# How do you solve the following linear system 3x-7y=1, -3x+5y=1? ?

Mar 7, 2017

$x = - 2$ and $y = - 1$

#### Explanation:

$3 x - 7 y = 1$
$- 3 x + 5 y = 1$

From the first equation, determine a temporary value for $3 x$.

$3 x - 7 y = 1$

Add $7 y$ to each side.

$3 x = 7 y + 1$

In the second equation, substitute $3 x$ with $\textcolor{red}{\left(7 y + 1\right)}$.

$- 3 x + 5 y = 1$

$- \textcolor{red}{\left(7 y + 1\right)} + 5 y = 1$

Open the brackets and simplify. The product of a negative and a positive is a negative.

$- 7 y - 1 + 5 y = 1$

$- 2 y - 1 = 1$

Add $1$ to both sides.

$- 2 y = 2$

Divide both sides by $- 2$.

$y = - 1$

In the first equation, substitute $y$ with $\textcolor{b l u e}{- 1}$.

$3 x - 7 y = 1$

$3 x - 7 \textcolor{b l u e}{\left(- 1\right)} = 1$

Open the brackets and simplify. The product of two negatives is a positive.

$3 x + 7 = 1$

Subtract $7$ from each side.

$3 x = - 6$

Divide both sides by $3$.

$x = - 2$