Free math worksheets

- A solution would be a point shared by both lines. - Compare the directions of the two lines and whether they ever meet.

1. The two lines have the same slope, so they are parallel. 2. They have different y-intercepts, so they are distinct lines. 3. Parallel distinct lines never intersect, so the system has no solution.

No solution; the two distinct lines are parallel and never intersect.

- Compare the coefficients and constants in the two equations. - Check whether one entire equation is a nonzero multiple of the other.

1. The second equation is exactly \(2\) times the first equation. 2. Therefore, both equations represent the same line. 3. Every solution of one equation is also a solution of the other, so the system has infinitely many solutions.

Infinitely many solutions; the two equations represent the same line.

- A solution of a two-line system is a point shared by both lines. - Count how many points the two lines have in common on the graph.

1. The two lines have different slopes, so they are not parallel and do not coincide. 2. The graph shows that they cross at one point. 3. Therefore, the system has exactly one solution.

Exactly one solution, because the two lines intersect at one point.

- A solution of a system must satisfy every equation in the system. - The first coordinate is the \(x\)-value and the second is the \(y\)-value. - Satisfying only one equation is not enough.

1. Substitute \((2, 1)\) into the first equation of part a): \(3(2)+1=7\). Substitute into the second: \(2-2(1)=0\). Both equations are true, so \((2,1)\) is a solution. 2. Substitute \((3,4)\) into the first equation of part b): \(4(3)-3(4)=0\), not \(2\). 3. Although the second equation is true because \(2(3)+4=10\), the ordered pair does not satisfy both equations, so it is not a solution of the system.

a) Yes, \((2, 1)\) is a solution. b) No, \((3, 4)\) is not a solution.

- A point lies on a line when its coordinates satisfy the line equation. - An intersection point must lie on both lines. - Test the point in each equation.

1. An intersection point must satisfy both equations. 2. Test \(P(4, 2)\) in line \(g\): \(1.5(4) - 4 = 6 - 4 = 2\). The point lies on \(g\). 3. Test the point in line \(h\): \(-4 + 6 = 2\). The point also lies on \(h\). 4. Therefore, \(P(4, 2)\) is the intersection point, and the claim is true.

The claim is true because \(P(4, 2)\) satisfies both equations.

- Rewrite the equations in comparable forms. - What does it mean graphically when two equations are equivalent? - Identical left sides with different constants create a contradiction. - Can distinct parallel lines intersect?

1. In part a), \(y = 2x - 5\) is equivalent to \(4x - 2y = 10\). The equations represent the same line, so the system has infinitely many solutions. 2. In part b), the left sides are identical but the constants are different. This would require \(7=-2\), which is impossible. 3. Therefore, the equations in part b) represent distinct parallel lines, so the system has no solution.

a) Infinitely many solutions b) No solution

- Compare the slopes in the two equations. - Then compare the y-intercepts. - What does having the same slope but different intercepts mean for two lines?

1. Both lines have slope \(3\), but their y-intercepts are different: \(-4\) and \(2\). 2. Therefore, the lines are distinct and parallel. 3. Distinct parallel lines do not intersect, so no ordered pair satisfies both equations. 4. Algebraically, setting the equations equal gives \(3x-4=3x+2\), which simplifies to the contradiction \(-4=2\).

The lines are distinct and parallel. There is no ordered pair that satisfies both equations, so the system has no solution.

- Does either line ever cross the other? - What does an intersection represent for a system? - Distinguish between lines that coincide and lines that remain separate.

1. The graph shows two distinct parallel lines. 2. Distinct parallel lines never intersect. 3. Therefore, there is no ordered pair that satisfies both equations, so the solution set is \(\emptyset\).

The solution set is \(\emptyset\) because the two lines are distinct and parallel.

- Isolate \(y\) in each equation. - What does it mean when two slope-intercept equations are identical? - How many points do coincident lines share?

1. Rewrite equation (I) as \(y=\frac{2}{3}x-2\). 2. Rewrite equation (II) as \(y=\frac{2}{3}x-2\). 3. The slope-intercept forms are identical, so the equations represent the same line. 4. Therefore, the system has infinitely many solutions and the two lines coincide.

a) Both equations become \(y = \frac{2}{3}x - 2\). b) The system has infinitely many solutions because the two lines coincide.

- Compare the slopes of the two lines. What does it mean for the number of solutions when the slopes are equal? - If two lines never intersect, can any ordered pair satisfy both equations?

