Aimathic – Solve and graph inequalities

Solve and graph inequalities

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Difficulty: 1 – 5

5509779

The graph shows an inequality solution set on the x-axis. A drawn ray continues beyond the graph boundary it reaches. a) Write the solution as an inequality. b) Write the solution in interval notation.

Hints

- Decide whether the endpoint itself belongs to the solution set. - Use the direction of the ray to decide whether the solutions are greater than or less than the endpoint. - In interval notation, infinity always uses a parenthesis.

Solution

a) The open endpoint means 2 is not included, and the ray extends to values less than 2. Therefore, $x<2$. b) Values extend without bound to the left and stop before 2, so the interval is $(-\infty, 2)$.

Answer

a) $x<2$ b) $(-\infty, 2)$

5548709

Solve $x+4\le1$. Then describe its number-line graph by stating the boundary point, whether the point is open or closed, and the shading direction.

Hints

- Isolate $x$ with one balancing step. - Does the inequality symbol include the boundary value? - On a number line, where are values smaller than the boundary located?

Solution

1. Subtract $4$ from both sides: $x\le-3$. 2. Equality is included, so the boundary point at $-3$ is closed. 3. Values less than $-3$ lie to the left, so the graph is shaded left from $-3$.

Answer

$x\le-3$; closed point at $-3$, shaded to the left.

5131809

Solve each inequality over the real numbers. Give each solution in set-builder notation. a) $5x+12<2x-3$ b) $4-3x\ge16+x$

Hints

- What happens to the inequality sign when multiplying or dividing by a negative number? - Move variable terms to one side and constants to the other. - Solve an inequality like an equation, while applying the sign-reversal rule.

Solution

a) Subtract $2x$ and $12$: $3x<-15$. Divide by $3$: $x<-5$. b) Subtract $x$ and $4$: $-4x\ge12$. Divide by $-4$ and reverse the inequality: $x\le-3$.

Answer

a) $\{x\in\mathbb{R}\mid x<-5\}$ b) $\{x\in\mathbb{R}\mid x\le-3\}$

5154609

Solve the inequality over the real numbers and give the solution set. $0.4x+\frac{3}{5}>1.8-0.2x$

Hints

- Convert the fraction to a decimal if that makes the arithmetic easier. - Move all variable terms to one side. - Isolate $x$ using inverse operations.

Solution

1. Since $\frac{3}{5}=0.6$, rewrite the inequality as $0.4x+0.6>1.8-0.2x$. 2. Add $0.2x$: $0.6x+0.6>1.8$. 3. Subtract $0.6$: $0.6x>1.2$. 4. Divide by $0.6$: $x>2$.

Answer

$\{x\in\mathbb{R}\mid x>2\}$

5240919

Find the solution set of each inequality over the real numbers. a) $4x-11<17$ b) $15-2x\ge21$ c) $3(x+4)\le5x+2$ d) $\frac{x-2}{3}>2$

Hints

- Use balancing steps just as you would for equations. - Watch for the one part where isolating the variable requires division by a negative number. - Distribute before collecting variable terms when parentheses are present.

Solution

1. In (a), add $11$: $4x<28$. Divide by $4$: $x<7$. 2. In (b), subtract $15$: $-2x\ge6$. Divide by $-2$ and reverse the inequality: $x\le-3$. 3. In (c), distribute: $3x+12\le5x+2$. Then $10\le2x$, so $x\ge5$. 4. In (d), multiply by the positive number $3$: $x-2>6$. Add $2$: $x>8$.

Answer

a) $x<7$ b) $x\le-3$ c) $x\ge5$ d) $x>8$

5241039

Write each solution set in interval notation. a) $x > -3$ and $x < 6$ b) $11 \ge x > 4$ c) $x$ is strictly between $-7.5$ and $-2$. d) $15 > x > 8.4$

Hints

- Use parentheses for an excluded endpoint. - Use a bracket for an included endpoint. - Write the smaller endpoint first. - Translate each compound inequality before writing the interval.

Solution

a) The endpoints are excluded: $-3 < x < 6$, so the interval is $(-3, 6)$. b) $4$ is excluded and $11$ is included: $4 < x \le 11$, so the interval is $(4, 11]$. c) Both endpoints are excluded: $-7.5 < x < -2$, so the interval is $(-7.5, -2)$. d) Both endpoints are excluded: $8.4 < x < 15$, so the interval is $(8.4, 15)$.

Answer

a) $(-3, 6)$ b) $(4, 11]$ c) $(-7.5, -2)$ d) $(8.4, 15)$

5509789

The two panels show solution-set rays on an x-axis. Each ray continues to the edge of its panel. a) For panel a), write the solution as an inequality and in interval notation. b) For panel b), write the solution as an inequality and in interval notation. c) Explain how the endpoint marker and ray direction determine the inequality symbol.

Hints

- First decide whether each endpoint is included or excluded. - Then use the direction of each ray to decide whether the values are less than or greater than the endpoint. - Match inclusion at a finite endpoint with a bracket in interval notation.

