# How do you solve for x in 8ax-7a^2=19a^2-5ax?

May 31, 2018

$x = 2 a$

#### Explanation:

$8 a x - 7 {a}^{2} = 19 {a}^{2} - 5 a x$

Collecting like terms;

Note: When crossing over a Negative value or unknown in a algebraic expression, the sign changes the the Positive value, also when crossing over a Positive value or uknown the sign changes to a Negative value, (Vice-Versa)

$8 a x + 5 a x = 19 {a}^{2} + 7 {a}^{2}$

Simplifying;

$13 a x = 26 {a}^{2}$

Dividing both sides by the coefficient of $x$;

$\frac{13 a x}{13 a} = \frac{26 {a}^{2}}{13 a}$

$\frac{\cancel{13 a} x}{\cancel{13 a}} = \frac{26 {a}^{2}}{13 a}$

$x = \frac{26 {a}^{2}}{13 a}$

$x = \frac{2 a \times 13 a}{13 a}$

$x = \frac{2 a \times \cancel{13 a}}{\cancel{13 a}}$

$x = \frac{2 a}{1}$

$x = 2 a$

May 31, 2018

$x = 2 a$

#### Explanation:

$\text{collect terms in x together on the left side and other}$
$\text{terms on the right side}$

$\text{add "5ax" to both sides}$

$8 a x + 5 a x - 7 {a}^{2} = 19 {a}^{2} \cancel{- 5 x} \cancel{+ 5 x}$

$13 a x - 7 {a}^{2} = 19 {a}^{2}$

$\text{add "7a^2" to both sides}$

$13 a x = 19 {a}^{2} + 7 {a}^{2}$

$13 a x = 26 {a}^{2}$

$\text{divide both sides by } 13 a$

$\frac{\cancel{13 a} x}{\cancel{13 a}} = \frac{26 {a}^{2}}{13 a} \Rightarrow x = 2 a$