1. Line \(g\) has \(y\)-intercept \(1\) and slope \(0.5\), so its equation is \(y = 0.5x + 1\). 2. Line \(h\) has \(y\)-intercept \(-2\) and slope \(0.5\), so its equation is \(y = 0.5x - 2\). 3. The lines have the same slope but different \(y\)-intercepts, so they are distinct parallel lines. 4. Because the lines have no point in common, the system has no solution. Its solution set is \(\emptyset\).

(I) \(y = 0.5x + 1\); (II) \(y = 0.5x - 2\); solution set: \(\emptyset\)

- Which line is higher at \(x = 0\)? - How do the y-values change as you move to the right, based on the slopes? - Which quadrant has positive x-values and negative y-values?

1. At \(x = 0\), the first line has y-value \(-2\) and the second has y-value \(-7\), so the first line is above the second. 2. The first line has negative slope \(-3\), while the second has positive slope \(2\). Moving to the right, the first line decreases and the second increases, so they must meet at a positive x-value. 3. At that positive x-value, the first equation gives \(y = -3x - 2 < 0\). 4. Therefore, the intersection has \(x > 0\) and \(y < 0\), so it lies in Quadrant IV.

The intersection point lies in Quadrant IV.

- Compare the slopes in each system. - If the lower line at \(x = 0\) is going to catch the upper line for positive \(x\), which line must rise faster? - What happens when the lower line has the smaller slope instead?

1. In both systems, at \(x = 0\) the first line is at \(y = 4\) and the second line is at \(y = -1\), so the first line starts above the second. 2. In System A, the second line has slope \(3\), which is greater than \(0.5\). It rises faster, so it catches the first line for a positive x-value. 3. In System B, the second line has slope \(0.2\), which is less than \(0.5\). Moving right, it falls farther behind the first line, so the intersection must occur for a negative x-value. 4. Therefore, System B has an intersection with a negative x-coordinate.

System B

- Rewrite each equation in slope-intercept form. - What must be true about the slopes of distinct parallel lines? - What does parallelism mean for the number of solutions of the system?

1. Rewrite equation (I): \(y = 2x + 4\). Its slope is \(m_1 = 2\). 2. Rewrite equation (II): \(-2y = -4x + 6\), so \(y = 2x - 3\). Its slope is \(m_2 = 2\). 3. The lines have the same slope but different y-intercepts, \(4\) and \(-3\). Therefore, they are distinct parallel lines. 4. Distinct parallel lines do not intersect, so the corresponding system has no solution.

Both lines have slope \(2\). Because their y-intercepts are different, the lines are parallel and have no intersection point. The system has no solution.

- Rewrite the given equation in slope-intercept form. - What relationship between two distinct lines makes a system have no solution? - Which part of \(y=mx+b\) should stay the same, and which part should change?

1. Rewrite the given equation in slope-intercept form: \(0.5x - 0.25y = 1\) is equivalent to \(y = 2x - 4\). 2. A system has no solution when its two lines are distinct and parallel, so the second line must also have slope \(2\) but a different y-intercept. 3. One valid choice is \(y = 2x - 8\), which is equivalent to \(0.5x - 0.25y = 2\). 4. The two lines have the same slope and different y-intercepts, so they never intersect.

One possible equation is \(0.5x - 0.25y = 2\). It represents a distinct parallel line, so the system has no solution.

- A solution must satisfy both equations. - Substitute the given coordinates into the equation containing \(k\). - Solve the resulting one-variable equation.

1. Because \((5, 2)\) is a solution, it must satisfy equation (II). 2. Substitute \(x = 5\) and \(y = 2\): \(3(5) - 2k = 11\). 3. Simplify: \(15 - 2k = 11\), so \(-2k = -4\). 4. Divide by \(-2\): \(k = 2\).

\(k = 2\)

- What must be true about the slopes of two parallel lines? - How many points do two distinct parallel lines have in common? - For part b), graph both lines carefully in the same coordinate plane.

1. A line parallel to \(g\) must have slope \(\frac{1}{2}\) but a different y-intercept. One possible equation is \(y = \frac{1}{2}x - 2\). 2. The two lines are distinct and parallel, so the system in part a) has no solution. 3. For part b), graph \(y = \frac{1}{2}x + 1\) and \(y = -x + 4\) on the same coordinate plane. The lines intersect at \((2,2)\). 4. Check the graph reading: \(\frac{1}{2}\cdot2+1=2\) and \(-2+4=2\).

a) One possible equation is \(y = \frac{1}{2}x - 2\). The system has no solution. b) The solution of the new system is \((2, 2)\).

