# How do you solve the system of equations x+ 4y = 4 and - 5x + 3y = 3?

May 23, 2018

$x = 0$
$y = 1$

#### Explanation:

$x + 4 y = 4$
$- 5 x + 3 y = 3$

Solving by Substitution

First, let's use one of these equations and simplify for the equation of a value. I think $x$ will be a good one to solve first because it looks easy to simplify. Let's begin:

$x + 4 y = 4$

Subtract $4 y$ from both sides to get an equation for $x$. You should now have:

$x = - 4 y + 4$

This will be our $x$ value that we'll be substituting in the second equation. Let's plug this term in:

$- 5 x + 3 y = 3$
#-5(-4y + 4) + 3y = 3

Distribute. $- 5 y \cdot - 4 y$ becomes $20 y$ because two positives make a negative, and $- 5 \cdot 4$ will become $- 20$ because only one negative is present.

$20 y - 20 + 3 y = 3$

Combine like terms.

$23 y - 20 = 3$

Now, it's a two-step equation. Add $20$ to both sides to cancel out $- 20$ in order to get the division step. You should now have:

$23 y = 23$

Divide by $23$ to isolate for $y$.

$y = 1$

Now knowing what $y$ is, return to your simplified equation for the value of $x$ and substitute the value of $y$ for $y$:

$x = - 4 y + 4$
$x = - 4 \left(1\right) + 4$
$x = - 4 + 4$
$x = 0$