How do you solve $3- 2( 5x - 4) > - 9$ and $3x > - 4( x - 2)$ ?

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Answer 1 · 2018-03-17T18:47:32

$3-2(5x-4)> -9$ ----------------- $3x> -4(x-2)$
$x<2$ ----------------------------------- $x>8/7$

Ok so the first one $3-2(5x-4)$
First we distribute the $-2$ giving us
$3-10x+8> -9$

We now combine like terms on the left side
$11-10x> -9$

Subtract 11 on both sides giving us
$-10x> -20$

We now divide by $-10$ on both sides.
REMEMBER: When dividing by a negative the sign always turn to the other side.

$(-10x)/(-10)< -20/(-10)$

$x<2$

Ok now the second one.
First we distribute the $-4$ giving us
$3x> -4x+8$

Now we add $4x$ on both sides giving us
$7x>8$

Now we divide $7$ on both sides
$(7x)/7>8/7$

$x>8/7$

How do you solve 3- 2( 5x - 4) > - 93- 2( 5x - 4) > - 9 and 3x > - 4( x - 2)3x > - 4( x - 2)?