# How do you solve the system using the elimination method for 3(x-y)=15 and 2x+7=7?

Jul 22, 2015

$\left(x , y\right) = \left(0 , - 5\right)$
$\textcolor{w h i t e}{\text{XXXX}}$by the elimination method (although the "elimination" has already been done)

#### Explanation:

Given
[1]$\textcolor{w h i t e}{\text{XXXX}}$$3 \left(x - y\right) = 15$
[2]$\textcolor{w h i t e}{\text{XXXX}}$$2 x + 7 = 7$

Since $y$ has already been eliminated in [2]
if we simplify [2] as
[3]$\textcolor{w h i t e}{\text{XXXX}}$$2 x = 0$
$\Rightarrow$
[4]$\textcolor{w h i t e}{\text{XXXX}}$$x = 0$

We can then substitute $x = 0$ back into [1] to get
[5]$\textcolor{w h i t e}{\text{XXXX}}$$3 \left(0 - y\right) = 15$
or, after simplification:
[6]$\textcolor{w h i t e}{\text{XXXX}}$$y = - 5$