# How do you solve 2x+6y=32 and 5x-2y=12 using substitution?

Mar 11, 2018

$\therefore$ color(red)(x=4 and y=4

#### Explanation:

$2 x + 6 y = 32$

$\implies x + 3 y = 16$

Multiplying both sides by 2

$x = 16 - 3 y$

Given that,

5x−2y=12

Replacing $x = 16 - 3 y$

$5 \left(16 - 3 y\right) - 2 y = 12$

$80 - 15 y - 2 y = 12$

$80 - 17 y = 12$

$17 y = 80 - 12$

$17 y = 68$

$y = \frac{68}{17}$

color(magenta)(y=4

$\therefore x = 16 - 3 y$

$x = 16 - 3 \times 4$

$x = 16 - 12$

color(magenta)(x=4

$\therefore$ color(red)(x=4 and y=4

~Hope this helps! :)

Mar 11, 2018

$x = 4 , y = 4$

#### Explanation:

$2 x + 6 y = 32$
$5 x - 2 y = 12$

In order to use substitution, rearrange one of the equations

$2 x + 6 y = 32$
$2 x = - 6 y + 32$ (Subtracted 6y from both sides)
$x = - 3 y + 16$ (Divided both sides by 2)

Now that we rearrange the equation, we plug it into $5 x - 2 y = 12$

$5 \left(- 3 y + 16\right) - 2 y = 12$
$- 15 y + 80 - 2 y = 12$ (Distributed 5 to the terms in bracket)
$- 17 y + 80 = 12$ (Combined like terms)
$- 17 y = - 68$ (Subtract 80 from both sides)
$y = 4$ (Divided both sides by -17)

Now that we have the y term, we can plug it into either equation to find x

$2 x + 6 \left(4\right) = 32$
$2 x + 24 = 32$ (Multiplied 6*4)
$2 x = 8$ (Subtracted 24 from both sides)
$x = 4$ (Divided both sides by 2)

Check your answer y plugging in y=4 and x=4 into either equation.