# How do you solve the following linear system:  3x=y+1, 5x + 2y = -4 ?

May 4, 2018

$x = - \frac{2}{11}$

$y = - \frac{17}{11}$

#### Explanation:

Two ways: substitution method is easier to show, but elimination is fine too.

$3 x = y + 1 \text{ } \left(1\right)$
$5 x + 2 y = - 4 \text{ } \left(2\right)$

$3 x - 1 = y$

Substitute this $3 x - 1$ into $y$ in equation $\left(2\right)$ because $y$ is equal to $3 x - 1$.

$5 x + 2 \left(3 x - 1\right) = - 4$

$5 x + 6 x - 2 = - 4$

$11 x = - 2$

$x = - \frac{2}{11}$

Plug $x$ value back into one of the two original equations (you can plug into any other ones, but use the original because you don't know if you made an error rearranging)

$3 x = y + 1$

$3 \left(- \frac{2}{11}\right) = y + 1$

$3 \left(- \frac{2}{11}\right) - 1 = y$

$\left(- \frac{6}{11}\right) - 1 = y$

$\left(- \frac{6}{11}\right) - \left(\frac{11}{11}\right) = y$

or you can do $- 1 \frac{6}{11}$ and then change to an improper fraction. You still get the same answer.

$- \frac{17}{11} = y$

$y = - \frac{17}{11}$

You can verify this if both sides of the equation are the same value-then your x and y values are correct (I plugged back into equation 1 again)

$3 x = y + 1$

$3 \left(- \frac{2}{11}\right) = - \frac{17}{11} + 1$

$- \frac{6}{11} = \left(- \frac{17}{11}\right) + \left(\frac{11}{11}\right)$

$- \frac{6}{11} = - \frac{6}{11}$

Good luck. Make sure to ask your teacher. Never be shy to ask. Be forward with your own learning as you are the one who has to put your learning in your own hands.