Question 75·Hard·Linear Inequalities in One or Two Variables

Set up two cases: one with $2x - 3 \ge 0$ and one with $2x - 3 < 0$. In each case, replace $|2x - 3|$ by either $2x - 3$ or $-2x + 3$, then solve the resulting linear inequality $> x + 4$.

In the case where you get an inequality like $-2x + 3 > x + 4$, combine like terms and remember that dividing both sides by a negative number flips the direction of the inequality sign.

After finding the solution from each case, think about whether you should take the intersection or the union of those solutions. For a ">" absolute value inequality, do the solutions usually form a single interval or two separate intervals?

In Desmos, enter y = abs(2x - 3) to graph the absolute value function.

On a new line, enter y = x + 4. Adjust the viewing window so you can clearly see where the two graphs intersect and how they compare across the x-axis.

Click the intersection points of the two graphs and note their x-coordinates. Then look at where the graph of $y = |2x - 3|$ is above the graph of $y = x + 4$; those x-values are the solutions to $|2x - 3| > x + 4$. Translate those regions on the x-axis into interval notation or inequalities.

The expression $|2x - 3|$ depends on the sign of $2x - 3$.

• If $2x - 3 \ge 0$, then $|2x - 3| = 2x - 3$.
• If $2x - 3 < 0$, then $|2x - 3| = -(2x - 3) = -2x + 3$.

So we will solve the inequality $|2x - 3| > x + 4$ in two cases:

1. $2x - 3 \ge 0$
2. $2x - 3 < 0$.

First find when $2x - 3 \ge 0$:

In this case, $|2x - 3| = 2x - 3$, so the inequality becomes

Solve it:

In Case 1, we also need $x \ge 3/2$. Since every $x > 7$ already satisfies $x \ge 3/2$, the solution for this case is

• $x > 7$.

Now find when $2x - 3 < 0$:

In this case, $|2x - 3| = -2x + 3$, so the inequality becomes

Solve it carefully (watch the sign when dividing by a negative):

In Case 2, we also need $x < 3/2$. Every $x < -1/3$ is already less than $3/2$, so the solution for this case is

• $x < -1/3$.

From Case 1, we got $x > 7$. From Case 2, we got $x < -1/3$.

Because $|2x - 3| > x + 4$ holds in either case, we take the union of these two solution sets:

• All real numbers less than $-1/3$, or
• All real numbers greater than $7$.

So the solution set is $x < -1/3$ or $x > 7$, which corresponds to answer choice A.