# How do you solve the system using the elimination method for 2a + 3b = 6 and 5a + 2b – 4 = 0?

Jul 22, 2015

$\left(a , b\right) = \left(0 , 2\right)$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$(using elimination method)

#### Explanation:

[1]$\textcolor{w h i t e}{\text{XXXX}}$$2 a + 3 b = 6$
[2]$\textcolor{w h i t e}{\text{XXXX}}$$5 a + 2 b - 4 = 0$

Rewite [2] into standard form:
[3]$\textcolor{w h i t e}{\text{XXXX}}$$5 a + 2 b = 4$

Multiply [1] by 2 and [3] by 3 to make the coefficients of $b$ the same in both.
[4]$\textcolor{w h i t e}{\text{XXXX}}$$4 a + 6 b = 12$
[5]$\textcolor{w h i t e}{\text{XXXX}}$$15 a + 6 b = 12$

Subtract [4] from [5]
[6]$\textcolor{w h i t e}{\text{XXXX}}$$11 a = 0$
Divide by 11
[7]$\textcolor{w h i t e}{\text{XXXX}}$$a = 0$

Substitute $0$ for $a$ in [1]
[8]$\textcolor{w h i t e}{\text{XXXX}}$$2 \left(0\right) + 3 b = 6$
Simplify
[9]$\textcolor{w h i t e}{\text{XXXX}}$$b = 2$