# How do you solve the system of equations x+ 2y = 3 and - 2x + y = 1?

Jul 30, 2017

Pick either one of the equations and make either $x$ or $y$ the subjects of the formula and substitute the value of either $y$ or $x$ into the other equation.

#### Explanation:

I.e. $\text{ "x=3-2y" }$ (from first equation)

Now substitute $x$ above into second equation:

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

$- 6 + 4 y + y = 1$

$- 6 + 5 y = 1$

$5 y = 7$

Therefore

$y = \frac{7}{5}$

Now substitute the $y$-value into the equation to find $x$:

$- 2 x + \frac{7}{5} = 1$

$- 2 x = \frac{5}{5} \left(= 1\right) - \frac{7}{5}$

$x = \frac{1}{5}$

Thus, $x$ and $y$ for the 2nd equation $= \frac{1}{5}$ and $= \frac{7}{5}$.

You must again substitute either into the 1st equation to find those values. Just remember to check the answers!