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Linear Inequalities in One Variable

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Question 1Question →

Let pp and qq be constants. The inequality p(x−5)<q(x+4)p(x - 5) < q(x + 4) has the solution set x>2x > 2. If p−q=−3p - q = -3, what is the value of p+qp + q?

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Answer: -1

Answer

-1
The correct answer is −1-1. Expanding the inequality p(x−5)<q(x+4)p(x - 5) < q(x + 4) gives px−5p<qx+4qpx - 5p < qx + 4q. Grouping the xx terms yields (p−q)x<5p+4q(p - q)x < 5p + 4q. Since p−q=−3p - q = -3, which is negative, dividing both sides by p−qp - q reverses the inequality to x>5p+4qp−qx > \frac{5p + 4q}{p - q}. Comparing this to the given solution x>2x > 2, the boundary value must satisfy 5p+4q−3=2\frac{5p + 4q}{-3} = 2, which simplifies to 5p+4q=−65p + 4q = -6. Using the system of equations p−q=−3p - q = -3 and 5p+4q=−65p + 4q = -6, we find p=−2p = -2 and q=1q = 1. Therefore, the value of p+qp + q is −2+1=−1-2 + 1 = -1.

Step-by-Step Solution

1
Expand both sides of the inequality and group the terms with xx on one side.
(p−q)x<5p+4q(p - q)x < 5p + 4q
Expanding p(x−5)<q(x+4)p(x - 5) < q(x + 4) gives px−5p<qx+4qpx - 5p < qx + 4q. Subtracting qxqx and adding 5p5p to both sides yields px−qx<5p+4qpx - qx < 5p + 4q, which factors to (p−q)x<5p+4q(p - q)x < 5p + 4q.
2
Determine the direction of the inequality when isolating xx.
x>5p+4qp−qx > \frac{5p + 4q}{p - q}
Since we are given p−q=−3p - q = -3, the quantity p−qp - q is negative. Dividing both sides of the inequality by a negative value reverses the inequality symbol from << to >>.
3
Equate the resulting boundary expression to the boundary of the given solution set x>2x > 2.
5p + 4q = -6
The boundary value of the solution set is 22, so we set 5p+4qp−q=2\frac{5p + 4q}{p - q} = 2. Substituting p−q=−3p - q = -3 gives 5p+4q−3=2\frac{5p + 4q}{-3} = 2, which simplifies to 5p+4q=−65p + 4q = -6.
4
Solve the system of linear equations for pp and qq.
p=−2p = -2 and q=1q = 1
We have the system of equations p−q=−3p - q = -3 and 5p+4q=−65p + 4q = -6. From the first equation, p=q−3p = q - 3. Substituting this into the second equation gives 5(q−3)+4q=−6  ⟹  9q−15=−6  ⟹  9q=9  ⟹  q=15(q - 3) + 4q = -6 \implies 9q - 15 = -6 \implies 9q = 9 \implies q = 1. Substituting q=1q = 1 back into p=q−3p = q - 3 gives p=−2p = -2.
5
Calculate the value of p+qp + q.
-1
Adding the values of pp and qq gives p+q=−2+1=−1p + q = -2 + 1 = -1.

Key Concept

Solving linear inequalities in one variable with variable coefficients and applying sign reversal rules.
Question 2Question →

If −2(x−5)≥4-2(x - 5) \geq 4, what is the maximum possible value of xx?

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Answer: 3

Answer

3
Distributing the −2-2 on the left side of the inequality −2(x−5)≥4-2(x - 5) \geq 4 yields −2x+10≥4-2x + 10 \geq 4. Subtracting 1010 from both sides gives −2x≥−6-2x \geq -6. Dividing both sides by −2-2 and reversing the inequality sign results in x≤3x \leq 3. The maximum possible value is therefore 3.

Step-by-Step Solution

1
Distribute −2-2 to the terms inside the parentheses.
−2x+10≥4-2x + 10 \geq 4
Applying the distributive property simplifies the inequality.
2
Subtract 1010 from both sides of the inequality.
−2x≥−6-2x \geq -6
This isolates the variable term on the left side.
3
Divide both sides of the inequality by −2-2 and reverse the inequality sign.
x≤3x \leq 3
Dividing both sides of an inequality by a negative number reverses the direction of the inequality.

Key Concept

Solving linear inequalities in one variable using the distributive property and division by negative numbers.
Question 3Question →

Which of the following represents all possible values of xx that satisfy the inequality −4(x+3)≤8-4(x + 3) \leq 8?

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Answer: x≥−5x \geq -5

Answer

x≥−5x \geq -5
To solve −4(x+3)≤8-4(x + 3) \leq 8, first distribute the −4-4 to obtain −4x−12≤8-4x - 12 \leq 8. Next, add 1212 to both sides to isolate the variable term, resulting in −4x≤20-4x \leq 20. Finally, divide both sides by −4-4 and reverse the inequality sign because of division by a negative number, which yields x≥−5x \geq -5.

