# How do you solve this system of equations y= \frac { - 3} { 4} x , x - 4y = 32?

Aug 13, 2017

$\left(x , y\right) = \left(8 , - 6\right)$

#### Explanation:

Given
[1]$\textcolor{w h i t e}{\text{XXX}} y = \frac{- 3}{4} x$
[2]$\textcolor{w h i t e}{\text{XXX}} x - 4 y = 32$

Using [1] we can substitute $\frac{- 3}{4} x$ for $y$ in [2] to get
[3]$\textcolor{w h i t e}{\text{XXX}} x - 4 \times \frac{- 3}{4} x = 32$

Simplifying [4]
[5]$\textcolor{w h i t e}{\text{XXX}} x - \left(- 3\right) x = 32$

Continuing the simplification:
[6]$\textcolor{w h i t e}{\text{XXX}} 4 x = 32$

Dividing both sides of [6] by $4$
[7]$\textcolor{w h i t e}{\text{XXX}} x = 8$

Using [7] we can substitute $8$ for $x$ in [1] to get
[8]$\textcolor{w h i t e}{\text{XXX}} y = \frac{- 3}{4} \times 8$

Simplifying [8]
[9]$\textcolor{w h i t e}{\text{XXX}} y = - 6$

Aug 13, 2017

$\left(x , y\right) \to \left(8 , - 6\right)$

#### Explanation:

$\textcolor{red}{y} = - \frac{3}{4} x \to \left(1\right)$

$x - 4 \textcolor{red}{y} = 32 \to \left(2\right)$

$\text{substitute "y=-3/4x" in } \left(2\right)$

$\Rightarrow x - \left(4 \times - \frac{3}{4}\right) = 32$

$\Rightarrow x + 3 x = 32$

$\Rightarrow 4 x = 32$

$\text{divide both sides by 4}$

$\Rightarrow x = 8$

$\text{substitute this value in } \left(1\right)$

$y = - \frac{3}{4} \times 8 = - 6$

$\textcolor{b l u e}{\text{As a check}}$

$\text{substitute these values in } \left(2\right)$

$8 + 24 = 32 \leftarrow \text{ True}$

$\Rightarrow \text{point of intersection } = \left(8 , - 6\right)$ graph{(y+3/4x)(y-1/4x+8)((x-8)^2+(y+6)^2-0.06)=0 [-12.49, 12.48, -6.25, 6.24]}

Aug 13, 2017

Substitution.

$x = 8$
$y = - 6$

#### Explanation:

There are many ways to solve systems of equations. For this system:
$\textcolor{g r e e n}{y} = - \frac{3}{4} x$ (Eq. 1)
$x - 4 y = 32$ (Eq. 2),
it would be easiest to solve it with substitution since Equation (Eq.) 1 is already solved for $y$. This means we can simply plug in the $y$ value in the second equation.

This is Eq 2:

$x - 4 \textcolor{g r e e n}{y} = 32$

If we plug in Eq. 1 into Eq. 2, we get:

$x - 4 \left(- \frac{3}{4} x\right) = 32$
$x + 3 x = 32$
$4 x = 32$
$x = 8$

Now we solved for the first variable. To solve for $y$, all we do is plug in our value of $x$ back into Eq. 1:

$y = - \frac{3}{4} x$
$y = - \frac{3}{4} \left(8\right) = - 6$

So, the solution to the system of equations is:

$x = 8$
$y = - 6$

To check this answer, you can plug in the $x$ and $y$ values into Eq.1 and Eq. 2 respectively to see if the equation solves correctly:

Eq 1 verification by plugging in the $x$ and $y$ values:
$y = - \frac{3}{4} x$
$- 6 = - \frac{3}{4} \left(8\right) = - 6$ (so it works)
Eq 2 verification by plugging in the $x$ and $y$ values:
$x - 4 y = 32$
$8 - 4 \left(- 6\right) = 8 + 24 = 32$ (so it works)