intercept \(4\). Equation (II) becomes \(y = x - 2\), with slope \(1\) and y-intercept \(-2\). 2. Diagram a shows these two lines. In diagram b, the rising line has y-intercept \(2\), so it does not represent equation (II). 3. In diagram a, the lines intersect at \((3, 1)\). 4. Check: \(3 + 1 = 4\) and \(3 - 1 = 2\).

Diagram a is correct. The intersection is \((3, 1)\).

- Pick one variable to eliminate and look for a common multiple of its two coefficients. - Each entire equation must be multiplied by the chosen scale factor. - Use opposite coefficients before adding the equations. - Check the final ordered pair in both original equations.

1. Multiply the first equation by \(3\): \(9x+6y=39\). 2. Multiply the second equation by \(-2\): \(-4x-6y=-24\). 3. Add the scaled equations: \(5x=15\), so \(x=3\). 4. Substitute into \(3x+2y=13\): \(9+2y=13\), so \(y=2\). 5. The solution is \((3, 2)\).

One valid setup is \(9x+6y=39\) and \(-4x-6y=-24\). Adding gives \(5x=15\). Therefore, the solution is \((3, 2)\).

- Look for an equation that already has one variable isolated. - Also look for coefficients that already cancel when the equations are added or subtracted. - Choose the method that requires fewer algebraic transformations for each system. - Verify each ordered pair in both equations of its system.

1. In part a), substitution is efficient because one equation already isolates \(y\). Substitute \(y=2x+1\) into \(3x+y=16\): \(3x+2x+1=16\), so \(x=3\) and \(y=7\). 2. In part b), elimination is efficient because the \(x\)-coefficients are already opposites. Add the equations: \(8y=24\), so \(y=3\). 3. Substitute \(y=3\) into \(4x+3y=17\): \(4x+9=17\), so \(x=2\).

a) Substitution; \((3, 7)\) b) Elimination; \((2, 3)\)

- Put parentheses around the entire second left-hand side before subtracting. - What happens to every sign inside an expression when that expression is subtracted? - After correcting the elimination step, solve the one-variable equation and substitute back.

1. The subtraction is \((4x+6y)-(4x-y)=26-5\). 2. Subtracting \(-y\) means adding \(y\), so the left side is \(7y\), not \(5y\). 3. The corrected equation is \(7y=21\), so \(y=3\). 4. Substitute into \(4x-y=5\): \(4x-3=5\), so \(x=2\). 5. The solution is \((2, 3)\).

The student treated subtracting \(-y\) as subtracting \(y\). The correct elimination step gives \(7y=21\), so the solution is \((2, 3)\).

- How do you find a point shared by two function graphs? - Which points on the x-axis form the base of the triangle? - Which coordinate of the intersection gives the triangle's height above the x-axis?

1. Set the functions equal: \(2x-4=-x+5\). Then \(3x=9\), so \(x=3\). Substituting gives \(y=2\), so \(S=(3,2)\). 2. For \(f\), solve \(2x-4=0\), giving \(x=2\). For \(g\), solve \(-x+5=0\), giving \(x=5\). 3. The base lies on the x-axis from \(x=2\) to \(x=5\), so its length is \(3\). The height is the y-coordinate of \(S\), which is \(2\). Therefore, the area is \(\frac{1}{2}\cdot3\cdot2=3\) square units.

a) \(S=(3, 2)\) b) The zero of \(f\) is \(x=2\), and the zero of \(g\) is \(x=5\). c) \(3\) square units

- Find each vertex by solving a pair of line equations. - Because one side is horizontal, how can you find its length and the corresponding height? - Use the triangle area formula once you know the base and height.

1. For \(g_1\) and \(g_2\), solve \(\frac{1}{2}x+2=-x+5\), giving \(x=2\) and \(y=3\), so \(A=(2,3)\). 2. For \(g_1\) and \(g_3\), solve \(-1=\frac{1}{2}x+2\), giving \(x=-6\), so \(B=(-6,-1)\). 3. For \(g_2\) and \(g_3\), solve \(-1=-x+5\), giving \(x=6\), so \(C=(6,-1)\). 4. Segment \(BC\) is horizontal with length \(6-(-6)=12\). The vertical height from \(A\) to \(y=-1\) is \(3-(-1)=4\). Thus the area is \(\frac{1}{2}\cdot12\cdot4=24\) square units.

