Aimathic – Linear inequalities in two variables

Linear inequalities in two variables

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Difficulty: 1 – 5

5442549

The graph represents $3x-2y>6$ and shows three marked points. For each point, state whether it is a solution. If it is not a solution, state whether it lies on the boundary.

Hints

- A point is a solution only if its coordinates make the inequality true. - Equality identifies points on the boundary line. - A strict inequality does not include its boundary.

Solution

1. At $A=(4,2)$, $3x-2y=12-4=8$, and $8>6$, so $A$ is a solution. 2. At $B=(2,0)$, $3x-2y=6$, so $B$ lies on the boundary and is not a solution because the inequality is strict. 3. At $C=(0,0)$, $3x-2y=0$, and $0\not>6$, so $C$ is not a solution and does not lie on the boundary.

Answer

A: Point $A$ is a solution. B: Point $B$ lies on the boundary and is not a solution. C: Point $C$ is not a solution and does not lie on the boundary.

5442599

The graph represents a linear inequality and shows a marked point $P$. Write the inequality and state whether $P$ is a solution.

Hints

- A vertical boundary corresponds to an inequality involving only $x$. - Use the shaded side to choose between greater than and less than. - A dashed boundary is not included in the solution set.

Solution

1. The boundary is the vertical line $x=-3$. 2. The boundary is dashed, so equality is excluded. 3. The shaded side contains x-values greater than $-3$, so the inequality is $x>-3$. 4. Point $P=(-3, 5)$ lies on the excluded boundary, so it is not a solution.

Answer

The inequality is $x>-3$. Point $P$ is not a solution.

5442639

Translate the statement “$y$ is greater than or equal to the value that is $3$ less than twice $x$” into a linear inequality. Then state whether its boundary is included.

Hints

- Build the expression on the right side in the order stated. - Match “greater than or equal to” with the correct comparison symbol. - The comparison determines whether equality belongs to the region.

Solution

1. “Twice $x$” is $2x$, and “$3$ less than” gives $2x-3$. 2. “Greater than or equal to” gives $y\ge2x-3$. 3. Equality is allowed, so the boundary line is included.

Answer

$y\ge2x-3$, with a solid boundary.

5442669

Does the entire x-axis belong to the solution set of $2x+5y\ge2x$? Explain by simplifying the inequality.

Hints

- Remove identical terms from both sides first. - Identify the resulting horizontal half-plane. - Every point on the x-axis has the same y-coordinate.

Solution

1. Subtract $2x$ from both sides: $5y\ge0$. 2. Divide by $5$: $y\ge0$. 3. Every point on the x-axis has $y=0$, so the entire x-axis is included as the boundary.

Answer

Yes. The inequality simplifies to $y\ge0$, so the x-axis is included.

5442779

Compare the solution sets of $y<3x+2$ and $y\le3x+2$. What points are in the second set but not the first?

Hints

- Compare the boundary and shaded side separately. - The only difference between the symbols is equality. - Identify the points where equality holds.

Solution

1. Both inequalities describe the half-plane below the same boundary line. 2. The strict inequality excludes the boundary, while the inclusive inequality contains it. 3. Therefore, the exact difference is every point on $y=3x+2$.

Answer

The points in the second set but not the first are exactly the boundary points $y=3x+2$.

5442559

Write the inequality represented by the shaded graph in slope-intercept form.

Hints

- Use two clear grid points on the boundary to find its slope. - A solid boundary means equality is included. - Compare the shaded side with the boundary’s y-values.

Solution

1. The boundary passes through $(0, -4)$ and $(3, 0)$, so its slope is $\frac{0-(-4)}{3-0}=\frac{4}{3}$. 2. The boundary equation is $y=\frac{4}{3}x-4$. 3. The boundary is solid, so equality is included. The shaded region is above the line, so the inequality uses $\ge$.

Answer

$y\ge\frac{4}{3}x-4$.

5442569

Elena rewrites $-2x+5y>15$ as $y>-\frac{2}{5}x+3$. Identify the algebra error, write the correct inequality, and describe the boundary.

Hints

- Move the x-term before dividing. - Adding a term to both sides does not change its sign arbitrarily. - Check the slope by reversing the algebra after isolating the variable.