Solution

a) The filled endpoint at $-1$ is included, and the ray extends left. The solution is $x\le-1$, or $(-\infty, -1]$. b) The open endpoint at 3 is excluded, and the ray extends right. The solution is $x>3$, or $(3, \infty)$. c) A filled endpoint includes the boundary, so the inequality uses $\le$ or $\ge$. An open endpoint excludes the boundary, so it uses $<$ or $>$. A leftward ray represents values less than the endpoint; a rightward ray represents values greater than the endpoint.

Answer

a) $x\le-1$; $(-\infty, -1]$ b) $x>3$; $(3, \infty)$ c) Filled means the endpoint is included; open means it is excluded. Leftward means less than, and rightward means greater than.

5548719

Solve $-2x + 5 \ge 11$. Then choose the panel that correctly graphs the solution on a number line. A colored segment that reaches a panel edge continues beyond the visible boundary.

Hints

- Solve the inequality before comparing the graphs. - What happens to an inequality sign when you divide by a negative number? - Decide whether the endpoint should be open or closed and whether the solution extends left or right.

Solution

1. Subtract $5$ from both sides: $-2x \ge 6$. 2. Divide by $-2$. Because you divide by a negative number, reverse the inequality sign: $x \le -3$. 3. The correct graph has a closed point at $-3$ and extends to the left. 4. Therefore, the correct panel is $b)$.

Answer

$b)$

5125979

Rectangle A has side lengths $x$ inches and $x + 6$ inches. Rectangle B has side lengths $x$ inches and $3x$ inches. For which values of $x$ is the perimeter of Rectangle A greater than the perimeter of Rectangle B? Assume $x > 0$.

Hints

- Write a perimeter expression for each rectangle. - Use an inequality to compare the two perimeters. - Solve the inequality even though the variable appears on both sides. - Include the restriction that a side length must be positive.

Solution

1. The perimeter of Rectangle A is $2(x + x + 6) = 4x + 12$. 2. The perimeter of Rectangle B is $2(x + 3x) = 8x$. 3. Compare the perimeters: $4x + 12 > 8x$. 4. Subtract $4x$: $12 > 4x$. 5. Divide by $4$: $3 > x$, or $x < 3$. 6. Combine this result with $x > 0$. The solution set is $0 < x < 3$.

Answer

The perimeter of Rectangle A is greater when $0 < x < 3$.

5128879

Consider the functions $f(x) = -x + 2$ and $g(x) = 0.5x - 1$. a) Find both function values at $x = -2$ and decide which graph is higher there. b) Find the intersection point $S$ algebraically. c) For which real values of $x$ is the graph of $g$ above the graph of $f$? Justify your answer.

Hints

- Compare the two function values at the same x-value. - At an intersection point, what must be true about the two function values? - To decide when one graph is above the other, compare the two expressions with an inequality.

Solution

1. At $x=-2$, $f(-2)=4$ and $g(-2)=-2$, so the graph of $f$ is higher there. 2. At an intersection, $-x+2=0.5x-1$. Solving gives $x=2$, and substitution gives $y=0$, so $S=(2,0)$. 3. For $g$ to be above $f$, solve $0.5x-1>-x+2$. This gives $1.5x>3$, so $x>2$.

Answer

a) $f(-2) = 4$ and $g(-2) = -2$; the graph of $f$ is higher. b) $S = (2, 0)$ c) $x > 2$

5131609

A line has equation $y=-2x+6$. A point $S$ has coordinates $(x, 10)$. Find all values of $x$ for which $S$ lies above the line.

Hints

- Which inequality symbol represents “above”? - Substitute the known coordinate into the inequality. - What happens to an inequality sign when dividing by a negative number?

Solution

1. A point is above the line when its y-coordinate is greater than the line's output: $10>-2x+6$. 2. Subtract $6$: $4>-2x$. 3. Divide by $-2$ and reverse the inequality: $-2<x$, or $x>-2$.

Answer

The point lies above the line when $x>-2$.

5131619

Consider the line $f(x)=\frac{2}{3}x-1$. a) Point $A=(6,y_A)$ must lie strictly above the line. What condition must $y_A$ satisfy? b) Point $B=(x_B,3)$ must lie strictly below the line. What condition must $x_B$ satisfy?

Hints

- For part (a), compare the point's y-coordinate with the line's y-value at the same x-coordinate. - For part (b), translate “below the line” into an inequality between $3$ and $f(x_B)$. - Check whether the inequality direction changes during the algebra in part (b).

Solution

1. At $x=6$, the line has value $f(6)=3$. For $A$ to be strictly above the line, $y_A>3$. 2. For $B$ to be strictly below the line, $3<\frac{2}{3}x_B-1$. 3. Solving the inequality gives $4<\frac{2}{3}x_B$, then $6<x_B$. Thus $x_B>6$.

Answer

a) $y_A > 3$ b) $x_B > 6$

5131819

Solve the inequality over the real numbers. Give the answer in interval notation. $\frac{3}{4}x-2>\frac{1}{2}(x+4)$

Hints

- Multiplying by a common denominator can eliminate fractions. - Distribute the factor to every term inside the parentheses. - How is a strict “greater than” condition written in interval notation?