Step-by-Step Solution

1
Distribute the coefficient outside the parentheses to the terms inside.
−4x−12≤8-4x - 12 \leq 8
Multiplying −4-4 by xx and −4-4 by 33 simplifies the left side of the inequality.
2
Isolate the variable term by adding 1212 to both sides of the inequality.
−4x≤20-4x \leq 20
Adding 1212 eliminates the constant term on the left side: 8+12=208 + 12 = 20.
3
Divide both sides of the inequality by −4-4 and reverse the inequality sign.
x≥−5x \geq -5
Dividing by a negative number reverses the direction of the inequality symbol: 20/−4=−520 / -4 = -5.

Key Concept

Solving linear inequalities in one variable using the distributive property and division by a negative number.
Estimated Time:45s
Question 4Question →

A gym membership costs $35\$35 per month plus an additional $5\$5 per fitness class attended. If a member wants to spend at most $65\$65 in a single month, what is the maximum number of fitness classes the member can attend?

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Answer: 6

Answer

The maximum number of fitness classes the member can attend is 66.
Let cc represent the number of classes. The inequality representing the budget constraint is 35+5c≤6535 + 5c \le 65. Subtracting 3535 from both sides of the inequality yields 5c≤305c \le 30. Dividing both sides by 55 yields c≤6c \le 6. Since the number of classes must be a whole number, the maximum number of classes the member can attend is 66.

Step-by-Step Solution

1
Set up the inequality representing the situation.
35+5c≤6535 + 5c \le 65
The cost of the membership is a fixed $35\$35 plus $5\$5 per class cc, and this total must not exceed $65\$65.
2
Subtract 3535 from both sides of the inequality to isolate the variable term.
5c≤305c \le 30
Isolating the term with the variable allows us to solve for cc.
3
Divide both sides by 55 to find the solution range for the number of classes.
c≤6c \le 6
Dividing by the coefficient of the variable gives the upper limit for the number of classes.

Key Concept

Solving one-variable linear inequalities in a real-world context.
Question 5Question →

Which of the following inequalities is equivalent to −3(x−4)<15-3(x - 4) < 15?

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Answer: x>−1x > -1

Answer

The inequality x>−1x > -1
The correct answer is obtained by dividing both sides of the inequality −3(x−4)<15-3(x - 4) < 15 by −3-3, which reverses the inequality symbol to yield x−4>−5x - 4 > -5. Adding 44 to both sides isolates the variable, resulting in x>−1x > -1. Alternatively, distributing the −3-3 first yields −3x+12<15-3x + 12 < 15, which simplifies to −3x<3-3x < 3, and dividing by −3-3 with a sign flip yields x>−1x > -1.

Step-by-Step Solution

1
Divide both sides of the inequality −3(x−4)<15-3(x - 4) < 15 by −3-3.
x−4>−5x - 4 > -5
Dividing an inequality by a negative number reverses the direction of the inequality symbol.
2
Add 44 to both sides of the inequality to isolate xx.
x>−1x > -1
Adding a constant to both sides preserves the inequality and isolates the variable xx.

Key Concept

Solving linear inequalities in one variable involving distribution and negative coefficients.
Question 6Question →

In a certain video game, a player earns 15 points for completing a level and loses 3 points for each hint they use. If a player wants to score at least 6 points on a level, what is the maximum number of hints they can use?

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Answer: 3

Answer

The maximum number of hints the player can use is 3.
To find the maximum number of hints, we construct the inequality representing the player's score: 15−3h≥615 - 3h \geq 6, where hh is the number of hints used. Subtracting 15 from both sides gives −3h≥−9-3h \geq -9. Dividing both sides by −3-3 and reversing the inequality sign yields h≤3h \leq 3. This means the player can use at most 3 hints to achieve a score of at least 6 points. Thus, the maximum number of hints is 3.

Step-by-Step Solution

1
Set up the inequality representing the score requirement.
15−3h≥615 - 3h \geq 6
The starting score is 15, and 3 points are lost for each hint hh. The final score must be at least (greater than or equal to) 6.
2
Subtract 15 from both sides of the inequality.
−3h≥−9-3h \geq -9
This isolates the variable term on the left side of the inequality.
3
Divide both sides by -3 and flip the inequality sign.
h≤3h \leq 3
Dividing both sides of an inequality by a negative number reverses the direction of the inequality sign.

Key Concept

Solving multi-step linear inequalities with negative coefficients
Estimated Time:45s
Question 7Question →

For which values of xx is the inequality 2−3(x+4)≥112 - 3(x + 4) \geq 11 true?