- Write both equations in slope-intercept form. - When do two equations represent the same line? - What conditions make two lines parallel? - Could these two lines ever intersect once?

1. Solve equation (I) for \(y\): \(6x - 3y = 12\) becomes \(y = 2x - 4\). 2. For infinitely many solutions, equation (II), \(y = 2x + k\), must represent the same line. Therefore, \(k = -4\). 3. For \(k \ne -4\), the two lines have the same slope \(2\) but different \(y\)-intercepts. They are distinct parallel lines, so the system has no solution.

For \(k = -4\), the system has infinitely many solutions. For every \(k \ne -4\), it has no solution.

- When do two equations represent the same line? - What coefficient relationships make two lines parallel? - Scale one equation so that its \(y\)-coefficient matches the other.

1. Multiply equation (I) by \(-2\): \(-10x+4y=-20\). 2. For infinitely many solutions, equation (II) must be identical to this scaled equation, so \(a=-10\) and \(b=-20\). 3. For no solution, the variable coefficients must remain proportional while the constant is different. 4. Therefore, \(a=-10\) and \(b\ne-20\).

a) \(a = -10\) and \(b = -20\) b) \(a = -10\) and \(b \ne -20\)

- In \(y = mx + b\), which value gives the slope? - When do two lines fail to intersect? - Which solving method is efficient when both equations are solved for \(y\)?

1. A system of two distinct parallel lines has no solution. Equation (I) has slope \(3\), so equation (II) is parallel when \(k=3\). 2. Their y-intercepts, \(-4\) and \(2\), are different, so when \(k=3\) the lines are distinct and the system has no solution. 3. When \(k=1\), set the expressions equal: \(3x-4=x+2\). Then \(2x=6\), so \(x=3\). 4. Substitute into equation (I): \(y=5\). Therefore the solution is \((3,5)\).

a) \(k = 3\) b) \((3, 5)\)

- Write both equations in slope-intercept form. - What does it mean when two lines have the same slope? - Compare their \(y\)-intercepts. - How many points do distinct parallel lines share?

1. Rewrite equation (I) in slope-intercept form: \(6x - 2y = 4\) becomes \(y = 3x - 2\). 2. The two lines have the same slope, \(3\), but different \(y\)-intercepts, \(-2\) and \(5\). 3. Therefore, the lines are distinct and parallel. They have no intersection point, so the system has no solution.

The system has no solution. The lines have \(0\) intersection points.

- Multiply the entire first equation by \(2\). - Compare the result with each second equation. - Must the constants also be proportional?

1. In Example 1, multiplying the first equation by \(2\) gives \(2x + 4y = 10\), exactly the second equation. The system has infinitely many solutions. 2. In Example 2, doubling the first equation would give \(2x + 4y = 10\), not \(12\). The equations represent distinct parallel lines, so the system has no solution. 3. The rule is incomplete. The constant must be multiplied by the same factor as the variable coefficients for the equations to be equivalent.

The student’s rule is incomplete. Infinitely many solutions occur only when every term, including the constant, is scaled by the same factor. Otherwise, the system may have no solution.

- Compare the slopes of the functions first. - What happens when two distinct lines have the same slope? - When solving the equation with fractions, multiplying both sides by a common denominator may help.

1. The functions \(f\) and \(g\) have the same slope, \(\frac{2}{3}\), but different y-intercepts, \(2\) and \(-5\). Therefore, their graphs are distinct parallel lines and do not intersect. 2. For \(g\) and \(h\), set the expressions equal: \(\frac{2}{3}x-5=-x+7\). 3. Then \(\frac{5}{3}x=12\), so \(x=\frac{36}{5}=7.2\). 4. Substitute into \(h\): \(y=-7.2+7=-0.2\). The intersection is \((7.2,-0.2)\).

a) The graphs are parallel because they have the same slope, \(\frac{2}{3}\), but different y-intercepts. b) The intersection is \((7.2, -0.2)\).

- Connect the number of solutions to the relative positions of two lines. - Compare slopes and intercepts. - Rewrite standard-form equations in slope-intercept form when useful.

1. In system (1), the slopes are equal but the y-intercepts differ. The lines are distinct and parallel, so there is no solution. 2. In system (2), the slopes are different. The lines intersect once, so there is exactly one solution. 3. In system (3), rewrite the second equation as \(y=2x+5\). The equations are identical, so there are infinitely many solutions.

