Solving Linear Inequalities on the SAT
Inequalities work almost exactly like equations — with one rule that the SAT tests relentlessly, because it's so easy to forget.
Reading the concept isn't enough. The score comes from practicing the real Bluebook-style format and finding the exact mistakes you keep making.
What the SAT tests here
You isolate the variable the same way you would in an equation. The catch: whenever you multiply or divide both sides by a negative number, the inequality sign flips direction. Compound inequalities (two bounds at once) get handled together, not as two separate problems.
How to solve one, step by step
Example: solve $-2x + 3 > 11$.
- Subtract 3 from both sides: $-2x > 8$.
- Divide by $-2$ and flip the sign: $x < -4$.
The mistakes that cost points
- Forgetting to flip the sign. Dividing or multiplying by a negative reverses the inequality. Skip it and your whole answer is backwards.
- Splitting a compound inequality. Treating $-3 < 2x \leq 8$ as two unrelated problems loses the connection between the bounds.
- Flipping for a negative term that isn't a multiplier. Only flip when you multiply or divide by a negative — not when you simply add or subtract one.
Practice questions
Try these the way you would on test day, then open the solution to check your method.
Easy
The inequality $-2x + k > 4$ has the solution $x < 3$. What is the value of $k$?
- A$k = 10$
- B$k = 2$
- C$k = 6$
- D$k = -2$
Show solution
Answer: A, $k = 10$. Substitute the boundary condition: when $x = 3$, the inequality becomes an equality. So $-2(3) + k = 4 \Rightarrow -6 + k = 4 \Rightarrow k = 10$. Verify the direction: $-2x + 10 > 4 \Rightarrow -2x > -6 \Rightarrow x < 3$ ✓. The flip occurs when dividing by $-2$, confirming $k = 10$.
Medium
If $px + 9 > 1$ and $p > 0$, which of the following describes all possible values of $x$?
- A$x < -\dfrac{8}{p}$
- B$x < \dfrac{8}{p}$
- C$x > -\dfrac{8}{p}$
- D$x > \dfrac{8}{p}$
Show solution
Answer: C, $x > -\dfrac{8}{p}$. Subtract $9$ from both sides: $px > 1 - 9 = -8$. Divide both sides by $p$. Since $p > 0$, the inequality stays the same: $x > -\dfrac{8}{p}$. The numerator stays as $-8$. Do not drop the negative sign.
Hard
If $px + 18 > 6$ and $p < 0$, which of the following describes all possible values of $x$?
- A$x < \dfrac{12}{p}$
- B$x > \dfrac{12}{p}$
- C$x < -\dfrac{12}{p}$
- D$x > -\dfrac{12}{p}$
Show solution
Answer: C, $x < -\dfrac{12}{p}$. Subtract $18$ from both sides: $px > 6 - 18 = -12$. Divide both sides by $p$. Since $p < 0$, flip the inequality: $x < -\dfrac{12}{p}$. The numerator stays as $-12$. Do not drop the negative sign.
Find your exact gaps
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