Solution

1. Add $2x$ to both sides: $5y>2x+15$. 2. Divide by $5$: $y>\frac{2}{5}x+3$. 3. Elena changed the sign of the x-term incorrectly. The boundary $y=\frac{2}{5}x+3$ is dashed because equality is excluded.

Answer

The correct inequality is $y>\frac{2}{5}x+3$, with a dashed boundary.

5442579

A museum gallery can supply at most $2400$ watts to temporary display lights. Each spotlight uses $180$ watts, and each light strip uses $75$ watts. a) Write the power constraint for $s$ spotlights and $l$ light strips, including the count restrictions. b) Determine whether $9$ spotlights and $8$ light strips are allowed.

Hints

- Add the power used by each type of light. - The phrase “at most” determines whether equality is included. - Counts cannot be negative or fractional.

Solution

1. The total power is $180s+75l$, so the power constraint is $180s+75l\le2400$. 2. Because lights are counted, $s$ and $l$ must be nonnegative whole numbers. 3. For $(s, l)=(9, 8)$, the total is $180\cdot9+75\cdot8=2220$. 4. Since $2220\le2400$, the plan is allowed.

Answer

a) $180s+75l\le2400$, with $s,l\in\{0,1,2,\ldots\}$. b) Yes; the plan uses $2220$ watts.

5442609

Write an inequality for the exact complement of the solution set of $x+2y\ge4$. Describe what happens to the boundary line.

Hints

- Negating an inclusive comparison requires both changing direction and changing strictness. - The geometric dividing line does not move. - Decide which region owns the boundary points.

Solution

1. The original region contains all points where $x+2y$ is at least $4$, including equality. 2. Its complement consists of points where $x+2y<4$. 3. The boundary line stays $x+2y=4$, but it is excluded from the complement and therefore dashed.

Answer

The complement is $x+2y<4$, with the same boundary drawn dashed.

5442619

For which real values of $k$ is the point $(2, -1)$ a solution of $kx+3y\le7$?

Hints

- Point membership can be tested by replacing both variables with their coordinates. - The result is an inequality in the parameter alone. - Preserve the comparison while isolating the parameter.

Solution

1. Substitute $(2, -1)$: $2k+3\cdot(-1)\le7$. 2. Simplify: $2k-3\le7$, so $2k\le10$. 3. Therefore, $k\le5$.

Answer

$k\le5$.

5442629

A point satisfying $2x+y<11$ has $x=4$ and an integer y-coordinate. What is the greatest possible value of $y$?

Hints

- Fixing one coordinate turns the condition into a one-variable inequality. - Pay attention to the strict endpoint. - Choose the largest integer that remains below the bound.

Solution

1. Substitute $x=4$: $2\cdot4+y<11$. 2. Then $8+y<11$, so $y<3$. 3. The greatest integer less than $3$ is $2$.

Answer

The greatest possible integer value is $y=2$.

5442649

Diego says the graph represents $y>-4x+1$. Which part of the graph is correct, which part is incorrect, and why?

Hints

- Treat the shaded side and boundary inclusion as separate decisions. - Compare the shaded y-values with the boundary values. - A strict inequality excludes equality.

Solution

1. The shaded side is correct because $y$ must be greater than the boundary value. 2. The boundary is incorrect because the graph uses a solid line. 3. Since the inequality is strict, the boundary must be dashed.

Answer

The shading above the line is correct. The solid boundary is incorrect; it must be dashed.

5442659

Determine whether $2x-y\le5$ and $6x-3y\le15$ have the same solution set. Justify your answer algebraically.

Hints

- Look for a common factor in every term of one inequality. - Dividing by a positive number preserves the comparison direction. - Equivalent inequalities have identical boundaries and shaded sides.

Solution

1. Divide $6x-3y\le15$ by the positive number $3$. 2. The result is $2x-y\le5$, with the inequality direction unchanged. 3. Therefore, the inequalities are equivalent and describe the same half-plane with the boundary included.

Answer

Yes. The inequalities are equivalent and have the same solution set.

5442679

Describe exactly which points on the y-axis satisfy $3x-y<6$. State whether the endpoint is included.