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Answer: x≤−7x \leq -7

Answer

The correct inequality is x≤−7x \leq -7.
Subtracting 2 from both sides of 2−3(x+4)≥112 - 3(x + 4) \geq 11 gives −3(x+4)≥9-3(x + 4) \geq 9. Dividing both sides by the negative number −3-3 requires reversing the inequality sign, yielding x+4≤−3x + 4 \leq -3. Subtracting 4 from both sides completes the isolation of xx, yielding the correct solution x≤−7x \leq -7.

Step-by-Step Solution

1
Subtract 2 from both sides of the inequality.
−3(x+4)≥9-3(x + 4) \geq 9
This isolates the term containing the parentheses on one side of the inequality.
2
Divide both sides of the inequality by −3-3 and reverse the direction of the inequality sign.
x+4≤−3x + 4 \leq -3
Dividing an inequality by a negative number requires reversing the direction of the inequality symbol to maintain equivalence.
3
Subtract 4 from both sides of the inequality.
x≤−7x \leq -7
This isolates the variable xx on one side of the inequality, providing the final solution set.

Key Concept

Solving multi-step linear inequalities in one variable, including distributing negative constants and reversing the inequality direction when dividing by a negative number.
Question 8Question →

A local delivery service charges a flat fee of 1212 dollars plus 1.501.50 dollars per mile to deliver a package. If a customer wants to spend no more than 3030 dollars for a package delivery, what is the maximum number of miles the delivery service can travel?

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Answer: 12

Answer

The maximum number of miles the delivery service can travel is 12.
The total delivery cost is the sum of the flat fee (1212 dollars) and the rate per mile (1.501.50 dollars multiplied by mm miles), which is represented by 1.50m+121.50m + 12. Since the total cost cannot exceed 3030 dollars, the inequality is 1.50m+12≤301.50m + 12 \leq 30. Subtracting 1212 from both sides gives 1.50m≤181.50m \leq 18. Dividing both sides by 1.501.50 yields m≤12m \leq 12. Thus, the maximum distance the delivery service can travel is 1212 miles.

Step-by-Step Solution

1
Set up the inequality representing the delivery cost constraint.
1.50m+12≤301.50m + 12 \leq 30
The total cost of the delivery is the flat fee of 1212 dollars plus 1.501.50 dollars per mile, mm, which must be less than or equal to the budget of 3030 dollars.
2
Subtract 12 from both sides of the inequality.
1.50m≤181.50m \leq 18
To isolate the variable term, subtract the constant flat fee from both sides of the inequality.
3
Divide both sides of the inequality by 1.50.
m≤12m \leq 12
Dividing by the per-mile rate calculates the maximum distance constraint on the variable mm.

Key Concept

Setting up and solving a one-variable linear inequality to determine a maximum boundary value in context.
Question 9Question →

Which of the following represents all possible values of xx that satisfy the inequality −2(x−5)>6-2(x - 5) > 6?

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Answer: x<2x < 2

Answer

The inequality is satisfied when x<2x < 2.
Dividing both sides of the inequality −2(x−5)>6-2(x - 5) > 6 by −2-2 yields x−5<−3x - 5 < -3, where the inequality sign is reversed because of division by a negative number. Adding 55 to both sides of x−5<−3x - 5 < -3 results in x<2x < 2.

Step-by-Step Solution

1
Divide both sides of the inequality −2(x−5)>6-2(x - 5) > 6 by −2-2.
x−5<−3x - 5 < -3
Dividing both sides of an inequality by a negative number requires reversing the direction of the inequality sign.
2
Add 55 to both sides of the inequality to isolate xx.
x<2x < 2
To solve for xx, eliminate the constant term −5-5 on the left side of the inequality.

Key Concept

Solving linear inequalities in one variable, including reversing the inequality sign when multiplying or dividing by a negative number.
Question 10Question →

Let cc be a constant such that c<0c < 0. If c(x−3)≤4(x+c)c(x - 3) \leq 4(x + c), which of the following inequalities must be true?

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Answer: x≥7cc−4x \geq \frac{7c}{c - 4}

Answer

x≥7cc−4x \geq \frac{7c}{c - 4}
The correct answer is obtained by first distributing terms on both sides to get cx−3c≤4x+4ccx - 3c \leq 4x + 4c. Rearranging the terms to isolate the variable xx on one side yields (c−4)x≤7c(c - 4)x \leq 7c. Since cc is a negative constant (c<0c < 0), the coefficient (c−4)(c - 4) must also be negative. Dividing both sides of the inequality by this negative coefficient requires reversing the direction of the inequality sign, which yields the final result.

