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.

5442649

A student 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.

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 $y=3x+2$.

5442819

A student 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.

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

A student 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. The student 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 $y=2$.

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 $y=-x+2$. A student 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

A student 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. The student 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 $\$835$.

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

A student classified four points for the inequality $y\ge2x-1$: <table><tr><th>Point</th><th>Student'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

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

5442949

A student 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 the stated classifications cannot both be correct.

Answer

The 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.

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.

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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