Hints

- Use the coordinate condition that defines the y-axis. - Isolate the remaining coordinate carefully. - Strictness determines whether the boundary point belongs to the set.

Solution

1. Points on the y-axis have $x=0$. 2. Substitute: $-y<6$. 3. Multiply by $-1$ and reverse the comparison: $y>-6$. 4. Thus the solution is the part of the y-axis above $(0, -6)$, excluding the endpoint.

Answer

All points $(0, y)$ with $y>-6$; the endpoint $(0, -6)$ is excluded.

5442719

The graph represents $y>x$ and shows two points that are reflections of each other across the boundary. State whether each point is a solution, and explain what the pair shows geometrically.

Hints

- Test each ordered pair directly in the inequality. - Swapping coordinates reflects a point across $y=x$. - Compare the two points’ positions relative to the boundary.

Solution

1. For $P=(2, 5)$, $5>2$, so $P$ is a solution. 2. For $Q=(5, 2)$, $2>5$ is false, so $Q$ is not a solution. 3. Reflection across the boundary $y=x$ exchanges the two half-planes.

Answer

Point $P$ is a solution, and point $Q$ is not. Reflection across $y=x$ swaps the two sides of the boundary.

5442739

A boundary is $y=-x+2$. Aisha tests $(1,1)$ to decide which side should be shaded and gets equality. Explain why this test point is inconclusive. If the region is known to contain $(0,0)$ and exclude the boundary, write the inequality.

Hints

- Check whether the proposed test point lies on the dividing line. - A useful test point must lie strictly on one side. - Use the included point to choose between the two strict inequalities.

Solution

1. The point $(1,1)$ lies on the boundary because $1=-1+2$. 2. A boundary point gives equality for both possible sides, so it cannot identify which half-plane is intended. 3. Test $(0,0)$: $0<2$, so the side containing the origin is $y<-x+2$. 4. The boundary is excluded, matching the strict inequality.

Answer

The test is inconclusive because $(1,1)$ is on the boundary. The region is $y<-x+2$.

5442759

Min-jun rewrites $4x-2y\ge8$ as $y\ge2x-4$. Identify the error, give the correct form, and state the shaded side.

Hints

- Isolate the term containing $y$ before dividing. - Check the sign of the coefficient used in the final division. - The final comparison determines the shaded side.

Solution

1. Subtract $4x$: $-2y\ge8-4x$. 2. Divide by $-2$ and reverse the comparison: $y\le2x-4$. 3. Min-jun failed to reverse the inequality when dividing by a negative number. The region is on or below the solid boundary.

Answer

The correct inequality is $y\le2x-4$; shade on or below the solid line.

5442769

A point satisfying $2x+y>0$ has $x=1$, and its y-coordinate is an integer from $-3$ through $5$. How many possible points are there?

Hints

- Fix the given x-coordinate first. - Solve the resulting strict inequality for y. - Count only integers that also lie in the stated interval.

Solution

1. Substitute $x=1$: $2+y>0$. 2. Thus $y>-2$. 3. In the stated range, the valid integer values are $-1,0,1,2,3,4,5$, giving $7$ points.

Answer

There are $7$ possible points.

5442789

A hallway mural must cover at least $60$ square feet. Each large panel covers $8$ square feet, and each small panel covers $3$ square feet. a) Write the coverage constraint for $L$ large panels and $S$ small panels, including the count restrictions. b) If $5$ large panels are used, what is the least whole number of small panels needed?

Hints

- Add the coverage contributed by each panel size. - Panel counts must be nonnegative whole numbers. - Round in the direction that still meets the minimum.

Solution

1. The coverage requirement is $8L+3S\ge60$. 2. Because panels are counted, $L$ and $S$ must be nonnegative whole numbers. 3. With $L=5$, $40+3S\ge60$. 4. Then $3S\ge20$, so $S\ge\frac{20}{3}$. The least whole number satisfying the requirement is $7$.

Answer

a) $8L+3S\ge60$, with $L,S\in\{0,1,2,\ldots\}$. b) At least $7$ small panels.