Solution

1. Distribute on the right: $\frac{3}{4}x-2>\frac{1}{2}x+2$. 2. Subtract $\frac{1}{2}x$ and add $2$: $\frac{1}{4}x>4$. 3. Multiply by $4$: $x>16$.

Answer

$(16, \infty)$

5131829

Solve the inequality over the real numbers. Pay close attention to nested grouping symbols and signs. $2-[3x-(x+5)]\le4(2-x)+1$

Hints

- Work from the innermost grouping symbols outward. - How do signs change when a minus sign is directly before brackets? - Combine like terms on each side before moving terms across the inequality.

Solution

1. Simplify inside the brackets: $2-[3x-x-5]\le8-4x+1$. 2. Combine terms: $2-[2x-5]\le9-4x$. 3. Distribute the negative sign: $2-2x+5\le9-4x$, so $7-2x\le9-4x$. 4. Add $4x$ and subtract $7$: $2x\le2$. 5. Divide by $2$: $x\le1$.

Answer

$\{x\in\mathbb{R}\mid x\le1\}$

5131869

A metalworker has a $16$-foot iron rod to make the frame of a rectangular gate. The width must be exactly twice the height. To minimize waste, the frame may use as much of the rod as possible but cannot exceed its total length. Find the greatest possible whole-number height, in inches.

Hints

- Write the perimeter of a rectangle in terms of its height. - Express all lengths in the same unit. - Which inequality represents “cannot exceed”?

Solution

1. Let $h$ be the height in inches. Then the width is $2h$. 2. The perimeter is $2(h+2h)=6h$. 3. Convert the rod length: $16\,\text{ft}=192\,\text{in}$. The material constraint is $6h\le192$. 4. Divide by $6$: $h\le32$. The greatest whole-number height is $32$ inches.

Answer

The greatest possible height is $32$ inches.

5131879

For a school project, a student will build the wireframe of a right triangular prism with equilateral triangular bases. There are $120$ inches of wire available. The prism's height must be exactly $10$ inches longer than the side length $a$ of each triangular base. Find all possible positive integer values of $a$, in inches, if the wire supply cannot be exceeded.

Hints

- Count the edges in both triangular bases and the lateral edges. - Write the total wire length using only one variable. - Remember that a side length must be positive.

Solution

1. The two triangular bases have $6$ edges of length $a$, and the prism has $3$ lateral edges of height $h$. 2. Since $h=a+10$, the total wire length is $6a+3(a+10)=9a+30$. 3. The constraint is $9a+30\le120$. Subtract $30$: $9a\le90$. Divide by $9$: $a\le10$. 4. Since $a$ is a positive integer, $a\in\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$.

Answer

The possible side lengths are all whole numbers from $1$ inch through $10$ inches.

5131889

A shipping service limits rectangular packages with square ends: the package length $l$ plus the girth, the perimeter of the square end, may be at most $165$ inches. A seller uses packages whose length is exactly three times the square side length $x$. For stability, the package length must be at least $48$ inches. Which integer values of $x$, in inches, satisfy both requirements?

Hints

- There are separate maximum-size and minimum-length conditions. - The girth is the perimeter of the square end. - Solve both inequalities and find their overlap.

Solution

1. The length is $l=3x$, and the girth is $4x$. 2. The size limit gives $3x+4x\le165$, so $7x\le165$ and $x\le\frac{165}{7}\approx23.57$. 3. The minimum length gives $3x\ge48$, so $x\ge16$. 4. Combining the conditions gives $16\le x\le23.57$. The integer values are $16, 17, 18, 19, 20, 21, 22, 23$.

Answer

The possible integer side lengths are $16$ inches through $23$ inches.

5132019

Two phone plans have different monthly charges. The Smart plan costs a $\$4.95$ base fee plus $\$0.12$ per call minute. The Power plan costs a $\$9.95$ base fee plus $\$0.08$ per minute. a) Find the monthly cost of each plan for $80$ minutes. b) For what call durations is the Power plan less expensive than the Smart plan?

Hints

- Write a cost expression for each plan. - How can “Power is less expensive” be written as an inequality? - Find when the two plans have equal costs, then compare beyond that point. - Include both fixed and variable costs.

Solution

a) The cost functions are $C_S(x)=4.95+0.12x$ and $C_P(x)=9.95+0.08x$. At $80$ minutes, $C_S(80)=\$4.95+\$0.12 \cdot 80=\$14.55$, and $C_P(80)=\$9.95+\$0.08 \cdot 80=\$16.35$. b) Solve $9.95+0.08x<4.95+0.12x$. Subtract $4.95$ and $0.08x$: $5.00<0.04x$. Divide by $0.04$: $x>125$.

Answer

a) Smart costs $\$14.55$, and Power costs $\$16.35$. b) Power is less expensive for more than $125$ minutes.

5132039

A car-sharing company offers two daily plans. - Plan A: a $\$24$ daily fee plus $\$0.25$ per mile. - Plan B: a flat $\$44$ daily fee with unlimited miles. a) For what daily mileages is Plan A less expensive than Plan B? b) A customer plans to drive $100$ miles. Which plan should the customer choose? Compare the costs. c) What should Plan A's daily fee be so that both plans cost exactly the same for $100$ miles, if the per-mile charge remains $\$0.25$?

