# How do you solve (3x+1)^2+(4x-3)^2=(5x-1)^2?

##### 1 Answer
Apr 26, 2017

$x = \frac{9}{8}$

#### Explanation:

Given:

${\left(3 x + 1\right)}^{2} + {\left(4 x - 3\right)}^{2} = {\left(5 x - 1\right)}^{2}$

Multiply out both sides to get:

$9 {x}^{2} + 6 x + 1 + 16 {x}^{2} - 24 x + 9 = 25 {x}^{2} - 10 x + 1$

Simplifying the left hand side, this becomes:

$\textcolor{red}{\cancel{\textcolor{b l a c k}{25 {x}^{2}}}} - 18 x + 10 = \textcolor{red}{\cancel{\textcolor{b l a c k}{25 {x}^{2}}}} - 10 x + 1$

Add $- 25 {x}^{2} + 18 x - 1$ to both sides to get:

$9 = 8 x$

Divide both sides by $8$ to get:

$x = \frac{9}{8}$