5442809

An event crew has a $\$900$ equipment budget. Delivery costs $\$75$, each stage riser costs $\$85$, and each microphone stand costs $\$25$. a) Write the budget constraint for $r$ risers and $m$ stands, including the count restrictions. b) Is a plan with $6$ risers and $10$ stands within budget?

Hints

- Include both variable costs and the fixed delivery charge. - A budget is an upper inclusive limit. - Counts must be nonnegative whole numbers.

Solution

1. The total cost is $85r+25m+75$, so the budget constraint is $85r+25m+75\le900$. 2. Because risers and stands are counted, $r$ and $m$ must be nonnegative whole numbers. 3. The proposed plan costs $85\cdot6+25\cdot10+75=510+250+75=835$. 4. Since $835\le900$, the plan is within budget.

Answer

a) $85r+25m+75\le900$, with $r,m\in\{0,1,2,\ldots\}$. b) Yes; the total cost is $\$835$.

5442819

Fatima rewrites $2x+y\ge6$ correctly as $y\ge-2x+6$ but then shades below the line. Explain the mismatch and give the correct graph description.

Hints

- Read the final comparison as a statement about y-values. - Larger y-values appear above a line. - Treat boundary style separately from shading direction.

Solution

1. The isolated form compares $y$ directly with the boundary value $-2x+6$. 2. The symbol $\ge$ means y-values on or above the boundary. 3. The boundary is solid because equality is included, and the region must be shaded above, not below.

Answer

Use a solid line $y=-2x+6$ and shade on or above it.

5442829

For which real values of $a$ is the origin a solution of $y<ax+3$? Explain why the answer does not depend on the slope.

Hints

- Test the origin directly. - Notice what happens to any term multiplied by zero. - The intercept controls the boundary’s value at the y-axis.

Solution

1. Substitute the origin: $0<a(0)+3$. 2. This simplifies to $0<3$, which is true for every real $a$. 3. At $x=0$, the slope term vanishes, so whether the origin is a solution depends only on the intercept.

Answer

The origin is a solution for every real value of $a$.

5442839

Rewrite $5x+2y<3x+10$ with $y$ isolated. Then describe the boundary and shaded side.

Hints

- Collect variable terms before isolating y. - Dividing by a positive number keeps the comparison direction. - The final y-comparison gives the shading direction.

Solution

1. Subtract $3x$: $2x+2y<10$. 2. Subtract $2x$: $2y<10-2x$. 3. Divide by $2$: $y<-x+5$. 4. The boundary is dashed, and the solution set lies below it.

Answer

$y<-x+5$; use a dashed boundary and shade below.

5442849

A theater crew has at most $12$ gallons of paint. Each large scenic flat uses $0.8$ gallon, and each small prop uses $0.25$ gallon. a) Write the paint constraint for $f$ flats and $p$ props, including the count restrictions. b) If $10$ flats are painted, what is the greatest whole number of props that can also be painted?

Hints

- Add the paint used by both kinds of objects. - Counts must be nonnegative whole numbers. - Use the remaining paint to bound the whole-number prop count.

Solution

1. The paint constraint is $0.8f+0.25p\le12$. 2. Because flats and props are counted, $f$ and $p$ must be nonnegative whole numbers. 3. With $f=10$, $8+0.25p\le12$. 4. Then $0.25p\le4$, so $p\le16$. The greatest whole number is $16$.

Answer

a) $0.8f+0.25p\le12$, with $f,p\in\{0,1,2,\ldots\}$. b) At most $16$ props.

5442909

The boundary of the half-plane $y\le mx+b$ must pass through the origin. What value must $b$ have? Does this condition restrict $m$? Explain.

Hints

- A point on the boundary makes the corresponding equation true. - Substitute both coordinates of the origin. - Notice which parameter disappears after the substitution.

Solution

1. A boundary point satisfies the equation $y=mx+b$. 2. Substituting the origin gives $0=m\cdot0+b$, so $b=0$. 3. The term containing $m$ is zero at the origin, so any real value of $m$ is possible.

Answer

$b=0$, and $m$ may be any real number.

