# How do you solve the following system: 2x+7y=1, 5x - 7y = 12 ?

Aug 13, 2017

See a solution process below:

#### Explanation:

Step 1) Solve each equation for $7 y$:

Equation 1:

$2 x + 7 y = 1$

$- \textcolor{red}{2 x} + 2 x + 7 y = - \textcolor{red}{2 x} + 1$

$0 + 7 y = - 2 x + 1$

$7 y = - 2 x + 1$

Equation 2:

$5 x - 7 y = 12$

$- \textcolor{red}{5 x} + 5 x - 7 y = - \textcolor{red}{5 x} + 12$

$0 - 7 y = - 5 x + 12$

$- 7 y = - 5 x + 12$

$\textcolor{red}{- 1} \times - 7 y = \textcolor{red}{- 1} \left(- 5 x + 12\right)$

$7 y = 5 x - 12$

Step 2) Because the left side of each equation is $7 y$ we can equate the right side of each equation and solve for $x$:

$- 2 x + 1 = 5 x - 12$

$\textcolor{red}{2 x} - 2 x + 1 + \textcolor{b l u e}{12} = \textcolor{red}{2 x} + 5 x - 12 + \textcolor{b l u e}{12}$

$0 + 13 = \left(\textcolor{red}{2} + 5\right) x - 0$

$13 = 7 x$

$\frac{13}{\textcolor{red}{7}} = \frac{7 x}{\textcolor{red}{7}}$

$\frac{13}{7} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{7}}} x}{\cancel{\textcolor{red}{7}}}$

$\frac{13}{7} = x$

$x = \frac{13}{7}$

*Step 3) Substitute $\frac{13}{7}$ for $x$ in the solution to either equation in Step 1 and calculate $y$:

$7 y = - 2 x + 1$ becomes:

$7 y = \left(- 2 \times \frac{13}{7}\right) + 1$

$7 y = - \frac{26}{7} + 1$

$7 y = - \frac{26}{7} + \left(\frac{7}{7} \times 1\right)$

$7 y = - \frac{26}{7} + \frac{7}{7}$

$7 y = - \frac{19}{7}$

$\frac{7 y}{\textcolor{red}{7}} = \frac{- \frac{19}{7}}{\textcolor{red}{7}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{7}}} y}{\cancel{\textcolor{red}{7}}} = - \frac{19}{49}$

$y = - \frac{19}{49}$

The Solution Is: $x = \frac{13}{7}$ and $y = - \frac{19}{49}$ or $\left(\frac{13}{7} , - \frac{19}{49}\right)$