(1) No solution (2) Exactly one solution (3) Infinitely many solutions

- Can you find other pairs whose difference is \(5\)? - A unique solution requires a second line that intersects the first at the desired point. - Can the same expression \(a - b\) equal \(5\) and \(8\) at the same time?

1. The equation \(a - b = 5\) has infinitely many solutions, such as \((10, 5)\), \((6, 1)\), and \((0, -5)\). Thus, \((10, 5)\) is only one solution. 2. A second equation that intersects the first at \((10, 5)\) can make the solution unique. For example, \(a + b = 15\). Together, the two equations have the single solution \((10, 5)\). 3. The equations \(a - b = 5\) and \(a - b = 8\) cannot both be true. Subtracting them gives the contradiction \(0 = 3\), so the system has no solution.

1. The equation has infinitely many solutions, so \((10, 5)\) is only one of them. 2. One possible additional equation is \(a + b = 15\). 3. The system has no solution.

- Multiply the first equation by a factor that removes the decimals. - Compare the resulting left sides of the two equations. - Subtract the equations and interpret the statement that remains. - Ask whether one identical linear expression can equal two different constants.

1. Multiply the first equation by \(5\): \(2x - 3y = 6\). 2. Compare it with the second equation, \(2x - 3y = 5\). 3. Subtracting the equations gives \(0 = 1\), a contradiction. 4. Therefore, the system has no solution.

The system has no solution.

- Solve one equation for a variable when substitution is convenient. - Simplify each equation by dividing by a common factor. - How many intersections can two lines have? - Check the solution to part a in both original equations.

1. For part a), solve the second equation for \(y\): \(y=5x-3\). 2. Substitute into the first equation: \(3x+2(5x-3)=7\), so \(13x=13\) and \(x=1\). Then \(y=2\), so the solution set is \(\{(1,2)\}\). 3. For part b), divide the first equation by \(3\) and the second by \(2\). Both become \(2x-3y=4\). 4. The equations represent the same line, so there are infinitely many solutions: \(\{(x,y)\mid 2x-3y=4\}\).

a) \(\{(1, 2)\}\) b) Infinitely many solutions: \(\{(x, y) \mid 2x - 3y = 4\}\)

- Clear fractions or decimals before comparing or solving. - A contradiction means the solution set is empty. - Choose an efficient solving method after simplifying. - Track signs carefully.

1. For part a), multiply the first equation by \(6\): \(3x+4y=30\). 2. This contradicts \(3x+4y=20\), so the system has no solution. 3. For part b), multiply both equations by \(10\): \(4x-3y=11\) and \(5x+2y=8\). 4. Multiply the first new equation by \(2\) and the second by \(3\): \(8x-6y=22\) and \(15x+6y=24\). 5. Add to get \(23x=46\), so \(x=2\). Then \(5(2)+2y=8\), so \(y=-1\). The solution set is \(\{(2,-1)\}\).

a) No solution; the solution set is empty. b) The solution set is \(\{(2, -1)\}\).

- Relate the number of solutions to the positions of the lines. - Compare slopes first, then intercepts. - Equivalent equations describe the same line. - When both equations are solved for \(y\), set their expressions equal.

1. In part a), the lines have the same slope but different y-intercepts, so the system has no solution. 2. In part b), the slopes are different, so there is exactly one solution. Set the expressions equal: \(-x+4=2x-2\). Then \(3x=6\), so \(x=2\) and \(y=2\). 3. In part c), the second equation is twice the first, so the equations represent the same line and there are infinitely many solutions.

a) No solution b) Exactly one solution, \((2, 2)\) c) Infinitely many solutions

- Solve one equation for a variable and substitute. - Be careful with signs when solving equation (II) for \(y\). - Compare the slopes in the two slope-intercept forms. - How do different slopes determine the number of intersections?

1. From equation (I), \(y = -\frac{1}{2}x + 2\). Substitute into equation (II): \(3x - 2\left(-\frac{1}{2}x + 2\right) = 12\). This gives \(4x - 4 = 12\), so \(x = 4\). Then \(y = 0\), so the solution is \((4, 0)\). 2. The slope-intercept forms are \(y = -0.5x + 2\) and \(y = 1.5x - 6\). 3. The slopes are different, so the two lines intersect at exactly one point. Therefore, the system has exactly one solution.