Hints

- Does Plan B's cost change with mileage? - Evaluate Plan A at $100$ miles and compare. - In part c, use an unknown for the new daily fee. - “Exactly the same cost” means set the expressions equal.

Solution

a) The cost functions are $C_A(x)=24+0.25x$ and $C_B(x)=44$, with $x\ge0$. Solve $24+0.25x<44$: $0.25x<20$, so $x<80$. Thus, Plan A is less expensive for $0\le x<80$. b) At $100$ miles, Plan A costs $\$24+\$0.25 \cdot 100=\$49$, while Plan B costs $\$44$. Plan B is less expensive. c) Let $F$ be the new daily fee. Solve $F+0.25 \cdot 100=44$: $F+25=44$, so $F=\$19$.

Answer

a) Plan A is less expensive for $0\le x<80$ miles. At $80$ miles, the costs are equal. b) Plan B; Plan A costs $\$49$, while Plan B costs $\$44$. c) The daily fee should be $\$19$.

5132169

Solve each inequality over the real numbers. Give each answer in interval notation. a) $4x-9\le2x+5$ b) $18-3x>33$ c) $\frac{2}{3}x+4<2(x-1)$

Hints

- Watch for any step that multiplies or divides both sides by a negative number. - Decide whether each boundary value is included before choosing a bracket or parenthesis. - Put all variable terms on one side before isolating the variable.

Solution

1. In (a), $4x-9\le2x+5$ gives $2x\le14$, so $x\le7$. 2. In (b), $18-3x>33$ gives $-3x>15$. Dividing by $-3$ reverses the inequality, so $x<-5$. 3. In (c), $\frac{2}{3}x+4<2x-2$. Rearranging gives $6<\frac{4}{3}x$, so $x>\frac{9}{2}=4.5$.

Answer

a) $(-\infty, 7]$ b) $(-\infty, -5)$ c) $(4.5, \infty)$

5132179

Consider the inequalities $6x+12\le36$ and $-3x\ge-12$. Determine whether they have the same solution set. Give each solution in interval notation and justify your conclusion.

Hints

- Solve the inequalities separately. - Pay attention when dividing the second inequality by a negative number. - Compare the resulting intervals.

Solution

1. Solve the first inequality: $6x+12\le36$, so $6x\le24$ and $x\le4$. Its solution is $(-\infty, 4]$. 2. Solve the second inequality: $-3x\ge-12$. Dividing by $-3$ reverses the inequality, giving $x\le4$. Its solution is also $(-\infty, 4]$. 3. The intervals are identical, so the inequalities have the same solution set.

Answer

Yes. Both solution sets are $(-\infty, 4]$.

5137579

A family compares two monthly streaming options. - Service A: a $\$7.50$ monthly fee plus $\$2.00$ per rented movie. - Service B: a $\$18.50$ monthly unlimited plan. a) Starting with how many movies is Service B less expensive? b) If Service A lowers its per-movie charge to $\$1.50$, starting with how many movies is Service B less expensive?

Hints

- Write an inequality comparing the monthly totals. - Movie counts must be whole numbers. - How does lowering Service A's per-movie price affect the break-even point?

Solution

a) Solve $\$18.50<\$7.50+\$2.00x$. This gives $11<2x$, so $x>5.5$. Since movie counts are whole numbers, Service B is less expensive starting at $6$ movies. b) Solve $\$18.50<\$7.50+\$1.50x$. This gives $11<1.50x$, so $x>7.333\ldots$. Therefore, Service B is less expensive starting at $8$ movies.

Answer

a) $6$ movies b) $8$ movies

5139469

Let $T_1(x)=\frac{1}{2}x+4$ and $T_2(x)=2x-5$. Find all real numbers $x$ for which $T_1(x)>T_2(x)$. Give the solution set.

Hints

- Write an inequality that compares the two expressions. - Move variable terms to one side and constants to the other. - You may use fractions or decimals consistently.

Solution

1. Write the inequality $\frac{1}{2}x+4>2x-5$. 2. Subtract $\frac{1}{2}x$: $4>\frac{3}{2}x-5$. 3. Add $5$: $9>\frac{3}{2}x$. 4. Multiply by $\frac{2}{3}$: $6>x$, so $x<6$.

Answer

$\{x\in\mathbb{R}\mid x<6\}$

5139479

Solve each inequality over the real numbers. Simplify completely and interpret what happens when the variable terms cancel. a) $2(3x-4)<6x-5$ b) $4-(x+5)\ge10-x$

Hints

- Simplify both sides before trying to isolate $x$. - What does it mean if all variable terms cancel? - Decide whether the remaining statement is always true or always false.

Solution

a) Distribute: $6x-8<6x-5$. Subtract $6x$: $-8<-5$, which is always true. Therefore, every real number is a solution. b) Simplify: $4-x-5\ge10-x$, so $-1-x\ge10-x$. Add $x$: $-1\ge10$, which is false. Therefore, there are no solutions.

