How do you solve $- 4 (x - 1) \leq - 3 (x - 5)$ $-4( x - 1) \leq - 3( x - 5)$ ?

Question

570 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-24T13:37:28

$- 11 \leq x$ $-11 <= x$

First we expand the brackets

$- 4 (x - 1) \leq - 3 (x - 5) \to$ $-4(x-1) <= -3(x-5) to$

$- 4 x + 4 \leq - 3 x + 15$ $-4x + 4 <= -3x + 15$

Then we add $4 x$ $4x$ to both sides

$4 \leq x + 15$ $4 <= x + 15$

Finally we take $15$ $15$ from both sides

$- 11 \leq x$ $-11 <= x$

How do you solve −4(x−1)≤−3(x−5)-4( x - 1) \leq - 3( x - 5)?