Last time, we learned how to isolate variables like $x$ in an equation with addition, subtraction and multiplication.

Therefore, you should be able to solve the following equation for $x$:

$3x-5=10+9y$

(Use the vertical slides to go through this problem)

$3x-5=10+9y$

$3x=10+9y\r{+5}$

$3x-5=10+9y$

$3x=10+9y\r{+5}$

$3x=\g{15}+9y$

$3x-5=10+9y$

$3x=15+9y$

$$\frac{\r{3}x}{\r{3}}=\frac{15+9y}{3}$$

$3x-5=10+9y$

$3x=15+9y$

$$\frac{\r{3}x}{\r{3}}=\frac{15+9y}{3}$$

$x=5+3y$