Step-by-Step Solution

1
Distribute the constants on both sides of the inequality.
cx−3c≤4x+4ccx - 3c \leq 4x + 4c
This simplifies the parentheses so that variable terms can be grouped.
2
Group all terms with xx on the left side and terms with cc on the right side.
cx−4x≤7ccx - 4x \leq 7c
Isolating the variable terms on one side makes it possible to factor and solve for xx.
3
Factor out xx on the left side of the inequality.
(c−4)x≤7c(c - 4)x \leq 7c
This expresses the left side as a product of xx and a single coefficient.
4
Analyze the sign of the coefficient (c−4)(c - 4) given that c<0c < 0.
c−4<0c - 4 < 0
Since cc is less than 00, subtracting 44 from cc must result in a value less than −4-4, which is strictly negative.
5
Divide both sides of the inequality by (c−4)(c - 4) and flip the inequality symbol.
x≥7cc−4x \geq \frac{7c}{c - 4}
Dividing an inequality by a negative number requires reversing the direction of the inequality sign.

Key Concept

Solving linear inequalities by isolating the variable, and correctly reversing the inequality direction when multiplying or dividing by a negative variable parameter.
Question 11Question →

If the solution to the inequality a(2−3x)−4(x+3)≥12a(2 - 3x) - 4(x + 3) \ge 12, where aa is a constant, is x≤−1x \le -1, what is the value of aa?

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Answer: 4

Answer

The value of aa is 44.
To solve the inequality a(2−3x)−4(x+3)≥12a(2 - 3x) - 4(x + 3) \ge 12, we first expand it to get 2a−3ax−4x−12≥122a - 3ax - 4x - 12 \ge 12. Grouping the xx terms gives (−3a−4)x≥24−2a(-3a - 4)x \ge 24 - 2a. Since the inequality's solution is x≤−1x \le -1, the direction of the inequality must flip, which means the coefficient of xx, namely −3a−4-3a - 4, must be negative. Dividing both sides by this coefficient gives the boundary value of the inequality as 24−2a−3a−4\frac{24 - 2a}{-3a - 4}. Setting this boundary equal to −1-1 yields 24−2a=3a+424 - 2a = 3a + 4, which simplifies to 5a=205a = 20, or a=4a = 4. Since a=4a = 4 makes the coefficient −3(4)−4=−16-3(4) - 4 = -16 negative, the solution holds.

Step-by-Step Solution

1
Expand the inequality using the distributive property.
2a−3ax−4x−12≥122a - 3ax - 4x - 12 \ge 12
This allows us to separate and group the terms containing the variable xx and the constant terms.
2
Group like terms and isolate the variable terms on the left-hand side.
(−3a−4)x≥24−2a(-3a - 4)x \ge 24 - 2a
By combining the coefficients of xx and adding 12−2a12 - 2a to both sides, we prepare the inequality to solve for xx.
3
Determine the effect of dividing by the variable's coefficient.
Since the given solution is x≤−1x \le -1, the inequality sign flipped from ≥\ge to ≤\le. Therefore, the coefficient −3a−4-3a - 4 must be negative.
Multiplying or dividing both sides of an inequality by a negative number reverses the direction of the inequality sign.
4
Set the boundary value of the solution equal to −1-1 and solve for aa.
a=4a = 4
Setting the boundary of the inequality 24−2a−3a−4\frac{24 - 2a}{-3a - 4} equal to −1-1 allows us to find the specific constant aa that produces this solution set.

Key Concept

Solving linear inequalities in one variable involving parameters and sign flips.
Question 12Question →

For a constant kk, the inequality k−3x2>5x+34\frac{k - 3x}{2} > \frac{5x + 3}{4} has exactly 4 positive integer solutions for xx. If kk is an integer, how many possible values of kk are there?

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Answer: 6

Answer

The number of possible integer values for kk is 6.
Simplifying the inequality gives x<2k−311x < \frac{2k - 3}{11}. For this inequality to have exactly 4 positive integer solutions, the solution set must contain only the integers 1, 2, 3, and 4. This requires the boundary 2k−311\frac{2k - 3}{11} to satisfy 4<2k−311≤54 < \frac{2k - 3}{11} \le 5. Solving for kk gives 23.5<k≤2923.5 < k \le 29. The integers in this interval are 24, 25, 26, 27, 28, and 29, which total 6 values.