Answer

a) $\mathbb{R}$ b) $\varnothing$

5139739

Solve the inequality and write the solution in interval notation. $\frac{1}{2}(6x-4)\le2(x+5)+1$

Hints

- Simplify both sides before isolating $x$. - Distribute carefully. - Which bracket is used when an endpoint is included?

Solution

1. Distribute: $3x-2\le2x+10+1$. 2. Combine terms: $3x-2\le2x+11$. 3. Subtract $2x$: $x-2\le11$. 4. Add $2$: $x\le13$.

Answer

$(-\infty, 13]$

5139749

Consider the inequality $5(x-2)\ge3x+4$. a) Test whether $x=5$ is a solution. b) Solve the inequality and give the complete solution in interval notation.

Hints

- A test value is a solution only if it makes the inequality true. - Solve step by step as you would solve an equation. - Check whether part a agrees with the final solution set.

Solution

a) Substitute $x=5$: $5(5-2)\ge3 \cdot 5+4$, which becomes $15\ge19$. This is false, so $5$ is not a solution. b) Distribute: $5x-10\ge3x+4$. Subtract $3x$ and add $10$: $2x\ge14$. Divide by $2$: $x\ge7$, so the interval is $[7, \infty)$.

Answer

a) No, because $15\ge19$ is false. b) $[7, \infty)$

5154629

Solve the inequality over the real numbers. $\frac{2x+1}{3}-\frac{x-2}{2}\le2$

Hints

- Multiply by a common denominator to eliminate the fractions. - Pay attention to the minus sign before the second fraction. - Multiplying by a positive number does not reverse the inequality.

Solution

1. Multiply every term by the least common denominator $6$: $2(2x+1)-3(x-2)\le12$. 2. Distribute: $4x+2-3x+6\le12$. 3. Combine terms: $x+8\le12$. 4. Subtract $8$: $x\le4$.

Answer

$\{x\in\mathbb{R}\mid x\le4\}$

52275112

Consider $T(x) = \frac{-12}{x + 4}$. Solve each condition. Use interval notation for the solution sets in parts 2 and 3. 1. $T(x)$ is undefined. 2. $T(x) > 0$. 3. $T(x) < 0$.

Hints

- Start by identifying the value that makes the denominator zero. - For a quotient with a negative numerator, what sign must the denominator have to make the quotient positive? - Use the denominator's sign on each side of the excluded value to determine the sign of the quotient.

Solution

1. The expression is undefined when the denominator is zero: $x+4=0$, so $x=-4$. 2. Because the numerator is negative, the quotient is positive when the denominator is negative. Solve $x+4<0$ to get $x<-4$, so the solution is $(-\infty,-4)$. 3. The quotient is negative when the denominator is positive. Solve $x+4>0$ to get $x>-4$, so the solution is $(-4,\infty)$.

Answer

1. $\{-4\}$ 2. $(-\infty,-4)$ 3. $(-4,\infty)$

52275212

Consider $B(x) = \frac{8}{2x-10}$. a) State the domain of $B$. b) Solve $B(x)<0$. Write the solution in interval notation. c) Replace the numerator $8$ with $-8$. Describe how the positive and negative intervals change, and explain why.

Hints

- First find the x-value that makes the denominator zero. - With a positive numerator, which sign of the denominator makes the quotient negative? - Changing only the sign of the numerator changes the sign of every defined output but not the domain.

Solution

1. The denominator is zero when $2x-10=0$, so $x=5$. Therefore, the domain is $\mathbb{R}\setminus\{5\}$. 2. Because the numerator $8$ is positive, the quotient is negative when the denominator is negative. Solve $2x-10<0$ to get $x<5$, so the solution is $(-\infty,5)$. 3. Replacing $8$ by $-8$ multiplies every defined output by $-1$, so the signs reverse. The new expression is positive on $(-\infty,5)$ and negative on $(5,\infty)$. The excluded value $x=5$ does not change.

Answer

a) $\mathbb{R}\setminus\{5\}$ b) $(-\infty,5)$ c) The signs reverse: positive on $(-\infty,5)$ and negative on $(5,\infty)$.

52275412

Consider $Q(x)=\frac{x-5}{x^2+1}$. a) Explain why the domain is all real numbers. b) Determine where $Q(x)$ is negative, positive, or equal to zero. Write the positive and negative solution sets in interval notation. c) Explain why replacing the numerator $x-5$ with $x^2+5$ would make the expression never equal to zero.

Hints

- What is the smallest possible value of $x^2+1$? - If the denominator is always positive, which factor determines the sign of the quotient? - For the modified numerator, decide whether $x^2+5$ can ever equal zero for a real x-value.

Solution

1. For every real $x$, $x^2\ge0$, so $x^2+1\ge1$. The denominator is never zero, so the domain is $\mathbb{R}$. 2. Because the denominator is always positive, the sign of $Q(x)$ is determined by $x-5$. Thus $Q(x)<0$ for $x<5$, $Q(x)=0$ at $x=5$, and $Q(x)>0$ for $x>5$. 3. If the numerator were $x^2+5$, then it would be at least $5$ for every real $x$, so it could never be zero.

