How do you solve $\frac{2}{3} (n + 5) = \frac{3}{4} (n - 1)$ $\frac { 2} { 3} ( n + 5) = \frac { 3} { 4} ( n - 1)$ ?

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Answer 1 · 2018-04-12T12:59:46

$n = 49$ $n = 49$

You have to isolate the " $n$ $n$ " terms and bring them to the side of the equation.

But first you get rid of the fractions on both sides by multiplying the left hand side by 4 and the right hand side by 3.

$\frac{2}{3} (n + 5) = \frac{3}{4} (n - 1)$ $2/3 (n + 5) = 3/4 (n - 1)$

$4 \cdot 2 (n + 5) = 3 \cdot 3 (n - 1)$ $4*2 (n + 5) = 3*3 (n - 1)$

$8 (n + 5) = 9 (n - 1)$ $8 (n + 5) = 9 (n - 1)$

Multiply out the brackets and

$8 n + 40 = 9 n - 9$ $8 n + 40 = 9 n - 9$

And bring $n$ $n$ to the right hand side.

$40 + 9 = 9 n - 8 n$ $40 + 9 = 9 n - 8 n$

$n = 49$ $n = 49$

How do you solve 23(n+5)=34(n−1)\frac { 2} { 3} ( n + 5) = \frac { 3} { 4} ( n - 1)?