5442929

Maya classified four points for the inequality $y\ge2x-1$: <table><tr><th>Point</th><th>Maya's classification</th></tr><tr><td>$(0, -1)$</td><td>is a solution and is not on the boundary</td></tr><tr><td>$(2, 4)$</td><td>is a solution and is not on the boundary</td></tr><tr><td>$(-1, -4)$</td><td>is not a solution</td></tr><tr><td>$(3, 5)$</td><td>lies on the boundary</td></tr></table> Exactly one classification is incorrect. Identify it and give the correct classification.

Hints

- Compare each y-coordinate with the boundary value at the same x-coordinate. - Equality identifies a boundary point. - Check all four rows before deciding which single entry is wrong.

Solution

1. For $(0, -1)$, the boundary value is $2\cdot0-1=-1$, so the point lies on the boundary and is a solution. 2. For $(2, 4)$, $4>2\cdot2-1=3$, so it is a solution and is not on the boundary. 3. For $(-1, -4)$, $-4<2\cdot(-1)-1=-3$, so it is not a solution. 4. For $(3, 5)$, $5=2\cdot3-1$, so it lies on the boundary and is a solution.

Answer

Maya's classification of $(0, -1)$ is incorrect. It lies on the boundary and is a solution.

5442949

Ethan says that for a half-plane with boundary $x+y=4$, the point $(1, 3)$ is a solution and is not on the boundary, while $(2, 2)$ is not a solution. Explain why no linear inequality with that boundary can make both classifications correct.

Hints

- Check each point against the boundary equation before considering either side. - Points producing the same boundary equality cannot be placed on opposite sides. - Think about how a strict or inclusive symbol treats every point on one boundary line.

Solution

1. For $(1, 3)$, $x+y=1+3=4$. 2. For $(2, 2)$, $x+y=2+2=4$. 3. Both points lie on the stated boundary line. 4. A linear inequality with this boundary either includes both boundary points or excludes both, so Ethan's classifications cannot both be correct.

Answer

Ethan's classifications are impossible because both points lie on $x+y=4$. Any inequality with that boundary either includes both points or excludes both points.

5442959

Multiply $-3x+6y<9$ by $-2$ to write an equivalent inequality with a positive x-coefficient. Explain why the inequality symbol must change.

Hints

- Apply the same factor to every term on both sides. - Recall how order changes when numbers are multiplied by a negative value. - Check equivalence by testing one simple point in both inequalities.

Solution

1. Multiplying every term by $-2$ gives $6x-12y$ on the left and $-18$ on the right. 2. Multiplication by a negative number reverses the order, so the equivalent inequality is $6x-12y>-18$. 3. Reversing the symbol preserves the same set of ordered-pair solutions.

Answer

$6x-12y>-18$. The symbol reverses because every quantity is multiplied by a negative number.

5550439

The four panels use the same boundary line for $2x+y<6$, but they differ in boundary style and shading. Which panel correctly graphs the inequality? Explain both why the boundary has that style and why the shaded side is correct.

Hints

- First identify the boundary equation by replacing the inequality sign with equality. - Decide whether equality is allowed before choosing the boundary style. - Use either the isolated inequality or a test point to decide which side of the boundary belongs to the solution set.

Solution

1. Rewrite the inequality as $y<-2x+6$. The boundary is the line $y=-2x+6$. 2. Because the inequality is strict, points on the boundary are not included, so the line must be dashed. 3. The inequality requires y-values below the boundary line. Equivalently, the test point $(0,0)$ satisfies $0<6$, so the side containing the origin is shaded. 4. Panel b) has a dashed boundary and shades below the line.

Answer

Panel b). The boundary is dashed because the inequality is strict, and the region below $y=-2x+6$ is shaded.

5442589

A boundary line passes through $(0, 2)$ and $(4, 0)$. The boundary is included, and the solution set contains the origin. Write an inequality in slope-intercept form for the half-plane.

Hints

- Use the two boundary points to find the line’s rate of change. - A test point determines which side of the line is included. - Boundary inclusion determines whether the inequality is strict.

Solution

1. The boundary slope is $\frac{0-2}{4-0}=-\frac{1}{2}$, and the y-intercept is $2$, so the boundary is $y=-\frac{1}{2}x+2$. 2. Test the origin: $0\le2$ is true, so the origin lies on the $y\le-\frac{1}{2}x+2$ side. 3. The boundary is included, so the inequality includes equality.