Answer

a) The domain is $\mathbb{R}$. b) Negative on $(-\infty,5)$; positive on $(5,\infty)$; zero at $x=5$. c) $x^2+5$ is always positive, so the modified expression is never zero.

5240929

Consider the inequality over the real numbers: $\frac{x+1}{4}-\frac{x-2}{3}\ge\frac{1}{2}$ a) Find the solution set. b) What is the greatest integer solution?

Hints

- A positive common denominator can clear all three fractions without changing the inequality direction. - Keep the subtraction in front of the second numerator when you distribute. - Once the variable has a negative coefficient, consider what happens to the inequality direction when you isolate it.

Solution

1. Multiply every term by the positive least common denominator, $12$: $3(x+1)-4(x-2)\ge6$. 2. Distribute and combine: $3x+3-4x+8\ge6$, so $-x+11\ge6$. 3. Then $-x\ge-5$. Multiply by $-1$ and reverse the inequality to get $x\le5$. 4. The greatest integer satisfying $x\le5$ is $5$.

Answer

a) $x\le5$ b) $5$

52441012

Consider $T(x)=\frac{4x-12}{x^2+1}$. a) Find the zero of $T$. b) Explain why $T(x)>0$ for every $x>3$. c) Evaluate $T(-1)$.

Hints

- A rational expression is zero when its numerator is zero and its denominator is nonzero. - Decide what sign $x^2+1$ has for every real x-value. - In part (b), compare the sign of the numerator on the interval $x>3$.

Solution

1. The denominator $x^2+1$ is positive for every real $x$, so $T(x)=0$ exactly when the numerator is zero. Solve $4x-12=0$ to get $x=3$. 2. If $x>3$, then $4x-12>0$. The denominator is also positive, so $T(x)>0$. 3. Substitute $x=-1$: $T(-1)=\frac{4(-1)-12}{(-1)^2+1}=\frac{-16}{2}=-8$.

Answer

a) $x=3$ b) For $x>3$, both the numerator and denominator are positive, so $T(x)>0$. c) $T(-1)=-8$

5267599

Consider the inequality $\frac{2x-3}{5}\le\frac{x+2}{2}-1$. a) Solve it over the real numbers. b) Verify by substitution whether $x=0$ is a solution.

Hints

- Multiply by a common denominator to eliminate all fractions. - Combine like terms carefully. - Substitute the test value into the original inequality. - Why does multiplying by a positive common denominator preserve the inequality sign?

Solution

a) Multiply by $10$: $2(2x-3)\le5(x+2)-10$. Distribute and simplify: $4x-6\le5x$. Subtract $4x$: $-6\le x$, so $x\ge-6$. b) At $x=0$, the left side is $-\frac{3}{5}=-0.6$, and the right side is $0$. Since $-0.6\le0$, $0$ is a solution.

Answer

a) $\{x\in\mathbb{R}\mid x\ge-6\}$ b) Yes, because $-0.6\le0$.

5267609

Let $T_1(x)=(x+4)^2$ and $T_2(x)=x(x+10)-2$. a) Find all values of $x$ for which $T_1(x)<T_2(x)$. b) What is the smallest integer that satisfies the condition?

Hints

- Expand both expressions before comparing them. - What happens to the quadratic terms? - Pay attention to the strict inequality. - Which integer comes immediately after the boundary value?

Solution

a) Write $(x+4)^2<x(x+10)-2$. Expand: $x^2+8x+16<x^2+10x-2$. The quadratic terms cancel, leaving $8x+16<10x-2$. Then $18<2x$, so $x>9$. b) The smallest integer greater than $9$ is $10$.

Answer

a) $x>9$ b) $10$

5267619

Solve each inequality over the real numbers. 1) $\frac{x-3}{4}-\frac{x+1}{2}\le\frac{x}{8}$ 2) $7-2(3x-1)>5x+20$

Hints

- Eliminate fractions using a common denominator. - Distribute a negative factor carefully. - Reverse the inequality when dividing by a negative number. - Move variable terms to one side and constants to the other.

Solution

1) Multiply by $8$: $2(x-3)-4(x+1)\le x$. Distribute: $2x-6-4x-4\le x$, so $-2x-10\le x$. Then $-10\le3x$, giving $x\ge-\frac{10}{3}$. 2) Distribute: $7-6x+2>5x+20$, so $9-6x>5x+20$. Then $-11x>11$. Divide by $-11$ and reverse the inequality: $x<-1$.

Answer

1) $x\ge-\frac{10}{3}$ 2) $x<-1$

5267729

Find all values of $x$ that satisfy both inequalities. $1-\frac{2x-5}{3}\le\frac{3x+1}{2}$ and $2(3x-1)>5x-4$

Hints

- Solve each inequality separately. - The word “and” means take the intersection of the solution sets. - A number line can help compare the two sets. - Identify where both conditions are true.

Solution

1. For the first inequality, multiply by $6$: $6-2(2x-5)\le3(3x+1)$. 2. Simplify: $6-4x+10\le9x+3$, so $16-4x\le9x+3$. Then $13\le13x$, giving $x\ge1$. 3. For the second inequality, $6x-2>5x-4$, so $x>-2$. 4. The intersection of $x\ge1$ and $x>-2$ is $x\ge1$.

