How do you solve $\frac{4}{d + 8} = \frac{7}{d - 18}$ $\frac { 4} { d + 8} = \frac { 7} { d - 18}$ ?

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Answer 1 · 2017-11-25T17:25:51

$d = - \frac{128}{3} ≅ - 42.667$ $d=-(128)/(3)~=-42.667$

We multiply in both sides by $(d + 8) (d - 18)$ $(d+8)(d-18)$ so we end up with:
$(d - 18) 4 = 7 (d + 8)$ $(d-18)4=7(d+8)$

we multiply,
$4 d - 72 = 7 d + 56$ $4d-72=7d+56$

we isolate $d$ $d$ ,
$4 d - 7 d = 56 + 72$ $4d-7d=56+72$

$- 3 d = 128$ $-3d=128$

$d = - \frac{128}{3}$ $d=-(128)/(3)$

How do you solve \frac { 4} { d + 8} = \frac { 7} { d - 18}4d+8=7d−18\frac { 4} { d + 8} = \frac { 7} { d - 18}?