How do you solve the following system: 2x - 3y = 10 , 2x + 3y = 12?

Dec 14, 2015

$\textcolor{w h i t e}{\times} x = \frac{11}{2}$, and $y = \frac{1}{3}$

Explanation:

$\textcolor{w h i t e}{\times} 2 x - 3 y = 10 , \textcolor{w h i t e}{x} 2 x + 3 y = 12$

$\textcolor{w h i t e}{\times} 2 x - 3 y = 10 \iff 3 y = 2 x - 10$
$\textcolor{w h i t e}{\times} 2 x + 3 y = 12 \iff 3 y = - 2 x + 12$

$\implies 2 x - 10 = y$
$\implies 2 x - 10 = \textcolor{red}{- 2 x + 12}$

Add $\textcolor{red}{+ 2 x + 10}$ to both side:
$\textcolor{w h i t e}{\times} 2 x - 10 \textcolor{red}{+ 2 x + 10} = - 2 x + 12 \textcolor{red}{+ 2 x + 10}$
$\implies 4 x = 22$

Multiply both sides by $\textcolor{red}{\frac{1}{4}}$:
$\textcolor{w h i t e}{\times} \textcolor{red}{\frac{1}{4} \times} 4 x = \textcolor{red}{\frac{1}{4} \times} 22$
$\implies x = \frac{11}{2}$

$\textcolor{w h i t e}{\times} 3 y = 2 x - 10$
$\implies 3 y = 2 \times \textcolor{red}{\frac{11}{2}} - 10$
$\implies 3 y = 1$

$\implies \textcolor{red}{\frac{1}{3} \times} 3 y = \textcolor{red}{\frac{1}{3} \times} 1$
$\implies y = \frac{1}{3}$