Answer

$y\le-\frac{1}{2}x+2$.

5442689

Rewrite $\frac{x}{3}-\frac{y}{2}\ge1$ with integer coefficients. Then find the intercepts of its boundary and state the boundary style.

Hints

- Use a positive common multiple to clear both denominators. - Find intercepts by setting one coordinate equal to zero at a time. - The comparison symbol determines the line style.

Solution

1. Multiply by $6$, a positive number: $2x-3y\ge6$. 2. With $y=0$, $2x=6$, so the x-intercept is $(3, 0)$. 3. With $x=0$, $-3y=6$, so the y-intercept is $(0, -2)$. 4. Equality is included, so the boundary is solid.

Answer

The inequality is $2x-3y\ge6$. The boundary intercepts are $(3, 0)$ and $(0, -2)$, and the boundary is solid.

5442699

The region $y\le-x+2$ is translated upward $5$ units. a) Write the inequality for the translated region. b) Determine whether $(3, 4)$ belongs to the original region, the translated region, both, or neither.

Hints

- A vertical translation changes the constant term of the boundary. - Keep the same inequality direction because the shaded side moves with the line. - Test the point separately in both regions.

Solution

1. Translating upward $5$ units changes the boundary from $y=-x+2$ to $y=-x+7$, with the same inclusive side. The new inequality is $y\le-x+7$. 2. For the original region at $(3, 4)$, $4\le-3+2=-1$ is false. 3. For the translated region, $4\le-3+7=4$ is true. 4. The point belongs only to the translated region.

Answer

a) $y\le-x+7$ b) $(3, 4)$ belongs only to the translated region.

5442729

A boundary line has x-intercept $-3$ and y-intercept $6$. The boundary is excluded, and the solution set lies above the boundary. Write the inequality in slope-intercept form.

Hints

- Convert the intercepts into two ordered pairs. - Use those points to find the boundary equation. - Boundary exclusion and the specified side determine the comparison symbol.

Solution

1. The boundary passes through $(-3, 0)$ and $(0, 6)$. 2. Its slope is $\frac{6-0}{0-(-3)}=2$, so the boundary is $y=2x+6$. 3. The boundary is excluded and the solution set lies above the line, so the inequality is $y>2x+6$.

Answer

$y>2x+6$.

5442749

The line $y=mx+1$ is the boundary of $3x-y\le-1$. a) Find $m$. b) Rewrite the inequality with $y$ isolated and state the shaded side.

Hints

- The boundary comes from replacing the comparison with equality. - Isolate the vertical coordinate in both the equation and inequality. - A negative coefficient requires care with the comparison direction.

Solution

1. Replace the inequality with equality: $3x-y=-1$. 2. Solve for $y$: $-y=-1-3x$, so $y=3x+1$. Therefore, $m=3$. 3. Solving the inequality gives $-y\le-1-3x$, then $y\ge3x+1$ after reversing the comparison. 4. The region is on or above the solid boundary.

Answer

a) $m=3$ b) $y\ge3x+1$; shade on or above the line.

5442799

The point $(2, 3)$ lies on the boundary $ax-y=5$. The boundary is dashed, and the solution set does not contain the origin. a) Find $a$. b) Write the inequality for the solution set.

Hints

- Use equality at the boundary point to recover the coefficient. - Test the origin in the completed left side. - Choose the strict comparison that makes the test point false.

Solution

1. Substitute the boundary point: $2a-3=5$, so $2a=8$ and $a=4$. 2. The boundary is $4x-y=5$. 3. The origin gives $0$, which is less than $5$. To exclude the origin, choose $4x-y>5$. 4. The strict symbol matches the dashed boundary.

Answer

a) $a=4$ b) $4x-y>5$

5442859

For which real values of $k$ do both points $(0, 1)$ and $(2, 5)$ satisfy $y\le kx+3$?

Hints

- Test each point separately. - One condition may place no restriction on the parameter. - Intersect the parameter conditions from both points.

Solution

1. For $(0, 1)$, the condition is $1\le3$, which is always true. 2. For $(2, 5)$, the condition is $5\le2k+3$. 3. Thus $2\le2k$, so $k\ge1$. 4. Both points satisfy the inequality exactly when $k\ge1$.

