# How do solve the following linear system?:  4x-2y=2 , -2x+5y=-3 ?

Apr 24, 2017

See the entire solution process below:

#### Explanation:

Step 1) Solve the first equation for $y$:

$4 x - 2 y = 2$

$4 x - \textcolor{red}{4 x} - 2 y = 2 - \textcolor{red}{4 x}$

$0 - 2 y = 2 - 4 x$

$- 2 y = 2 - 4 x$

$\frac{- 2 y}{\textcolor{red}{- 2}} = \frac{2 - 4 x}{\textcolor{red}{- 2}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 2}}} y}{\cancel{\textcolor{red}{- 2}}} = \frac{2}{\textcolor{red}{- 2}} - \frac{4 x}{\textcolor{red}{- 2}}$

$y = - 1 + 2 x$

Step 2) Substitute $- 1 + 2 x$ for $y$ in the second equation and solve for $x$:

$- 2 x + 5 y = - 3$ becomes:

$- 2 x + 5 \left(- 1 + 2 x\right) = - 3$

$- 2 x + \left(5 \cdot - 1\right) + \left(5 \cdot 2 x\right) = - 3$

$- 2 x - 5 + 10 x = - 3$

$10 x - 2 x - 5 = - 3$

$\left(10 - 2\right) x - 5 = - 3$

$8 x - 5 = - 3$

$8 x - 5 + \textcolor{red}{5} = - 3 + \textcolor{red}{5}$

$8 x - 0 = 2$

$8 x = 2$

$\frac{8 x}{\textcolor{red}{8}} = \frac{2}{\textcolor{red}{8}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{8}}} x}{\cancel{\textcolor{red}{8}}} = \frac{1}{4}$

$x = \frac{1}{4}$

Step 3) Subsitute $\frac{1}{4}$ for $x$ in the solution to the first equation at the end of Step 1 and calculate $y$:

$y = - 1 + 2 x$ becomes:

$y = - 1 + \left(2 \cdot \frac{1}{4}\right)$

$y = - 1 + \frac{1}{2}$

$y = - \frac{1}{2}$

The solution is: $x = \frac{1}{4}$ and $y = - \frac{1}{2}$ or $\left(\frac{1}{4} , - \frac{1}{2}\right)$