How do you solve $\frac{n + 3}{2 n - 8} = \frac{2}{3}$ $(n+3)/(2n-8)=2/3$ ?

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Answer 1 · 2016-09-30T12:52:35

$n = 25$ $n=25$

$given : \frac{n + 3}{2 n - 8} = \frac{2}{3}$ $"given : "(n+3)/(2n-8)=2/3$

$3 \cdot (n + 3) = 2 \cdot (2 n - 8)$ $3*(n+3)=2*(2n-8)$

$3 n + 9 = 4 n - 16$ $3n+9=4n-16$

$9 + 16 = 4 n - 3 n$ $9+16=4n-3n$

$n = 25$ $n=25$

How do you solve (n+3)/(2n-8)=2/3n+32n−8=23(n+3)/(2n-8)=2/3?