Step-by-Step Solution

1
Multiply both sides of the inequality by 4.
2(k−3x)>5x+32(k - 3x) > 5x + 3
Clearing the denominators makes the inequality easier to solve.
2
Distribute the 2 on the left side.
2k−6x>5x+32k - 6x > 5x + 3
Expanding terms allows us to group variables.
3
Add 6x6x to both sides and subtract 3 from both sides.
11x<2k−311x < 2k - 3
Grouping xx on one side and parameter terms on the other side.
4
Divide by 11.
x<2k−311x < \frac{2k - 3}{11}
Isolating xx gives the boundary for the solution set.
5
Establish the range for the boundary term 2k−311\frac{2k - 3}{11}.
4<2k−311≤54 < \frac{2k - 3}{11} \le 5
For the inequality to contain exactly the positive integers 1, 2, 3, and 4, the boundary must lie strictly above 4 and less than or equal to 5.
6
Solve the compound inequality for kk.
23.5<k≤2923.5 < k \le 29
Performing algebraic operations on all parts of the inequality to isolate kk.
7
Identify and count all integer solutions for kk.
6 integers (24, 25, 26, 27, 28, 29)
Counting the integers in the range (23.5,29](23.5, 29] yields the final answer.

Key Concept

Solving linear inequalities in one variable with parameter constraints and identifying integer solution sets.
Estimated Time:3m 0s
Question 13Question →

In the inequality −2(3x−k)−5(x+1)>17-2(3x - k) - 5(x + 1) > 17, kk is an integer constant. If the maximum integer value of xx that satisfies the inequality is 22, what is the least possible value of kk?

Show answer & explanation

Answer: 23

Answer

The correct answer is 23.
By simplifying the inequality to x<2k−2211x < \frac{2k - 22}{11}, we establish that the upper bound of the solution interval must be strictly greater than 22 but less than or equal to 33 for 22 to be the maximum integer solution. Solving the resulting compound inequality 2<2k−2211≤32 < \frac{2k - 22}{11} \le 3 yields 22<k≤27.522 < k \le 27.5. The smallest integer within this range is 2323.

Step-by-Step Solution

1
Distribute the constants on the left side of the inequality.
−6x+2k−5x−5>17-6x + 2k - 5x - 5 > 17
Applying the distributive property simplifies the expression and removes the parentheses.
2
Combine like terms on the left side and isolate the xx term.
−11x+2k−5>17  ⟹  −11x>22−2k-11x + 2k - 5 > 17 \implies -11x > 22 - 2k
Grouping xx terms together and moving the constant terms to the other side prepares the inequality for division.
3
Divide both sides by −11-11 and flip the inequality sign.
x<22−2k−11  ⟹  x<2k−2211x < \frac{22 - 2k}{-11} \implies x < \frac{2k - 22}{11}
Dividing by a negative number reverses the direction of the inequality sign from greater-than (>>) to less-than (<<).
4
Set up the inequality for the maximum integer solution to be 22.
2<2k−2211≤32 < \frac{2k - 22}{11} \le 3
For 22 to be the largest integer satisfying x<Lx < L (where LL is the boundary), 22 must be strictly less than LL, and LL must be less than or equal to the next integer, 33.
5
Solve the compound inequality for the parameter kk.
22<2k−22≤33  ⟹  44<2k≤55  ⟹  22<k≤27.522 < 2k - 22 \le 33 \implies 44 < 2k \le 55 \implies 22 < k \le 27.5
Multiplying all parts by 11, adding 22, and dividing by 2 isolates kk.
6
Find the least integer value of kk in the interval (22,27.5](22, 27.5].
2323
The integers that satisfy 22<k≤27.522 < k \le 27.5 are 23,24,25,26,23, 24, 25, 26, and 2727. The least of these values is 2323.

Key Concept

Solving linear inequalities in one variable involving parameter bounds, negative coefficients, and integer solution constraints.
Question 14Question →

An online service provider offers two monthly subscription plans. Under Plan A, the customer pays a flat monthly fee of CC dollars. Under Plan B, the monthly cost, in dollars, is determined by the expression 1.5(20−x)−0.8(3x−5)1.5(20 - x) - 0.8(3x - 5), where xx is the number of premium features the customer uses. The provider wants Plan B to be strictly cheaper than Plan A for any customer who uses more than 4 premium features. If CC is an integer, what is the minimum possible value of CC?

Show answer & explanation

Answer: 19

Answer

19
To find the minimum integer value of CC, we simplify Plan B's cost expression to 34−3.9x34 - 3.9x and set up the inequality 34−3.9x<C34 - 3.9x < C. Solving for xx by dividing by −3.9-3.9 and reversing the inequality sign gives x>34−C3.9x > \frac{34 - C}{3.9}. For Plan B to be cheaper than Plan A for all customers using more than 4 features, the solution set x>34−C3.9x > \frac{34 - C}{3.9} must contain the interval x>4x > 4. This requires the boundary point to be at most 4, so 34−C3.9≤4\frac{34 - C}{3.9} \le 4. Solving this inequality yields C≥18.4C \ge 18.4. The smallest integer value greater than or equal to 18.418.4 is 19.