1. \((4, 0)\) 2. \(y = -0.5x + 2\) and \(y = 1.5x - 6\) 3. The slopes are different, so the system has exactly one solution.

- Translate each condition into an equation. - Factor the second equation. - Substitute the expression from the first equation. - Does the result produce a true statement or a contradiction?

1. Let \(x\) and \(y\) be the two numbers. The conditions give \(x + y = 12\) and \(4x + 4y = 50\). 2. Factor the second equation: \(4(x + y) = 50\). 3. Substitute \(x + y = 12\): \(4(12) = 50\), which gives the false statement \(48 = 50\). 4. Therefore, the system is inconsistent and no such pair exists.

No such pair exists because the conditions lead to the contradiction \(48 = 50\).

- Write one equation for each price statement. - Compare the coefficients and constants. - What does it mean when one equation is a multiple of the other? - Could more than one pair of ticket prices satisfy both statements?

1. Let \(x\) be the adult ticket price and \(y\) the child ticket price, in dollars. 2. The system is \(3x + 5y = 110\) and \(9x + 15y = 330\). 3. Multiplying the first equation by \(3\) gives the second equation exactly. 4. The equations contain the same information, so the system has infinitely many solutions. Therefore, the adult ticket price cannot be determined uniquely.

No. The second equation is three times the first, so there are infinitely many possible price pairs and the adult ticket price is not unique.

- Write the perimeter formula for a rectangle. - Express both requirements as equations. - Simplify the perimeter equation before comparing. - What does a contradiction mean for the construction?

1. Let \(l\) be the length and \(w\) the width, in centimeters. 2. The perimeter condition gives \(2(l + w) = 80\), so \(l + w = 40\). 3. The second condition gives \(l + w = 45\). 4. The same sum cannot equal both \(40\) and \(45\). The system has no solution, so the frame cannot be built.

No. A perimeter of \(80\,\text{cm}\) requires \(l + w = 40\,\text{cm}\), which contradicts \(l + w = 45\,\text{cm}\).

- Solve each second equation for \(y\). - Compare the two equations in each system. - When do equal slopes produce distinct parallel lines? - Can equal slopes also describe the same line?

1. In System A, \(x - 2y = -4\) becomes \(y = 0.5x + 2\). In System B, \(x - 2y = 6\) becomes \(y = 0.5x - 3\). 2. In System A, the equations are identical, so there are infinitely many solutions. In System B, the slopes match but the intercepts differ, so there is no solution. 3. The claim is incomplete. Equal slopes give no solution only when the lines have different intercepts. If the intercepts also match, the lines are identical and there are infinitely many solutions.

1. System A: \(y = 0.5x + 2\); System B: \(y = 0.5x - 3\) 2. System A has infinitely many solutions; System B has no solution. 3. Equal slopes give no solution only when the intercepts differ.

- 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): \(\emptyset\).

- Rewrite each equation in slope-intercept form. - Compare slopes and y-intercepts. - Identify parallel, intersecting, and coincident lines in the three graphs. - Relate each graph type to its number of solutions.

1. For system (1), rewrite the second equation as \(y=-x-1\). The lines are parallel and distinct, so the system matches graph B and has no solution. 2. For system (2), rewrite the second equation as \(y=-x+2\). The lines are identical, so the system matches graph A and has infinitely many solutions. 3. For system (3), rewrite the second equation as \(y=2x-1\). The lines intersect once at \((1,1)\), so the system matches graph C and has exactly one solution.

(1) Graph B; no solution (2) Graph A; infinitely many solutions (3) Graph C; exactly one solution

- A point on a line satisfies that line’s equation. - A solution of the entire system must satisfy every equation. - Count points that lie on all three lines, not pairwise intersections. - Would three pairwise intersections provide one point common to every line?

1. A solution of a system is an ordered pair that satisfies every equation at the same time. Graphically, it must lie on all three lines. 2. The graph shows that lines \(g\), \(h\), and \(j\) all pass through \(S(1, 2)\). 3. Therefore, the system has exactly one solution, \((1, 2)\). If the lines had three different pairwise intersection points, no single point would satisfy all three equations.

The statement is incorrect. The system has exactly one solution, \((1, 2)\), because that is the only point common to all three lines.

- A common solution must lie on every line, not merely on two of them. - Compare the three pairwise intersection points in the graph. - Is there one point where all three lines meet?