Answer

$k\ge1$.

5442889

A dashed boundary line passes through $(0, 4)$ and $(6, 0)$. The point $(0, 5)$ is a solution. Write the linear inequality in standard form.

Hints

- Use the two intercepts to determine the equation of the boundary line. - Test the given point to decide which comparison direction is correct. - Use the boundary style to decide whether equality belongs in the inequality.

Solution

1. The intercepts give the boundary equation $\frac{x}{6}+\frac{y}{4}=1$, which is equivalent to $2x+3y=12$. 2. At $(0, 5)$, the expression $2x+3y$ equals $15$, which is greater than $12$. 3. The dashed boundary means equality is excluded, so the inequality is $2x+3y>12$.

Answer

$2x+3y>12$

5442899

Rewrite $2(x-y)+3\le x+7$ in slope-intercept form. Then describe the boundary line and the side of the line that belongs to the solution set.

Hints

- Expand the grouped expression before combining like terms. - Isolate the y-term and notice the sign of its coefficient. - Use the final comparison to determine both the boundary style and the shaded side.

Solution

1. Expand and simplify: $2x-2y+3\le x+7$, so $x-2y\le4$. 2. Subtract $x$: $-2y\le4-x$. 3. Divide by $-2$ and reverse the inequality: $y\ge\frac{1}{2}x-2$. 4. The boundary $y=\frac{1}{2}x-2$ is solid, and the solution set is on or above it.

Answer

$y\ge\frac{1}{2}x-2$. The boundary is a solid line, and the solution set is on or above the line.

5442919

How many ordered pairs of integers $(x, y)$ satisfy all three conditions $0\le x\le4$, $0\le y\le3$, and $2x+y<7$? Show how you count them.

Hints

- Fix one integer coordinate at a time and find the allowable values of the other coordinate. - Use the given bounds so that only finitely many cases need checking. - Organize the count to avoid listing the same pair twice or overlooking an endpoint.

Solution

1. For $x=0$, all four values $y=0,1,2,3$ work. 2. For $x=1$, all four values $y=0,1,2,3$ work. 3. For $x=2$, the values $y=0,1,2$ work, giving three pairs. 4. For $x=3$, only $y=0$ works, giving one pair. 5. For $x=4$, no allowed value of $y$ works. The total is $4+4+3+1=12$.

Answer

There are $12$ ordered pairs.

5442969

Simplify $4x+2y\le3x-y+9$. Then give the x-intercept and y-intercept of its boundary and state which side of the boundary belongs to the solution set.

Hints

- Collect all variable terms on one side before finding intercepts. - Set one coordinate equal to zero at a time to locate the boundary intercepts. - Rewrite in terms of y to identify the selected side of the line.

Solution

1. Subtract $3x$ and add $y$ to get $x+3y\le9$. 2. Setting $y=0$ gives the x-intercept $(9, 0)$. 3. Setting $x=0$ gives the y-intercept $(0, 3)$. 4. Solving for $y$ gives $y\le-\frac{1}{3}x+3$, so the solid boundary and the side below it belong to the solution set.

Answer

The simplified inequality is $x+3y\le9$. The boundary intercepts are $(9, 0)$ and $(0, 3)$, and the solution set is on or below the solid boundary line.

5550449

Rewrite $3x-2y\le6$ as an inequality in $y$. Then choose the panel that correctly graphs it. Explain why the inequality sign changes while isolating $y$, and justify both the boundary style and the shaded side in your chosen panel.

Hints

- Isolate the term containing $y$ before dividing. - What special rule applies when an inequality is divided by a negative number? - After isolating $y$, read the inequality as a statement about which y-values are allowed relative to the boundary line.

Solution

1. Subtract $3x$: $-2y\le6-3x$. 2. Divide by $-2$. Because the divisor is negative, reverse the inequality: $y\ge\frac{3}{2}x-3$. 3. Equality is included, so the boundary line is solid. 4. The inequality requires points on or above the line. Panel d) shows a solid boundary and shades above it.

Answer

$y\ge\frac{3}{2}x-3$; panel d). Dividing by $-2$ reverses the inequality, the boundary is solid, and the region above the line is shaded.

