How do you solve by substitution x - 4y = -1  and 3x + 5y = 31?

Jul 13, 2015

$x = 7$
$y = 2$

Explanation:

We are given:

$x - 4 y = - 1$

and

$3 x + 5 y = 31$

We can do this a number of ways, but let's start with $x - 4 y = - 1$

Let's multiply $x - 4 y = - 1$ by $- 3$ to get its $x$ to look like the $x$ in $3 x + 5 y = 31$ but with an opposite sign:

$x - 4 y = - 1$

$\left(- 3\right) \left(x - 4 y\right) = \left(- 3\right) \left(- 1\right)$

$- 3 x + 12 y = 3$

Now, let's add $- 3 x + 12 y = 3$ to $3 x + 5 y = 31$ to get rid of those $x$'s:

$- 3 x + 12 y = 3$
$3 x + 5 y = 31$

$17 y = 34$

Now, divide both sides by $17$ to get:

$y = 2$

Now, just plug in the value for $y$ into $3 x + 5 y = 31$

$3 x + 5 y = 31$
$3 x + 5 \left(2\right) = 31$
$3 x + 10 = 31$

subtract both sides by $10$:

$3 x + 10 = 31$
$3 x = 21$

divide both sides by $3$:

$3 x = 21$
$x = 7$

So, $x = 7$ and $y = 2$

To check if our answers are correct, you can take either of the two given equations and plug in the values of $x$ and $y$.

Let's try this with $x - 4 y = - 1$

$x - 4 y = - 1$

$\left(7\right) - 4 \left(2\right) = - 1$

$7 - 8 = - 1$

$- 1 = - 1$