1. A point satisfies all three constraints only if it lies on all three graphed lines. 2. The graph shows three different pairwise intersections, forming a triangle; there is no single point shared by all three lines. 3. Therefore, \(0\) ordered pairs satisfy all three constraints, and the system has no solution.

There are \(0\) operating points that satisfy all three constraints. The system has no solution.

- A point where two lines intersect represents a common solution. - What does “parallel” imply about common points? - If two lines are exactly the same, which points belong to both lines? - The solution set contains every ordered pair that satisfies both equations.

1. In graph 1, the lines intersect at \((2,-1)\). This point satisfies both equations, so the solution set is \(\{(2,-1)\}\). 2. In graph 2, the lines are distinct and parallel. They have no common point, so the solution set is \(\emptyset\). 3. In graph 3, the lines are identical. Every point on the line satisfies both equations, so the system has infinitely many solutions. 4. The solution set for graph 3 is \(\{(x,y)\mid y=1.5x+1\}\).

a) \(\{(2, -1)\}\) b) The lines are parallel and have no intersection, so the solution set is \(\emptyset\). c) The system has infinitely many solutions. The solution set is \(\{(x, y) \mid y = 1.5x + 1\}\).

- What y-coordinate does a point on the x-axis have? - When are two distinct lines parallel? - Recall the signs of x- and y-coordinates in each quadrant.

1. For the intersection to lie on the x-axis, \(y=0\). From \(0=-x+6\), the intersection must have \(x=6\). Substitute into \(h\): \(0=6m+2\), so \(m=-\frac{1}{3}\). 2. For the lines to have no intersection, they must be distinct and parallel. Since \(g\) has slope \(-1\) and the y-intercepts are different, \(m=-1\). 3. If \(m=1\), solve \(-x+6=x+2\). Then \(4=2x\), so \(x=2\), and substitution gives \(y=4\). 4. Both coordinates are positive, so the intersection lies in Quadrant I.

a) \(m = -\frac{1}{3}\) b) \(m = -1\) c) Quadrant I

- Rewrite the first equation in slope-intercept form. - When do two lines have no common point? - Can changing only the slope make two different fixed intercepts equal? - What must match for two lines to have infinitely many points in common?

1. Rewrite line \(g\): \(2y=-5x+10\), so \(y=-2.5x+5\). Its slope is \(-2.5\), and its y-intercept is \(5\). 2. For the lines to be parallel, line \(h\) must have the same slope, so \(m=-2.5\). Because its y-intercept is \(3\), not \(5\), the lines are distinct and the system has no solution. 3. Infinitely many solutions would require equal slopes and equal y-intercepts. 4. The y-intercept of line \(h\) is fixed at \(3\), so no value of \(m\) can make the two lines identical.

a) \(m = -2.5\) b) No. The fixed y-intercepts, \(5\) and \(3\), are different.

- A specified solution must make both equations true. - Substitute \((3, 2)\) into the general form of equation (II). - What coefficient relationship creates distinct parallel lines?

1. The point \((3,2)\) satisfies equation (I) because \(2(3)+3(2)=12\). 2. To make it satisfy equation (II), substitute the coordinates: \(4(3)+2d=e\), so \(e=12+2d\). 3. To obtain a unique solution, equation (II) must not be a multiple of equation (I). Since doubling equation (I) gives \(4x+6y=24\), choose \(d\ne6\). For example, \(d=1\) gives \(e=14\). 4. For no solution, the variable coefficients must be proportional while the constants are not. Thus \(d=6\) and \(e\ne24\).

a) One possible choice is \(d = 1\) and \(e = 14\). In general, \(e = 12 + 2d\) with \(d \ne 6\). b) \(d = 6\) and \(e \ne 24\)

- How must slopes compare for two lines to intersect once? - When are two lines distinct and parallel? - When do two equations represent the same line?

1. Rewrite equation (I) as \(y=\frac{2}{3}x-2\). 2. For exactly one solution, choose a line with a different slope, such as \(y=x\). 3. For no solution, choose a distinct parallel line, such as \(2x-3y=0\). 4. For infinitely many solutions, choose an equivalent equation, such as \(2x-3y=6\).

a) One possible equation is \(y = x\). b) One possible equation is \(2x - 3y = 0\). c) One possible equation is \(2x - 3y = 6\).

- When do two lines have no common point? - What happens when two lines have different slopes? - What must match for two equations to represent the same line?