Step-by-Step Solution

1
Simplify the cost expression for Plan B
34−3.9x34 - 3.9x
To combine like terms and express Plan B's cost in standard linear form.
2
Set up the inequality stating Plan B is strictly cheaper than Plan A
34−3.9x<C34 - 3.9x < C
Plan B is cheaper than Plan A when its cost is less than CC dollars.
3
Solve the inequality for xx in terms of CC
x>34−C3.9x > \frac{34 - C}{3.9}
Isolating xx allows us to find the threshold number of premium features, remembering to reverse the inequality direction when dividing by the negative coefficient −3.9-3.9.
4
Relate the threshold condition to the given minimum number of premium features
34−C3.9≤4\frac{34 - C}{3.9} \le 4
For Plan B to be cheaper for any x>4x > 4, the solution interval x>34−C3.9x > \frac{34 - C}{3.9} must cover the entire interval x>4x > 4. Thus, the boundary point must be at most 4.
5
Solve the boundary inequality for CC
C≥18.4C \ge 18.4
Multiplying by 3.93.9 and isolating CC gives the lower bound for the cost of Plan A.
6
Find the minimum integer value for CC
19
Since CC must be an integer and at least 18.418.4, the smallest integer that satisfies this inequality is 19.

Key Concept

Solving linear inequalities in one variable with parameter constraints and real-world conditions.
Question 15Question →

For a constant aa, the inequality 5x−a(3−2x)≥45x - a(3 - 2x) \ge 4 has a solution set of the form x≤dx \le d, where dd is a constant. Which of the following must be true about the value of aa?

Show answer & explanation

Answer: a<−52a < -\frac{5}{2}

Answer

The value of aa must satisfy a<−52a < -\frac{5}{2}.
To find the correct range for aa, we first expand the inequality 5x−a(3−2x)≥45x - a(3 - 2x) \ge 4 using the distributive property, which yields 5x−3a+2ax≥45x - 3a + 2ax \ge 4. Grouping the xx terms gives (5+2a)x≥3a+4(5 + 2a)x \ge 3a + 4. The problem states that the solution set is of the form x≤dx \le d. Because the inequality sign flipped from greater-than-or-equal-to (≥\ge) to less-than-or-equal-to (≤\le), the coefficient of xx must be negative. Setting the coefficient 5+2a<05 + 2a < 0 and solving for aa gives a<−52a < -\frac{5}{2}.

Step-by-Step Solution

1
Expand the inequality to separate the terms.
5x−3a+2ax≥45x - 3a + 2ax \ge 4
Apply the distributive property to the term −a(3−2x)-a(3 - 2x), paying close attention to the signs: −a×3=−3a-a \times 3 = -3a and −a×(−2x)=2ax-a \times (-2x) = 2ax.
2
Group and factor the terms containing xx on the left side, and move the constant terms to the right side.
(5+2a)x≥3a+4(5 + 2a)x \ge 3a + 4
Factor out xx from the terms 5x5x and 2ax2ax to isolate the variable, and add 3a3a to both sides of the inequality.
3
Analyze the relationship between the coefficient of xx and the inequality sign of the solution set.
5+2a<05 + 2a < 0
The original inequality has a greater-than-or-equal-to sign (≥\ge), but the given solution set is of the form x≤dx \le d (less-than-or-equal-to). For the inequality sign to reverse when dividing both sides by the coefficient of xx, the coefficient (5+2a)(5 + 2a) must be negative.
4
Solve the inequality for aa.
a<−52a < -\frac{5}{2}
Subtract 5 from both sides to get 2a<−52a < -5, then divide both sides by 2.

Key Concept

Solving linear inequalities in one variable with symbolic coefficients and applying the inequality sign reversal rule when multiplying or dividing by a negative value.
Question 16Question →

In the inequality 3(4−2x)−5(x−c)≥−73(4 - 2x) - 5(x - c) \geq -7, where cc is a constant, the solution set consists of all values of xx such that x≤3x \leq 3. What is the value of cc?

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Answer: 2.8

Answer

2.8
To find the value of cc, first simplify the inequality 3(4−2x)−5(x−c)≥−73(4 - 2x) - 5(x - c) \geq -7 by expanding the terms using the distributive property, which yields 12−6x−5x+5c≥−712 - 6x - 5x + 5c \geq -7. Combining like terms gives −11x+12+5c≥−7-11x + 12 + 5c \geq -7. Next, isolate the variable term by subtracting 1212 and 5c5c from both sides to obtain −11x≥−19−5c-11x \geq -19 - 5c. Dividing both sides of the inequality by −11-11 and reversing the inequality sign results in x≤19+5c11x \leq \frac{19 + 5c}{11}. Given that the solution set consists of all values of xx such that x≤3x \leq 3, the boundary value 19+5c11\frac{19 + 5c}{11} must equal 33. Solving the equation 19+5c11=3\frac{19 + 5c}{11} = 3 gives 19+5c=3319 + 5c = 33, which simplifies to 5c=145c = 14 and results in c=2.8c = 2.8.