Answer

$x\ge1$

5267739

Find all integer values of $x$ that satisfy the system of inequalities. $\begin{cases}5x-3>2x+6\\\frac{2x+8}{4}\leq5\end{cases}$

Hints

- Solve each inequality separately. - Because both conditions must hold, take the intersection of the two solution sets. - Include only integers in the final answer.

Solution

1. Solve the first inequality: $5x-3>2x+6$, so $3x>9$ and $x>3$. 2. Solve the second inequality: $\frac{2x+8}{4}\leq5$, so $2x+8\leq20$, $2x\leq12$, and $x\leq6$. 3. The intersection is $3<x\leq6$. 4. The integers in this interval are $4$, $5$, and $6$.

Answer

$\{4, 5, 6\}$

5267749

Find all integer solutions of the system of inequalities. $\begin{cases}\frac{3x-5}{2}-\frac{x+1}{3}<2\\1-3(2-x)\geq-5\end{cases}$

Hints

- Multiply the first inequality by the least common denominator to eliminate the fractions. - Distribute carefully when a minus sign is in front of parentheses. - Find the overlap of the two solution sets, then keep only integer values.

Solution

1. Multiply the first inequality by $6$: $3(3x-5)-2(x+1)<12$. 2. Distribute and combine like terms: $9x-15-2x-2<12$, so $7x<29$ and $x<\frac{29}{7}$. 3. Solve the second inequality: $1-3(2-x)\geq-5$ becomes $-5+3x\geq-5$, so $x\geq0$. 4. The intersection is $0\leq x<\frac{29}{7}$. The integer solutions are $0$, $1$, $2$, $3$, and $4$.

Answer

$\{0, 1, 2, 3, 4\}$

5267799

Solve the system of inequalities over the real numbers. I: $\frac{2x+1}{3}-\frac{x-2}{6}\leq2$ II: $4-3(x+1)<2x+11$

Hints

- Use a common denominator to eliminate the fractions in inequality I. - Distribute the negative factor carefully in inequality II. - Reverse the inequality sign when dividing by a negative number. - Take the intersection because both inequalities must be true.

Solution

1. Multiply inequality I by $6$: $2(2x+1)-(x-2)\leq12$. 2. Simplify: $4x+2-x+2\leq12$, so $3x\leq8$ and $x\leq\frac{8}{3}$. 3. For inequality II, distribute and simplify: $4-3x-3<2x+11$, so $1-3x<2x+11$. 4. Subtract $1$ and $2x$: $-5x<10$. Divide by $-5$ and reverse the inequality: $x>-2$. 5. The intersection is $-2<x\leq\frac{8}{3}$.

Answer

$\left(-2, \frac{8}{3}\right]$

5267809

Determine whether any real number $x$ satisfies both inequalities. Give the solution set. I: $5-\frac{x+2}{2}<1$ II: $2(x-3)\leq\frac{x+4}{2}$

Hints

- Solve each inequality separately. - Reverse the inequality sign when dividing by a negative number. - Compare the two solution sets on a number line. - What does it mean if the two sets do not overlap?

Solution

1. Multiply inequality I by $2$: $10-(x+2)<2$. 2. Simplify: $8-x<2$, so $-x<-6$. Divide by $-1$ and reverse the inequality: $x>6$. 3. Multiply inequality II by $2$: $4(x-3)\leq x+4$. 4. Simplify: $4x-12\leq x+4$, so $3x\leq16$ and $x\leq\frac{16}{3}$. 5. No real number can satisfy both $x>6$ and $x\leq\frac{16}{3}$, because $\frac{16}{3}<6$.

Answer

There is no solution: $\varnothing$.

5280789

Find all values of $x$ that satisfy each condition. 1) The value of $4(x + 2)$ is at most $10$. 2) The value of $15 - 5x$ is nonnegative. 3) The value of $5x - 8$ is greater than the value of $2x + 1$.

Hints

- Translate “at most” and “nonnegative” into inequality symbols. - Include the boundary value when the wording allows equality. - Move variable terms to one side when comparing two expressions. - Reverse the sign when dividing by a negative coefficient.

Solution

1) “At most $10$” gives $4(x + 2) \le 10$. Distribute: $4x + 8 \le 10$. Then $4x \le 2$, so $x \le 0.5$. 2) “Nonnegative” gives $15 - 5x \ge 0$. Then $-5x \ge -15$. Divide by $-5$ and reverse the sign: $x \le 3$. 3) Write $5x - 8 > 2x + 1$. Then $3x > 9$, so $x > 3$.

Answer

1) $x \le 0.5$ 2) $x \le 3$ 3) $x > 3$

5509799

The graph shows two lines labeled $f$ and $g$. a) Read the y-intercept and slope of each line and use them to write formulas for $f(x)$ and $g(x)$. b) Use the graph to determine the x-values for which $f(x)>g(x)$. State what feature of the graph supports your answer. c) Verify the result algebraically using the formulas you found in part a).

Hints

- Use the grid to read each line's y-intercept and one-unit horizontal change. - “Above” means the function has the greater y-value at the same x-value. - Use the formulas recovered from the graph for the algebraic check.

