How do you solve the inequality $3 x + 2 \leq 1$ $3x + 2 ≤ 1$ ?

Question

6757 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-11T21:10:44

$x \leq - \frac{1}{3}$ $x<=-1/3$

$3 x + 2 \leq 1$ $3x+2<=1$

Begin by moving like terms to either side, so subtract 2 from each side.

$3 x \leq - 1$ $3x<=-1$

Now divide by 3 to isolate x, and you have an answer!

$x \leq - \frac{1}{3}$ $x<=-1/3$

How do you solve the inequality 3x+2≤13x + 2 ≤ 1?