a) \(A=(2, 3)\), \(B=(-6, -1)\), \(C=(6, -1)\) b) \(24\) square units

- Compare the equations of the diagonals in the two configurations. - How does the slope of diagonal \(AC\) change as \(C\) moves farther right? - What equation describes every point on diagonal \(BD\)?

a) With \(C=(6,6)\), diagonal \(AC\) is \(y=x\) and diagonal \(BD\) is \(y=-x+6\). Setting them equal gives \(x=3\) and \(y=3\), so \(S_1=(3,3)\). b) With \(C'=(12,6)\), diagonal \(AC'\) is \(y=0.5x\). Setting \(0.5x=-x+6\) gives \(1.5x=6\), so \(x=4\) and \(y=2\). Thus \(S_2=(4,2)\). c) Every intersection lies on \(BD\), where \(x=6-y\). The diagonal from \(A\) to any point to the right on \(y=6\) meets \(BD\) above the x-axis, so \(y>0\). Therefore, \(x=6-y<6\).

a) \(S_1 = (3, 3)\) b) \(S_2 = (4, 2)\) c) The intersection remains above the x-axis on \(BD\). Since \(x = 6 - y\) there and \(y > 0\), its x-coordinate always stays below \(6\).

- Write one speed equation for the tailwind and one for the headwind. - For part b), calculate the time for each \(20\)-mile leg separately. - Compare that sum with the time for \(40\) miles at the still-air speed.

1. Let \(v_c\) be the cyclist's speed with no wind and \(v_w\) the wind speed. Then \(v_c+v_w=32\) and \(v_c-v_w=24\). Adding gives \(2v_c=56\), so \(v_c=28\,\text{mph}\), and then \(v_w=4\,\text{mph}\). 2. With wind, the outbound time is \(\frac{20}{32}=\frac{5}{8}\) hour and the return time is \(\frac{20}{24}=\frac{5}{6}\) hour. The total is \(\frac{35}{24}\) hours, or \(87.5\) minutes. 3. With no wind, the \(40\)-mile round trip would take \(\frac{40}{28}=\frac{10}{7}\) hours, or approximately \(85.7\) minutes. Since \(87.5>85.7\), the claim is false.

a) The cyclist's speed with no wind is \(28\,\text{mph}\), and the wind speed is \(4\,\text{mph}\). b) The claim is false. The trip with wind takes \(87.5\) minutes, while the trip with no wind would take approximately \(85.7\) minutes.

- Write the area of the original rectangle. - Express the dimensions after each change. - Compare each new area with the original area. - The \(lw\) terms cancel after each equation is expanded.

1. The original area is \(lw\). 2. The first condition gives \((l + 2)(w - 1) = lw\). 3. Expand and simplify: \(lw - l + 2w - 2 = lw\), so \(-l + 2w = 2\). 4. The second condition gives \((l - 1)(w + 2) = lw + 12\). 5. Expand and simplify: \(lw + 2l - w - 2 = lw + 12\), so \(2l - w = 14\). 6. From \(-l + 2w = 2\), write \(l = 2w - 2\). 7. Substitute into \(2l - w = 14\): \(2(2w - 2) - w = 14\). 8. Simplify: \(3w - 4 = 14\), so \(w = 6\). Then \(l = 2 \cdot 6 - 2 = 10\).

The original rectangle is \(10\,\text{cm}\) long and \(6\,\text{cm}\) wide.

- Express all four side lengths using the common shorter length. - After finding the side lengths, drop perpendiculars from the endpoints of the shorter base. - How much of the longer base extends beyond the shorter base on each side? - Use a leg of the trapezoid as the hypotenuse of one of the resulting right triangles.

1. Let \(x\) be the common length of each leg and the shorter base. Then the longer base is \(3x\). The perimeter equation is \(x+x+x+3x=24\), so \(6x=24\) and \(x=4\). 2. The side lengths are therefore \(4\,\text{cm}\), \(4\,\text{cm}\), \(4\,\text{cm}\), and \(12\,\text{cm}\). 3. In an isosceles trapezoid, dropping perpendiculars from the shorter base splits the extra base length equally. Each horizontal leg of the resulting right triangles is \(\frac{12-4}{2}=4\,\text{cm}\). 4. If \(h\) is the trapezoid's height, the Pythagorean theorem gives \(h^2+4^2=4^2\). Thus \(h^2=0\), so \(h=0\). 5. A height of zero means the figure collapses into a line segment, so no nondegenerate trapezoid with positive area exists.

The longer base is \(12\,\text{cm}\), and the other three sides are \(4\,\text{cm}\) each. No nondegenerate trapezoid exists because the Pythagorean theorem gives height \(0\,\text{cm}\).

