# How do solve the following linear system?: -2x + 1 = 2y, 7x - 4y - 13 = 0 ?

Oct 8, 2017

Answer: $\left(\frac{15}{11} , - \frac{19}{22}\right)$

#### Explanation:

Solve by substitution: $- 2 x + 1 = 2 y$, $7 x - 4 y - 13 = 0$

First, we can solve for $y$ in the first equation* by dividing both sides by $2$:
$- 2 x + 1 = 2 y$

$y = \frac{- 2 x + 1}{2}$

We can now substitute this $y$ term into the second equation and solve for $x$:
$7 x - 4 y - 13 = 0$

$7 x - 4 \left(\frac{- 2 x + 1}{2}\right) = 13$

Notice that we can cancel a $2$ from the numerator and denominator:
$7 x - 2 \left(- 2 x + 1\right) = 13$

$7 x + 4 x - 2 = 13$

Combining $x$-terms and moving all constants to one side, we find the value of $x$:
$11 x = 15$

$x = \frac{15}{11}$

To solve for $y$, we plug in the $x$ value we just found into the first equation which we solved for $y$ and solve for the $y$ value which corresponds with $x = \frac{15}{11}$:
$y = \frac{- 2 x + 1}{2}$

$= \frac{- 2 \left(\frac{15}{11}\right) + 1}{2}$

$= \frac{\frac{- 30}{11} + 1}{2}$

$= \frac{\frac{- 30 + 11}{11}}{2}$

$= - \frac{19}{22}$

Therefore, the solution to the linear system of equations is $\left(\frac{15}{11} , - \frac{19}{22}\right)$

*This was chosen because we can see that if we solve for $y$ in the first equation and plug into the second equation, the $\frac{1}{2}$ resulting from dividing the first equation by $2$ would cancel out with the $4$ in the second equation when we substitute in (since $2$ is a factor of $4$). Solving for any other variable and substituting into the other equation would result in some unwanted fraction which would thus complicate the problem a bit more than necessary.