How do you solve $\frac { - 3} { x + 7} = \frac { 8} { x - 4}$ ?

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Answer 1 · 2016-11-06T04:53:21

$x=4$

$-3(x-4)=8(x-7)$
$-3x+12=8x+56$
$12-56=8x+3x$
$-44=11x$
$x=-44/11$
$x=-44/11$

here x can never be 0 or -7 as these can make denominators zero

Answer 2 · 2016-11-06T05:06:55

$x=-4$

$(-3)/(x+7)=8/(x-4)$

Cross multiply.

$-3(x-4)=8(x+7)$

Expand both sides.

$-3x+12=8x+56$

Add $3x$ to both sides.

$12=8x+56+3x$

$12=11x+56$

Subtract $56$ from both sides.

$12-56=11x$

$-44=11x$

Divide both sides by $11$ .

$-44/11=x$

$-4=x$

Switch sides.

$x=-4$

How do you solve \frac { - 3} { x + 7} = \frac { 8} { x - 4}\frac { - 3} { x + 7} = \frac { 8} { x - 4}?