- Use the equal-side property of a square. - Write an area equation for the original rectangle and the adjusted square. - Express one original dimension in terms of the other. - Substitute that relationship into the area equation.

1. Let \(L\) be the original length and \(W\) the original width, in centimeters. 2. Because the adjusted figure is a square, \(L - 3 = W + 2\), so \(L = W + 5\). 3. Equal areas give \(LW = (L - 3)(W + 2)\). 4. Substitute \(L = W + 5\): \((W + 5)W = (W + 2)(W + 2)\). 5. Expand: \(W^2 + 5W = W^2 + 4W + 4\). 6. Simplify: \(W = 4\). 7. Then \(L = 4 + 5 = 9\).

The original rectangle is \(9\,\text{cm}\) long and \(4\,\text{cm}\) wide.

- Choose one train’s speed and time as the two unknowns. - Express the other trains’ speeds and times relative to those unknowns. - The product of speed and time is the same route length for all three trains. - Expand the two distance equations; the product of the unknowns cancels.

1. Let \(v\) be the freight train’s speed in miles per hour and \(t\) its travel time in hours. 2. The passenger train’s speed and time are \(v + 20\) and \(t - 3\). The express train’s speed and time are \(v + 50\) and \(t - 5\). 3. Because all three trains travel the same distance, \(vt = (v + 20)(t - 3)\) and \(vt = (v + 50)(t - 5)\). 4. Expand the first equation and simplify: \(20t - 3v = 60\). 5. Expand the second equation and simplify: \(50t - 5v = 250\), or \(10t - v = 50\). 6. From \(10t - v = 50\), write \(v = 10t - 50\). 7. Substitute into \(20t - 3v = 60\): \(20t - 3(10t - 50) = 60\). Then \(-10t + 150 = 60\), so \(t = 9\). 8. The freight train’s speed is \(v = 10 \cdot 9 - 50 = 40\,\text{mph}\). The passenger train travels at \(60\,\text{mph}\), and the express train travels at \(90\,\text{mph}\). 9. The route length is \(40 \cdot 9 = 360\,\text{mi}\).

The freight train travels at \(40\,\text{mph}\), the passenger train at \(60\,\text{mph}\), and the express train at \(90\,\text{mph}\). The route is \(360\,\text{mi}\) long.

- Write the area of the original rectangle. - Make one area equation for each proposed change. - Expand and simplify; the product \(lw\) cancels in both equations. - Solve the resulting linear system.

1. Let \(l\) be the original length and \(w\) the original width, in feet. The original area is \(lw\). 2. The first condition gives \((l + 2)(w - 1) = lw\). 3. Expand and simplify: \(lw - l + 2w - 2 = lw\), so \(-l + 2w = 2\). 4. The second condition gives \((l - 3)(w + 3) = lw + 6\). 5. Expand and simplify: \(lw + 3l - 3w - 9 = lw + 6\), so \(l - w = 5\). 6. From \(l - w = 5\), write \(l = w + 5\). 7. Substitute into \(-l + 2w = 2\): \(-(w + 5) + 2w = 2\), so \(w = 7\). 8. Then \(l = 7 + 5 = 12\).

The garden bed was planned to be \(12\,\text{ft}\) long and \(7\,\text{ft}\) wide.

- Convert each copper-to-zinc ratio into a copper percentage. - Determine the total mass of copper required in the final alloy. - Write one equation for total mass and one for copper mass. - Solve the resulting system.

1. Alloy A is \(\frac{4}{4 + 1} = 0.80\), or \(80\%\), copper. Alloy B is \(\frac{1}{1 + 3} = 0.25\), or \(25\%\), copper. 2. Let \(x\) be the pounds of Alloy A and \(y\) the pounds of Alloy B. 3. The total mass gives \(x + y = 50\). 4. The copper mass gives \(0.80x + 0.25y = 0.47 \cdot 50\), or \(0.8x + 0.25y = 23.5\). 5. Solve the first equation for \(y\): \(y = 50 - x\). 6. Substitute: \(0.8x + 0.25(50 - x) = 23.5\). 7. Simplify: \(0.8x + 12.5 - 0.25x = 23.5\), so \(0.55x = 11\) and \(x = 20\). 8. Then \(y = 50 - 20 = 30\).

The foundry needs \(20\,\text{lb}\) of Alloy A and \(30\,\text{lb}\) of Alloy B.

