How do you solve $- \frac{8}{7} = \frac{n + 3}{5}$ $-\frac{8}{7} = \frac{n + 3}{5}$ ?

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Answer 1 · 2016-09-20T14:10:31

try to make denominators equal. then solve the equation...

$- \frac{8 \cdot 5}{7 \cdot 5} = \frac{(n + 3) \cdot 7}{5 \cdot 7}$ $-(8*5)/(7*5) = ((n+3)*7)/(5*7)$

$- \frac{40}{35} = \frac{7 n + 21}{35}$ $-40/35 = (7n+21)/35$

$- 40 = 7 n + 21$ $-40 = 7n + 21$

$7 n = - 61$ $7n = -61$

$n = - \frac{61}{7}$ $n=-61/7$

How do you solve −87=n+35-\frac{8}{7} = \frac{n + 3}{5}?