# How do you solve (3k-7)/5=16?

Dec 31, 2016

$\textcolor{b l u e}{\text{k=29}}$

#### Explanation:

color(green)("Multiply both sides by 5 to remove the denominator,"

$\cancel{5} \times \frac{3 k - 7}{\cancel{5}} = 16 \times 5$

5 cancels out on the left side and you're left with $3 k - 7$. And, when 16 gets multiplied by 5, you get 80 on the right side:

$3 k - 7 = 80$

$\textcolor{m a \ge n t a}{\text{Add 7 to both sides:}}$

$3 k - \cancel{7} = 80$
$\textcolor{w h i t e}{a a} + \cancel{7} \textcolor{w h i t e}{a a} + 7$

The 7's cancel out, so you only have $3 k$ left. And when you add 80+7 on the right side, you obtain 87:

$3 k = 87$

$\textcolor{red}{\text{Divide both sides by 3 to get k by itself:}}$

$\frac{\cancel{\text{3}} k}{\cancel{3}} = \frac{87}{3}$

Now you are just left with $k$ on the left side and 29 on the right side.

Thus,
$\textcolor{b l u e}{\text{k=29}}$