- Determine the total mass of salt needed in the final solution. - Account for the \(1\,\text{kg}\) of water when writing the total-mass equation. - Decide how much salt the pure water contributes. - Write one equation for total mass and one for salt mass.

1. Let \(x\) be the mass of Solution A and \(y\) the mass of Solution B, in kilograms. 2. The final mixture has a mass of \(10\,\text{kg}\), including the \(1\,\text{kg}\) of water, so \(x + y + 1 = 10\), or \(x + y = 9\). 3. The final mixture must contain \(0.12 \cdot 10 = 1.2\,\text{kg}\) of salt. Pure water contributes no salt, so \(0.05x + 0.20y = 1.2\). 4. From \(x + y = 9\), write \(x = 9 - y\). 5. Substitute: \(0.05(9 - y) + 0.20y = 1.2\). 6. Simplify: \(0.45 - 0.05y + 0.20y = 1.2\), so \(0.15y = 0.75\) and \(y = 5\). 7. Then \(x = 9 - 5 = 4\).

Add \(4\,\text{kg}\) of Solution A and \(5\,\text{kg}\) of Solution B.

- Determine what happens to the total mass when the same amount is removed from one alloy and added to the other. - Use the original mixture to find a relationship between the two masses. - Write a silver-mass equation for each scenario.

1. Let \(x\) be the planned mass of Alloy A and \(y\) the planned mass of Alloy B, in grams. 2. The original mixture gives \(0.60x + 0.90y = 0.70(x + y)\). 3. Simplify the first equation: \(0.20y = 0.10x\), so \(x = 2y\). 4. The adjusted mixture has the same total mass, so \(0.60(x - 100) + 0.90(y + 100) = 0.80(x + y)\). 5. Substitute \(x = 2y\): \(0.60(2y - 100) + 0.90(y + 100) = 0.80(3y)\). 6. Simplify: \(1.20y - 60 + 0.90y + 90 = 2.40y\), so \(2.10y + 30 = 2.40y\). 7. Therefore, \(30 = 0.30y\), so \(y = 100\), and \(x = 2 \cdot 100 = 200\).

The original plan used \(200\,\text{g}\) of Alloy A and \(100\,\text{g}\) of Alloy B.

- Write a cost rule for Company A. - Represent Company B’s unknown hourly rate and base fee with variables. - Use the \(4\)-hour and \(8\)-hour information to write two equations. - After finding Company B’s rule, compare the two cost expressions with an inequality.

1. Company A’s cost is \(A(t) = 10t + 40\), where \(t\) is the number of hours. 2. Let Company B’s cost be \(B(t) = mt + b\), where \(m\) is the hourly rate and \(b\) is the base fee. 3. At \(4\) hours, the costs are equal: \(4m + b = 10 \cdot 4 + 40 = 80\). 4. At \(8\) hours, Company B costs \(\$20.00\) more: \(8m + b = 10 \cdot 8 + 40 + 20 = 140\). 5. Subtract the first equation from the second: \(4m = 60\), so \(m = 15\). 6. Substitute into \(4m + b = 80\): \(4 \cdot 15 + b = 80\), so \(b = 20\). 7. Compare the costs: Company A is cheaper when \(10t + 40 < 15t + 20\). 8. Solve the inequality: \(20 < 5t\), so \(t > 4\).

Company B charges a \(\$20.00\) base fee and \(\$15.00\) per hour. Company A is less expensive for rentals longer than \(4\) hours.

- Represent the current capacity as rows times seats per row. - Write an equation for each renovation plan. - Expand carefully and cancel the original capacity term. - Solve the resulting linear system and check both plans.

1. Let \(x\) be the current number of rows and \(y\) the current number of seats per row. 2. Plan A gives \((x + 2)(y + 5) = xy + 200\). 3. Plan B gives \((x - 5)(y - 2) = xy - 150\). 4. Expand the first equation: \(xy + 5x + 2y + 10 = xy + 200\), so \(5x + 2y = 190\). 5. Expand the second equation: \(xy - 2x - 5y + 10 = xy - 150\), so \(2x + 5y = 160\). 6. Multiply \(5x + 2y = 190\) by \(5\): \(25x + 10y = 950\). Multiply \(2x + 5y = 160\) by \(-2\): \(-4x - 10y = -320\). 7. Add the equations: \(21x = 630\), so \(x = 30\). 8. Substitute into \(2x + 5y = 160\): \(2 \cdot 30 + 5y = 160\), so \(y = 20\).

The theater currently has \(30\) rows with \(20\) seats in each row.