# How do you solve 4x + y = 4 and 2x + 8y = 0 using substitution?

Apr 9, 2016

$\left(x , y\right) = \left(\frac{16}{15} , - \frac{4}{15}\right)$
(see below for solution method using substitution)

#### Explanation:

Given: $4 x + y = 4$

This implies: $y = 4 - 4 x$
So we can substitute $\left(4 - 4 x\right)$ for $y$
in the second given equation: $2 x + 8 y = 0$
to get:
$\textcolor{w h i t e}{\text{XXX}} 2 x + 8 \left(4 - 4 x\right) = 0$

Simplifying:
$\textcolor{w h i t e}{\text{XXX}} 2 x + 32 - 32 x = 0$

$\textcolor{w h i t e}{\text{XXX}} 30 x = 32$

$\textcolor{w h i t e}{\text{XXX}} x = \frac{16}{15}$

The we can substitute $\frac{16}{15}$ for $x$
in the first given equation: $4 x + y = 4$
to get:
$\textcolor{w h i t e}{\text{XXX}} 4 \cdot \left(\frac{16}{15}\right) + y = 4$

$\textcolor{w h i t e}{\text{XXX}} y = 4 - \frac{64}{15} = - \frac{4}{15}$

Apr 9, 2016

$\left(x , y\right) = \left(\frac{16}{15} , - \frac{4}{15}\right)$

#### Explanation:

Solve by substitution and elimination

color(blue)(4x+y=4

color(blue)(2x+8y=0

We can eliminate $4 x$ from the first equation by $2 x$ in the second equation (whole equation) with $- 2$ to get $- 4 x$

$\rightarrow - 2 \left(2 x + 8 y = 0\right)$

Use the distributive property

color(brown)(a(b+c=x)=ab+ac=ax

$\rightarrow - 4 x - 16 y = 0$

Now, add the above equation with the first equation to eliminate $4 x$

$\rightarrow \left(4 x + y = 4\right) + \left(- 4 x - 16 y = 0\right)$

$\rightarrow - 15 y = 4$

Divide both sides by $- 15$

$\rightarrow \frac{\cancel{- 15} y}{\cancel{- 15}} = \frac{4}{-} 15$

color(green)(rArry=-4/15

Now,substitute the value of $y$ to the first equation

$\rightarrow 4 x + \left(- \frac{4}{15}\right) = 4$

$\rightarrow 4 x - \frac{4}{15} = 4$

Add $\frac{4}{15}$ both sides

$\rightarrow 4 x - \frac{4}{15} + \frac{4}{15} = 4 + \frac{4}{15}$

$\rightarrow 4 x = \frac{64}{15}$

Divide both sides by $4$

$\rightarrow \frac{\cancel{4} x}{\cancel{4}} = \frac{64}{15} \div 4$

(I have written $\frac{64}{15} / 4$ as $\frac{64}{15} \div 4$)

Take the reciprocal and multiply

$\rightarrow x = {\cancel{64}}^{16} / 15 \cdot \frac{1}{\cancel{4}} ^ 1$

color(green)(rArrx=16/15

color(blue)(ul bar |(x,y)=(16/15,-4/15)|