1. For no solution, the lines must be distinct and parallel. Choose \(m=3\) and any \(n\ne-5\), such as \(n=0\). 2. For exactly one solution, choose a different slope, so \(m\ne3\). For example, \(m=1\) and \(n=-5\). 3. For infinitely many solutions, the equations must be identical. Thus \(m=3\) and \(n=-5\).

a) One example is \(m = 3\), \(n = 0\). b) One example is \(m = 1\), \(n = -5\). c) \(m = 3\), \(n = -5\)

- Solve both equations for \(y\). - Compare slopes and intercepts. - Which slope condition guarantees an intersection? - What conditions produce parallel or identical lines?

1. Rewrite equation (I) as \(y = -2x + 4\), and equation (II) as \(y = -ax + b\). For infinitely many solutions, the slope and intercept must match, so \(a = 2\) and \(b = 4\). 2. For no solution, the slopes must match but the intercepts must differ, so \(a = 2\) and \(b \ne 4\). 3. For exactly one solution, the slopes must differ, so \(a \ne 2\), with any value of \(b\).

1. \(a = 2\) and \(b = 4\) 2. \(a = 2\) and \(b \ne 4\) 3. \(a \ne 2\), with any \(b\)

- A point of intersection must satisfy both equations. - First find the \(y\)-value when \(x = 2\). - What slope relationship makes two lines parallel?

1. At \(x=2\), equation (I) gives \(y=1.5(2)-3=0\). 2. Substitute \((2,0)\) into equation (II): \(0=2m+2\), so \(m=-1\). 3. For no solution, the lines must be distinct and parallel. Equation (II) must therefore have slope \(1.5\). 4. Because the y-intercepts \(-3\) and \(2\) differ, \(m=1.5\) gives no solution.

a) \(m = -1\) b) \(m = 1.5\)

- Compare slopes and y-intercepts. - What does a shared y-intercept imply about possible intersections? - Use substitution or elimination for the specific value \(a = 1\).

1. Rewrite equation (I) as \(y=-0.5x+2\) and equation (II) as \(y=-\frac{a}{4}x+2\). 2. For infinitely many solutions, the slopes must match: \(-\frac{a}{4}=-0.5\), so \(a=2\). The y-intercepts already match, so the lines are identical. 3. Both lines always have y-intercept \(2\). Therefore, they can be identical or intersect at \((0,2)\), but they can never be distinct parallel lines. No value of \(a\) gives no solution. 4. When \(a=1\), solve \(3x+6y=12\) and \(x+4y=8\). This gives \(y=2\) and \(x=0\), so the solution is \((0,2)\).

a) \(a = 2\) b) No value of \(a\) gives no solution. c) \((0, 2)\)

- Simplify the first equation so its \(x\)-coefficient matches the second. - When do matching variable coefficients produce parallel or identical lines? - For part c, use elimination after substituting the parameter values.

1. Divide equation (1) by \(2\): \(2x+3y=6\). 2. With \(b=5\), the variable coefficients must match while the constants differ for no solution, so \(a=3\). 3. For infinitely many solutions, equation (2) must be identical to the simplified first equation, so \(a=3\) and \(b=6\). 4. With \(a=1\) and \(b=4\), the system is \(4x+6y=12\) and \(2x+y=4\). 5. Multiply the second equation by \(2\): \(4x+2y=8\). Subtract to get \(4y=4\), so \(y=1\). Then \(2x+1=4\), so \(x=1.5\).

a) \(a = 3\) b) \(a = 3\), \(b = 6\) c) The solution set is \(\{(1.5, 1)\}\).

- Rewrite the equations in comparable forms. - Which coefficients determine whether the slopes match? - How do constants distinguish identical from parallel lines? - What happens when the slopes differ?

1. Geometrically, one solution means the lines intersect once; no solution means they are distinct and parallel; infinitely many solutions means they are the same line. 2. Rewrite line \(g\) in standard form: \(3x-2y=-4\). 3. For line \(h\) to be distinct and parallel, its variable coefficients must match those of \(g\) while the constant differs. Thus \(a=2\) and \(b\ne-4\). 4. If \(a\ne2\), the lines are not parallel and intersect once. This includes \(a=0\), when \(h\) is the vertical line \(x=\frac{b}{3}\). 5. The value of \(b\) changes the location of the intersection but not whether it exists when \(a\ne2\).

a) One solution: one intersection; no solution: distinct parallel lines; infinitely many solutions: identical lines. b) \(a = 2\) and \(b \ne -4\) c) \(a \ne 2\); \(b\) changes the intersection’s location but not whether it exists.