5442709

Reflect the region $y>-x+1$ across the x-axis. Write the inequality for the reflected region and state the boundary style.

Hints

- An x-axis reflection changes the sign of the vertical coordinate. - Substitute the transformed coordinate before simplifying. - Multiplying by a negative number reverses the comparison.

Solution

1. Under reflection across the x-axis, a reflected point $(x,y)$ comes from $(x,-y)$ in the original region. 2. Substitute $-y>-x+1$. 3. Multiply by $-1$ and reverse the comparison: $y<x-1$. 4. The inequality remains strict, so the boundary is dashed.

Answer

The reflected region is $y<x-1$, with a dashed boundary.

5442869

Every point in the region $x+y\le5$ is dilated from the origin by a scale factor of $2$. Write an inequality for the image region.

Hints

- Relate an image point to its preimage under the dilation. - Substitute the preimage coordinates into the original inequality. - Clear the positive scale factor after substitution.

Solution

1. An image point $(X, Y)$ comes from $\left(\frac{X}{2}, \frac{Y}{2}\right)$ in the original region. 2. Substitute: $\frac{X}{2}+\frac{Y}{2}\le5$. 3. Multiply by $2$: $X+Y\le10$. 4. Renaming the image coordinates as $(x, y)$, the image region is $x+y\le10$.

Answer

$x+y\le10$.

5442879

A half-plane includes its boundary and has the classifications below. <table><tr><th>Point</th><th>Classification</th></tr><tr><td>$(2, 2)$</td><td>lies on the boundary</td></tr><tr><td>$(6, 0)$</td><td>lies on the boundary</td></tr><tr><td>$(0, 0)$</td><td>is a solution</td></tr><tr><td>$(4, 2)$</td><td>is not a solution</td></tr></table> Write an inequality for the half-plane.

Hints

- Use the boundary points to determine the line. - Boundary inclusion requires an inclusive comparison. - Use the solution and nonsolution points to confirm the correct direction.

Solution

1. The boundary through $(2, 2)$ and $(6, 0)$ has slope $\frac{0-2}{6-2}=-\frac{1}{2}$. 2. Its equation is $y=-\frac{1}{2}x+3$, or $x+2y=6$. 3. The origin gives $0\le6$, so the solution set is $x+2y\le6$. 4. The point $(4, 2)$ gives $8>6$, confirming that it is not a solution.

Answer

$x+2y\le6$.

5442939

The half-plane $2x-y\le6$ is reflected through the origin. Write an inequality for the reflected half-plane. Then determine whether $(4, 1)$ is a solution of the reflected inequality.

Hints

- Express the original coordinates in terms of a point after a half-turn about the origin. - Substitute both transformed coordinates before simplifying. - Test the given point only after the reflected inequality is established.

Solution

1. Reflection through the origin sends a reflected point $(x, y)$ back to $(-x, -y)$ in the original region. 2. Substitute $(-x, -y)$ into the original inequality: $2(-x)-(-y)\le6$. 3. Simplify to $-2x+y\le6$, which is equivalent to $2x-y\ge-6$. 4. At $(4, 1)$, $2x-y=7$, and $7\ge-6$, so the point is a solution.

Answer

The reflected inequality is $2x-y\ge-6$. The point $(4, 1)$ is a solution.

5442979

Compare the line $y=2x+b$ with the inequality $y>2x-5$. For each possible value of $b$, determine whether every point on the line is a solution, the line is the excluded boundary, or no point on the line is a solution.

Hints

- Compare the two parallel-line expressions for a general x-coordinate. - Cancel the common part before analyzing the parameter. - Treat the equality case separately because the inequality is strict.

Solution

1. Substitute the line expression into the inequality: $2x+b>2x-5$. 2. Cancel $2x$ to get $b>-5$. 3. If $b>-5$, every point on the line is a solution. 4. If $b=-5$, the line is the excluded boundary, so none of its points are solutions. 5. If $b<-5$, no point on the line is a solution.

Answer

If $b>-5$, every point is a solution. If $b=-5$, the line is the excluded boundary. If $b<-5$, no point is a solution.

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