# How do you solve 5x + 2y = 20 and x + 4y = 13?

Apr 23, 2018

$\left(x , y\right) \to \left(3 , \frac{5}{2}\right)$

#### Explanation:

$5 x + 2 y = 20 \to \left(1\right)$

$x + 4 y = 13 \to \left(2\right)$

$\text{from equation "(2)" we can express x in terms of y}$

$\Rightarrow x = 13 - 4 y \to \left(3\right)$

$\text{substitute "x=13-4y" in equation } \left(1\right)$

$5 \left(13 - 4 y\right) + 2 y = 20 \leftarrow \textcolor{b l u e}{\text{distribute}}$

$\Rightarrow 65 - 20 y + 2 y = 20$

$\Rightarrow - 18 y + 65 = 20 \leftarrow \textcolor{b l u e}{\text{subtract 65 from both sides}}$

$\Rightarrow - 18 y = - 45$

$\text{divide both sides by } - 18$

$\frac{\cancel{- 18} y}{\cancel{- 18}} = \frac{- 45}{- 18}$

$\Rightarrow y = \frac{45}{18} = \frac{5}{2}$

$\text{substitute "y=5/2" in equation } \left(3\right)$

$\Rightarrow x = 13 - \left(4 \times \frac{5}{2}\right) = 13 - 10 = 3$

$\text{the solution is } \left(x , y\right) \to \left(3 , \frac{5}{2}\right)$

Apr 23, 2018

x=3, y=5/2

#### Explanation:

(1) $5 x + 2 y = 20$
(2) $x + 4 y = 13$

First, multiply equation 1 by 2, this gives both equations the same y coefficient of 4. The equations are now:
(1) $10 x + 4 y = 40$
(2) $x + 4 y = 13$

Next subtract equation 2 from 1 (equation 1 - equation 2)
This makes
$\left(10 x - x\right) + \left(4 y - 4 y\right) = \left(40 - 13\right)$

Simplified: $9 x - 0 = 27$

Then divide both sides by 9 to solve for x

$\frac{9}{9} x = \frac{27}{9}$
$x = 3$

Then input x=3 back into equation 2

$\left(3\right) + 4 y = 13$

Subtract 3 from both sides:
$4 y = 10$

Divide both sides by 4 to solve for y

$y = \frac{10}{4}$ (simplified to $\frac{5}{2}$)

Therefore $x = 3$ and $y = \frac{5}{2}$