SAT Linear Inequalities Question 81 | Free SAT Prep | Aniko

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Question 81·Hard·Linear Inequalities in One or Two Variables

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The compound inequality

2 - 4k < 3k + 5 \le 8 - k

must hold simultaneously. Which of the following describes all possible values of $k$ that satisfy the compound inequality?

For compound inequalities of the form $A < B \le C$ , always break them into two separate inequalities and solve each one carefully. Pay special attention to when you divide by a negative number, since that flips the inequality sign. After solving both, remember that a chained inequality means both must hold simultaneously, so take the intersection (overlap) of the solution sets and keep track of whether each endpoint is strict ( $<$ or $>$ ) or inclusive ( $\le$ or $\ge$ ) when writing the final answer.

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Understand the compound inequality

Read the inequality $2 - 4k < 3k + 5 \le 8 - k$ as two separate inequalities joined together. What are those two inequalities?

Solve each inequality step by step

For each inequality, isolate $k$ using inverse operations (add/subtract first, then multiply/divide). Be careful with the sign of the coefficient of $k$ when you divide.

Watch for flipping the inequality sign

In the first inequality, when you divide by a negative number, what must you do to the inequality symbol? Make sure you apply this rule correctly.

Combine the two solution sets correctly

Once you have a solution from each inequality, remember that both must be true at the same time. Should you take the overlap (intersection) or combine them with an "or" (union)?

Desmos Guide

Enter each inequality in Desmos

In Desmos, use $x$ instead of $k$ . Type 2 - 4x < 3x + 5 on one line and 3x + 5 <= 8 - x on another line. Desmos will shade the solution set for each inequality along the $x$ -axis.

Look at the overlap of the shaded regions

Focus on where the shaded regions from the two inequalities overlap on the $x$ -axis. The $x$ -values in this overlapping region are the solutions that satisfy both inequalities at the same time.

Identify the boundary points and their types

Click on the boundary points of the overlapping region to see their exact coordinates. Note whether each endpoint is open (strict inequality) or closed (includes the endpoint); this tells you how to write the combined inequality for all possible $x$ that work.

Step-by-step Explanation

Separate the compound inequality into two parts

The compound inequality

2 - 4k < 3k + 5 \le 8 - k

means both of these must be true at the same time:

$2 - 4k < 3k + 5$
$3k + 5 \le 8 - k$

We will solve each inequality separately and then find the values of $k$ that satisfy both.

Solve the left inequality: $2 - 4k < 3k + 5$

Start with

2 - 4k < 3k + 5

Subtract $2$ from both sides:

-4k < 3k + 3

Subtract $3k$ from both sides:

-7k < 3

Divide both sides by $-7$ . Remember: dividing by a negative flips the inequality sign.

-7k < 3 \Rightarrow k > -\frac{3}{7}

So the first inequality tells us that $k$ must be greater than $-\frac{3}{7}$ .

Solve the right inequality: $3k + 5 \le 8 - k$

Now solve

3k + 5 \le 8 - k

Add $k$ to both sides:

4k + 5 \le 8

Subtract $5$ from both sides:

4k \le 3

Divide both sides by $4$ :

k \le \frac{3}{4}

So the second inequality tells us that $k$ must be less than or equal to $\frac{3}{4}$ .

Combine both conditions (intersection of solutions)

From the two inequalities we found:

From the first: $k > -\frac{3}{7}$
From the second: $k \le \frac{3}{4}$

Because the original compound inequality requires both conditions to be true at the same time, we take the overlap (intersection) of these two solution sets.

This means $k$ must be greater than $-\frac{3}{7}$ and less than or equal to $\frac{3}{4}$ :

-\frac{3}{7} < k \le \frac{3}{4}

This matches answer choice B.

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Desmos Guide

Step-by-step Explanation

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Desmos Guide

Step-by-step Explanation