- When do two equations describe the same line? - How can a line remain parallel but move to a different position? - A required solution must satisfy the new equation. - Choose a new line that is not parallel to equation (I).

1. For infinitely many solutions, choose an equivalent equation, such as \(2x-3y=4\). 2. For no solution, choose a distinct parallel line, such as \(6x-9y=10\). The lines have the same slope but different y-intercepts. 3. For the unique solution \((2,0)\), choose a nonparallel line through that point, such as \(x+y=2\). 4. The point \((2,0)\) satisfies both equations, and nonparallel lines intersect only once.

a) One possible equation is \(2x - 3y = 4\). b) One possible equation is \(6x - 9y = 10\). c) One possible equation is \(x + y = 2\).

- Compare the quantities and totals in the first two orders. - A scaled copy of an equation does not create a new independent condition. - Combine one original equation with the third-order equation. - Choose substitution or elimination to solve.

1. The second order is exactly half of the first order, including the total cost. Therefore, the two equations are dependent and provide only one independent condition, so they cannot determine two unit prices uniquely. 2. Let \(a\) be the price of one case of apple juice and \(w\) the price of one case of sparkling water. 3. Divide the first-order equation by \(2\) to get \(5a+4w=92\). The third order gives \(3a+5w=76\). 4. Multiply the first equation by \(3\) and the second by \(5\): \(15a+12w=276\) and \(15a+25w=380\). 5. Subtract to get \(13w=104\), so \(w=8\). Then \(5a+32=92\), so \(a=12\).

a) The second statement is only half of the first, so it adds no independent information. b) Apple juice costs \(\$12.00\) per case, and sparkling water costs \(\$8.00\) per case.

- Define variables for the two item counts. - Multiply each unit price by its quantity. - Try several nonnegative integer values for one variable. - What kind of second condition would create another independent equation?

1. Let \(x\) be the number of apples and \(y\) the number of granola bars. The equation is \(0.5x + y = 4.5\). 2. Nonnegative integer solutions include \((1, 4)\), \((3, 3)\), \((5, 2)\), \((7, 1)\), and \((9, 0)\). 3. One equation with two unknowns is underdetermined. In this context, several nonnegative integer pairs satisfy the equation. 4. A second independent condition is needed. For example, if the student bought \(6\) items total, then \(x + y = 6\), and the unique solution is \((3, 3)\).

1. \(0.5x + y = 4.5\) 2. Examples: \((1, 4)\), \((3, 3)\), and \((5, 2)\) 3. One equation in two unknowns has multiple possible solutions. 4. Example: “The student bought \(6\) items total.”

- Scale one equation so the variable coefficients can be compared. - When do two lines share every point? - How can the constant change while the slope stays the same? - What coefficient change will make the slopes different?

1. Multiply equation (II) by \(-3\): \(3x - 6y = 12\), which is exactly equation (I). Therefore, the system has infinitely many solutions. 2. The equations represent the same line. 3. Keep the left side of equation (II) but change the constant, for example to \(-x + 2y = 0\). This creates a distinct parallel line and no solution. 4. Change a variable coefficient so the lines are not parallel, for example \(-x + 3y = -4\). Then the lines intersect once.

1. Infinitely many solutions 2. The lines are identical. 3. One example is \(-x + 2y = 0\). 4. One example is \(-x + 3y = -4\).

- If two lines are identical, how are their equations related? - How many points do two coincident lines have in common? - You can write one equation in the form \(y = mx + b\) and obtain an equivalent second equation by rearranging or multiplying it.

1. The visible line passes through \((0, -1)\) and \((2, 2)\). Its slope is \(\frac{2 - (-1)}{2 - 0} = \frac{3}{2}\), and its \(y\)-intercept is \(-1\). 2. One equation for the line is \(y = \frac{3}{2}x - 1\). 3. An equivalent second equation is \(3x - 2y = 2\). 4. Because both equations describe the same line, every point on the line satisfies both equations. The system has infinitely many solutions. 5. The solution set can be written as \(\{(t, \frac{3}{2}t - 1) \mid t \in \mathbb{R}\}\).

(I) \(y = \frac{3}{2}x - 1\); (II) \(3x - 2y = 2\); solution set: \(\{(t, \frac{3}{2}t - 1) \mid t \in \mathbb{R}\}\)