Solution

1. The graph of $f$ has y-intercept $5$ and slope $-1$, so $f(x)=-x+5$. 2. The graph of $g$ has y-intercept $-1$ and slope $2$, so $g(x)=2x-1$. 3. The lines intersect at $x=2$. To the left of that x-value, the graph of $f$ lies above the graph of $g$, so $f(x)>g(x)$ for $x<2$. 4. Algebraically, solve $-x+5>2x-1$. Add $x$ and add $1$: $6>3x$. Divide by $3$: $x<2$, matching the graph.

Answer

a) $f(x)=-x+5$ and $g(x)=2x-1$ b) $x<2$; the graph of $f$ is above the graph of $g$ to the left of their intersection. c) $-x+5>2x-1$ gives $x<2$

5548729

Solve the compound inequality $3x+2<-7$ or $5-x\le1$. Give the solution in interval notation.

Hints

- Solve the two inequalities separately before combining their solution sets. - Watch the inequality direction when isolating $x$ in the second inequality. - The word “or” means values satisfying either condition are included.

Solution

1. Solve the first inequality: $3x+2<-7$ gives $3x<-9$, so $x<-3$. 2. Solve the second inequality: $5-x\le1$ gives $-x\le-4$, so dividing by $-1$ reverses the inequality and gives $x\ge4$. 3. Because the conditions are joined by “or,” take the union: $(-\infty,-3)\cup[4,\infty)$.

Answer

$(-\infty,-3)\cup[4,\infty)$

5548739

Noah solves $-4(2x-1)>12$ and writes: $-8x+4>12$, then $-8x>8$, so $x>-1$. a) Identify Noah's error. b) Give the correct solution and describe its number-line graph.

Hints

- Check Noah's work one line at a time and identify the first step that changes the solution set. - What special rule applies when an inequality is divided by a negative number? - Use the strict inequality to decide whether the boundary point is open or closed.

Solution

1. Noah's expansion and subtraction are correct through $-8x>8$. 2. Dividing an inequality by the negative number $-8$ requires reversing the inequality sign. 3. Therefore, $x<-1$. 4. The number-line graph has an open point at $-1$ and is shaded to the left.

Answer

a) Noah failed to reverse the inequality when dividing by $-8$. b) $x<-1$; open point at $-1$, shaded left.

5132189

Consider the inequality $2x+k>10$, where $k$ is a constant. Find $k$ so that the solution set is exactly $(7, \infty)$. Show your work.

Hints

- Solve the inequality for $x$ while leaving $k$ in the expression. - Match the resulting boundary with the desired interval endpoint. - Solve the resulting equation for $k$.

Solution

1. Solve for $x$: $2x+k>10$, so $x>\frac{10-k}{2}$. 2. For the solution boundary to be $7$, require $\frac{10-k}{2}=7$. 3. Then $10-k=14$, so $-k=4$ and $k=-4$.

Answer

$k=-4$

5225309

Let $a$ and $b$ be rational numbers. 1) Under what conditions does $ab = a$? Include every possible value of $a$. 2) Suppose $a > 0$. What condition on $b$ makes $ab < a$? 3) Suppose $a < 0$. What condition on $b$ makes $ab < a$?

Hints

- Factor the equation $ab-a=0$. - Consider the zero-product property. - Dividing an inequality by a positive number preserves its direction. - Dividing an inequality by a negative number reverses its direction.

Solution

1) Rewrite $ab = a$ as $ab-a=0$, then factor: $a(b-1)=0$. Therefore, either $a=0$, with any rational $b$, or $b=1$. 2) If $a>0$, divide $ab<a$ by the positive number $a$. The inequality direction is preserved, giving $b<1$. 3) If $a<0$, divide $ab<a$ by the negative number $a$. The inequality direction reverses, giving $b>1$.

Answer

1) $a=0$ with any rational $b$, or $b=1$ 2) If $a>0$, then $b<1$. 3) If $a<0$, then $b>1$.

5241079

In a two-digit number, the ones digit is exactly $4$ greater than the tens digit. The number is greater than $20$ and less than $60$. Let $t$ be the tens digit. Write a compound inequality in $t$ that represents the range condition, solve the inequality algebraically, and then use the digit constraint to list all possible two-digit numbers.

Hints

- Express the ones digit using the tens digit first. - Write the two-digit number as tens value plus ones value. - Translate “greater than $20$ and less than $60$” into one compound inequality. - After solving, enforce that the tens digit must be an integer digit.

Solution

1. The ones digit is $t+4$, so the number is $10t+(t+4)=11t+4$. 2. The range condition is $20<11t+4<60$. 3. Subtract $4$: $16<11t<56$. 4. Divide by $11$: $\frac{16}{11}<t<\frac{56}{11}$. 5. Since $t$ is a digit, $t\in\{2,3,4,5\}$. The corresponding ones digits are $6,7,8,9$. 6. The numbers are $26,37,48,59$.

Answer

Compound inequality: $20<11t+4<60$, so $\frac{16}{11}<t<\frac{56}{11}$. Possible numbers: $26,37,48,59$.

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