Step-by-Step Solution

1
Distribute the constants on the left side of the inequality.
12−6x−5x+5c≥−712 - 6x - 5x + 5c \geq -7
Applying the distributive property simplifies the parentheses.
2
Combine the variable terms.
−11x+12+5c≥−7-11x + 12 + 5c \geq -7
Grouping like terms simplifies the inequality.
3
Subtract 1212 and 5c5c from both sides of the inequality.
−11x≥−19−5c-11x \geq -19 - 5c
This isolates the term containing xx.
4
Divide both sides by −11-11 and reverse the direction of the inequality.
x≤19+5c11x \leq \frac{19 + 5c}{11}
Dividing by a negative number requires flipping the inequality sign to maintain equivalence.
5
Set the algebraic boundary 19+5c11\frac{19 + 5c}{11} equal to the given boundary value of 33 and solve for cc.
c=2.8c = 2.8
Since the solution set is x≤3x \leq 3, the boundary values must be equivalent.

Key Concept

Solving multi-step linear inequalities in one variable containing parameters.
Question 17Question →

For a constant kk, the inequality 2(3−x)−k(x+5)>42(3 - x) - k(x + 5) > 4 has the solution set x<−3x < -3. What is the value of kk?

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Answer: 44

Answer

4
The correct answer is 44. Expanding the inequality 2(3−x)−k(x+5)>42(3 - x) - k(x + 5) > 4 gives 6−2x−kx−5k>46 - 2x - kx - 5k > 4. Grouping the xx terms yields (−2−k)x>5k−2(-2 - k)x > 5k - 2. Because the given solution set is x<−3x < -3, the inequality sign must flip, indicating that the coefficient −2−k-2 - k is negative. Dividing by this negative coefficient yields x<5k−2−2−kx < \frac{5k - 2}{-2 - k}. Setting the boundary expression equal to the boundary of the solution set gives 5k−2−2−k=−3\frac{5k - 2}{-2 - k} = -3. Multiplying both sides by −2−k-2 - k results in 5k−2=6+3k5k - 2 = 6 + 3k. Subtracting 3k3k and adding 22 to both sides gives 2k=82k = 8, which simplifies to k=4k = 4.

Step-by-Step Solution

1
Distribute the constants in the inequality
6−2x−kx−5k>46 - 2x - kx - 5k > 4
To clear parentheses and prepare to group terms.
2
Group the xx terms and constant terms
(−2−k)x>5k−2(-2 - k)x > 5k - 2
To isolate the variable xx on one side of the inequality.
3
Divide by the coefficient of xx and flip the inequality direction
x<5k−2−2−kx < \frac{5k - 2}{-2 - k}
Since the solution set is x<−3x < -3, the direction of the inequality must flip from >> to <<, meaning the coefficient −2−k-2 - k must be negative.
4
Set the boundary value equal to −3-3 and solve for kk
k=4k = 4
The boundary of the solution set must be equal to −3-3. Solving 5k−2−2−k=−3\frac{5k - 2}{-2 - k} = -3 yields 5k−2=6+3k5k - 2 = 6 + 3k, which simplifies to 2k=82k = 8, so k=4k = 4.

Key Concept

Solving linear inequalities in one variable with symbolic coefficients, accounting for direction flips when dividing by negative quantities.
Question 18Question →

A shipping container has a maximum weight capacity of 24,15024,150 kilograms. The container is loaded with 1212 machinery units, each weighing 1,1501,150 kilograms. The remaining space will be filled with packing crates, each weighing 180180 kilograms. A safety regulation requires that a clearance weight of at least 15%15\% of the total loaded weight (the combined weight of the machinery units and the packing crates) must be left unused. What is the maximum number of packing crates that can be loaded into the container?

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Answer: 40

Answer

The maximum number of packing crates that can be loaded is 40.
By setting up the inequality representing the physical constraints, we find that the number of packing crates, xx, must satisfy x≤40x \le 40. Since the question asks for the maximum number of packing crates, the maximum value is 40.

Step-by-Step Solution

1
Calculate the constant weight of the machinery units.
The total weight of the 1212 machinery units is 12×1,150=13,80012 \times 1,150 = 13,800 kilograms.
This establishes the base weight that is already loaded in the container.
2
Define the variable and write the expression for the total loaded weight.
Let xx be the number of packing crates. The total loaded weight is 13,800+180x13,800 + 180x kilograms.
This represents the combined weight of the machinery and the crates in terms of the variable xx.
3
Set up the inequality representing the safety clearance requirement.
24,150−(13,800+180x)≥0.15(13,800+180x)24,150 - (13,800 + 180x) \ge 0.15(13,800 + 180x)
The unused weight capacity (maximum capacity minus loaded weight) must be at least 15%15\% of the loaded weight.
4
Solve the inequality for xx.
24,150≥1.15(13,800+180x)  ⟹  21,000≥13,800+180x  ⟹  7,200≥180x  ⟹  x≤4024,150 \ge 1.15(13,800 + 180x) \implies 21,000 \ge 13,800 + 180x \implies 7,200 \ge 180x \implies x \le 40.
Isolating xx gives the range of allowable values for the number of packing crates.

