Which of the following describes all real values of that satisfy the inequality

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SAT Linear Inequalities Question 54 | Free SAT Prep | Aniko

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Question 54·Hard·Linear Inequalities in One or Two Variables

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Which of the following describes all real values of $x$ that satisfy the inequality

|5 - 2x| \ge x + 1?

For absolute value inequalities like $|ax + b| \ge$ (expression), first identify where the inside of the absolute value equals zero to split the number line into cases. In each case, drop the absolute value by replacing $|ax + b|$ with either $ax + b$ or $-(ax + b)$ , solve the resulting linear inequality, and then combine the solutions. Remember: "greater than or equal" absolute value inequalities usually give two separated intervals (solutions on the ends of the number line), while "less than or equal" ones usually give one interval in the middle.

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Focus on the absolute value expression

Notice the absolute value $|5 - 2x|$ . Ask yourself: for what value of $x$ does $5 - 2x$ equal zero, and how does that split the number line into regions?

Write two inequalities without absolute value

Use the definition of absolute value: when $5 - 2x$ is nonnegative, $|5 - 2x| = 5 - 2x$ ; when $5 - 2x$ is negative, $|5 - 2x| = 2x - 5$ . Write the inequality separately in each case.

Solve and then combine

Solve each resulting linear inequality and then think about how to combine the solutions: do you get one continuous interval or two separate ranges of $x$ -values?

Desmos Guide

Graph both sides of the inequality

In Desmos, enter y1 = abs(5 - 2x) and y2 = x + 1. You will see the V-shaped graph of the absolute value and the straight line.

Find the intersection points

Use the Desmos intersection tool (tap on the intersection points or click where the graphs cross) to read off the $x$ -values where y1 = y2. These $x$ -values are the boundaries between where the inequality holds and where it does not.

Determine where the inequality is satisfied

Look along the graph and note for which $x$ -values the graph of y1 = abs(5 - 2x) is on or above the graph of y2 = x + 1. The solution set is the union of those $x$ -regions where the absolute value graph lies at or above the line.

Step-by-step Explanation

Locate the breakpoint for the absolute value

The inequality is

|5 - 2x| \ge x + 1.

The absolute value expression $|5 - 2x|$ changes form where its inside is zero:

5 - 2x = 0 \Rightarrow 2x = 5 \Rightarrow x = \frac{5}{2}.

So we will consider two cases:

Case 1: $x \le \frac{5}{2}$ (so $5 - 2x \ge 0$ )
Case 2: $x > \frac{5}{2}$ (so $5 - 2x < 0$ )

Case 1: When $x \le \dfrac{5}{2}$ , replace $|5 - 2x|$ with $5 - 2x$

For $x \le \frac{5}{2}$ , $5 - 2x$ is nonnegative, so $|5 - 2x| = 5 - 2x$ .

The inequality becomes

5 - 2x \ge x + 1.

Solve this step by step:

Subtract $x$ from both sides:

5 - 3x \ge 1

Subtract $5$ from both sides:

-3x \ge -4

Divide both sides by $-3$ (remember to flip the inequality sign):

x \le \frac{4}{3}.

This solution must also respect the case condition $x \le \frac{5}{2}$ , but since $\frac{4}{3} < \frac{5}{2}$ , the condition $x \le \frac{4}{3}$ already fits inside $x \le \frac{5}{2}$ . So all solutions from Case 1 satisfy $x \le \frac{4}{3}$ .

Case 2: When $x > \dfrac{5}{2}$ , replace $|5 - 2x|$ with $2x - 5$

For $x > \frac{5}{2}$ , $5 - 2x$ is negative, so $|5 - 2x| = 2x - 5$ .

The inequality becomes

2x - 5 \ge x + 1.

Solve this step by step:

Subtract $x$ from both sides:

x - 5 \ge 1

Add $5$ to both sides:

x \ge 6.

This solution automatically satisfies $x > \frac{5}{2}$ (because $6 > \frac{5}{2}$ ), so all solutions from Case 2 satisfy $x \ge 6$ .

Combine the solution sets and match to an answer choice

From the two cases, we have:

From Case 1: $x \le \frac{4}{3}$
From Case 2: $x \ge 6$

The overall solution set is the union of these:

x \le \frac{4}{3} \quad \text{or} \quad x \ge 6.

This corresponds to answer choice D.

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Step-by-step Explanation

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