# How do you solve 2 x + 3 y = -7 and 2 x - 7 y = 7 using substitution?

Jun 30, 2017

By arranging equations, you can get $x = - \frac{7}{5}$ and $y = - \frac{7}{5}$

#### Explanation:

Arrange the first equation:

$2 x = - 3 y - 7$

Now put this into the second equation:

$- 3 y - 7 - 7 y = 7$

$- 10 y - 7 = 7$

$- 10 y = 14$

$y = - \frac{14}{10}$ or

$y = - \frac{7}{5}$

Put this into the first original equation

$2 x + 3 \times \left(- \frac{7}{5}\right) = - 7$

$2 x - \frac{21}{5} = - 7$

$2 x = \frac{21}{5} - 7$

$2 x = \frac{21 - 35}{5}$

$2 x = - \frac{14}{5}$

$x = - \frac{14}{10}$

$x = - \frac{7}{5}$

Your solution is $x = - \frac{7}{5}$ and $y = - \frac{7}{5}$

Jun 30, 2017

See a solution process below:

#### Explanation:

Step 1) We can solve both equations for $2 x$ which is a common term:

Equation 1)

$2 x + 3 y = - 7$

$2 x + 3 y - \textcolor{red}{3 y} = - 7 - \textcolor{red}{3 y}$

$2 x + 0 = - 7 - 3 y$

$2 x = - 7 - 3 y$

Equation 2)

$2 x - 7 y = 7$

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

$2 x - 0 = 7 + 7 y$

$2 x = 7 + 7 y$

Step 2) Equate the right sides of each equation and solve for $y$:

$- 7 - 3 y = 7 + 7 y$

$- \textcolor{b l u e}{7} - 7 - 3 y + \textcolor{red}{3 y} = - \textcolor{b l u e}{7} + 7 + 7 y + \textcolor{red}{3 y}$

$- 14 - 0 = 0 + \left(7 + \textcolor{red}{3}\right) y$

$- 14 = 10 y$

$- \frac{14}{\textcolor{red}{10}} = \frac{10 y}{\textcolor{red}{10}}$

$- \frac{7}{5} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{10}}} y}{\cancel{\textcolor{red}{10}}}$

$- \frac{7}{5} = y$

$y = - \frac{7}{5}$

Step 3) Substitute $- \frac{7}{5}$ for $y$ in the solution to either equation in Step 1 and calculate $x$. I will choose the second equation:

$2 x = 7 + 7 y$ becomes:

$2 x = 7 + \left(7 \cdot - \frac{7}{5}\right)$

$2 x = 7 + \left(- \frac{49}{5}\right)$

$2 x = \left(7 \times \frac{5}{5}\right) - \frac{49}{5}$

$2 x = \frac{35}{5} - \frac{49}{5}$

$2 x = - \frac{14}{5}$

$\textcolor{red}{\frac{1}{2}} \times 2 x = \textcolor{red}{\frac{1}{2}} \times - \frac{14}{5}$

$1 x = - \frac{14}{10}$

$x = - \frac{7}{5}$

The solution is: $x = - \frac{7}{5}$ and $y = - \frac{7}{5}$ or $\left(- \frac{7}{5} , - \frac{7}{5}\right)$