Key Concept

Formulating and solving linear inequalities in one variable based on real-world constraints.
Question 19Question →

A researcher is monitoring the temperature, TT, in degrees Celsius, of a chemical reaction. The target temperature range is maintained by an automated cooling system that activates when the temperature satisfies the inequality −5(T−12)−8(3−2T)<14(T+3)-5(T - 12) - 8(3 - 2T) < 14(T + 3). Which of the following inequalities represents all possible values of TT for which the cooling system will activate?

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Answer: T>−2T > -2

Answer

The inequality representing all possible values of the temperature is T>−2T > -2.
The correct inequality is obtained by first expanding both sides of the inequality to get −5T+60−24+16T<14T+42-5T + 60 - 24 + 16T < 14T + 42. Simplifying the left side yields 11T+36<14T+4211T + 36 < 14T + 42. Subtracting 14T14T from both sides gives −3T+36<42-3T + 36 < 42, and subtracting 3636 from both sides gives −3T<6-3T < 6. Finally, dividing both sides by −3-3 and reversing the inequality sign results in T>−2T > -2.

Step-by-Step Solution

1
Distribute the constants on both sides of the inequality: −5(T−12)−8(3−2T)<14(T+3)-5(T - 12) - 8(3 - 2T) < 14(T + 3).
−5T+60−24+16T<14T+42-5T + 60 - 24 + 16T < 14T + 42
Remove the parentheses to prepare the inequality for simplification.
2
Combine the like terms on the left side of the inequality.
11T+36<14T+4211T + 36 < 14T + 42
Simplify the expression before isolating the variable.
3
Subtract 14T14T and 3636 from both sides of the inequality.
−3T<6-3T < 6
Group all variable terms on one side and constant terms on the other.
4
Divide both sides of the inequality by −3-3 and reverse the inequality sign.
T>−2T > -2
Dividing by a negative number reverses the inequality relationship.

Key Concept

Solving multi-step linear inequalities in one variable, distributing coefficients, combining like terms, and reversing the inequality sign when dividing by a negative number.
Estimated Time:2m 0s
Question 20Question →

The temperature, TT, in degrees Fahrenheit (∘F^\circ\text{F}), of a laboratory incubator mm minutes after a cooling cycle begins is modeled by the equation T=95−0.8(2m−5)T = 95 - 0.8(2m - 5). For a specific experiment, the incubator temperature must be at most 75∘F75^\circ\text{F}. Which of the following inequalities represents all possible values of mm for which the incubator temperature meets this requirement?

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Answer: m≥15m \geq 15

Answer

The correct inequality is m≥15m \geq 15.
The correct inequality is m≥15m \geq 15. To find this, set the temperature model 95−0.8(2m−5)95 - 0.8(2m - 5) to be less than or equal to 7575. Subtracting 9595 from both sides gives −0.8(2m−5)≤−20-0.8(2m - 5) \leq -20. Dividing both sides by −0.8-0.8 requires flipping the inequality sign, which yields 2m−5≥252m - 5 \geq 25. Adding 55 to both sides gives 2m≥302m \geq 30, and dividing by 22 yields m≥15m \geq 15.

Step-by-Step Solution

1
Set up the inequality based on the requirement that the temperature must be at most 75∘F75^\circ\text{F}.
95−0.8(2m−5)≤7595 - 0.8(2m - 5) \leq 75
The phrase 'at most' corresponds to a less-than-or-equal-to sign.
2
Subtract 9595 from both sides of the inequality.
−0.8(2m−5)≤−20-0.8(2m - 5) \leq -20
This begins the process of isolating the term containing the variable mm.
3
Divide both sides by −0.8-0.8 and flip the inequality sign.
2m−5≥252m - 5 \geq 25
Dividing both sides of an inequality by a negative number requires reversing the direction of the inequality sign.
4
Add 55 to both sides of the inequality.
2m≥302m \geq 30
This isolates the variable term 2m2m.
5
Divide both sides by 22 to solve for mm.
m≥15m \geq 15
This yields the solution set for the variable mm.

Key Concept

Solving multi-step linear inequalities in one variable, specifically expanding using the distributive property and reversing the inequality sign when multiplying or dividing by a negative number.
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