How do you solve the system 2x-3y=12 and x=4y+1?

Jun 2, 2018

$x = 9$
$y = 2$

Explanation:

$2 x - 3 y = 12$
$x = 4 y + 1$

Solving by Substitution

First, we're going to use an equation for the value of a variable in order to plug it into the opposite equation of the system. Because $x = 4 y + 1$ is already an equation for the value of a variable, we'll be using it. In the other equation of the system, plug in $x$'s value where $x$ is. So:

$2 \left(4 y + 1\right) - 3 y = 12$

Next, you'll be distributing. What this means is that you'll be multiplying the outside number, $2$, by the terms in the parentheses, $4 y$ and $1$. So:

$2 \cdot 4 y = 8 y$
$2 \cdot 1 = 2$

$8 y + 2 - 3 y = 12$

Combine like terms. $8 y - 3 y = 5 y$, so:

$5 y + 2 = 12$

This is a two-step equation. To solve it, subtract 2 from both sides to isolate for $y$. You should now have:

$5 y = 10$

Divide by $5$ to isolate for $y$:

$y = 2$

Plug the value of $y$ back into the equation for the value of $x$:

$x = 4 y + 1$
$x = 4 \left(2\right) + 1$
$x = 8 + 1$
$x = 9$

To truly prove that $x$ is 9 and $y$ is 2:

$2 x - 3 y = 12$
$2 \left(9\right) - 3 \left(2\right) = 12$
$18 - 6 = 12$
$12 = 12$