# How do you solve x = 4y and 2x + 3y = 44 using substitution?

Jun 1, 2018

You have $x = 4 y$, so, what you do is, substitute $4 y$ wherever you see $x$ in the second equation. This sequence looks like:

$2 x + 3 y = 44$

$2 \left(4 y\right) + 3 y = 44$

$8 y + 3 y = 44$

$11 y = 44$

$y = \frac{44}{11} = 4$

...now that you know that y = 4, you can plug this in to either equation and solve for x.

$x = 4 y = 4 \left(4\right)$

$x = 16$

CHECK YOUR WORK. Feed these values into each equation to verify that they are actually solutions.

$x = 4 y$
$16 = 4 \cdot 4$ CHECK

$2 x + 3 y = 44$

$\left(2 \cdot 16\right) + \left(3 \cdot 4\right) = 32 + 12 = 44$ CHECK

GOOD LUCK

Jun 1, 2018

$\left(x , y\right) \to \left(16 , 4\right)$

#### Explanation:

$x = 4 y \to \left(1\right)$

$2 x + 3 y = 44 \to \left(2\right)$

$\textcolor{b l u e}{\text{substitute "x=4y" into equation }} \left(2\right)$

$2 \left(4 y\right) + 3 y = 44$

$8 y + 3 y = 44$

$11 y = 44 \Rightarrow y = \frac{44}{11} = 4$

$\text{substitute "y=4" into equation } \left(1\right)$

$x = 4 \times 4 = 16$

$\text{point of intersection } = \left(16 , 4\right)$