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

Mar 5, 2018

$x = \frac{27}{19}$ and $y = - \frac{14}{19}$

#### Explanation:

Let's isolate $y$ in the second equation

$y = 3 x - 5$

Now we can solve for $x$ in the other equation

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

$4 x + 15 x - 25 = 2$

$19 x = 27$

$x = \frac{27}{19}$

Now we solve for $y$

$y = 3 x - 5$

$y = 3 \left(\frac{27}{19}\right) - 5$

$y = \frac{81}{19} - \frac{5}{1} \times \frac{19}{19}$

$y = \frac{81}{19} - \frac{95}{19}$

$y = - \frac{14}{19}$

To check our work, let's substitute $\frac{27}{19}$ for $x$ and $- \frac{14}{19}$ for $y$ in the first equation and solve.

$4 \left(\frac{27}{19}\right) + 5 \left(- \frac{14}{19}\right) = 2$

$\frac{108}{19} - \frac{70}{19} = \frac{38}{19}$

$\frac{38}{19} = 2 = 2$!

We were right!

Mar 5, 2018

$\textcolor{b l u e}{x = \frac{27}{19}} , \textcolor{p u r p \le}{y = - \frac{14}{19}}$

#### Explanation:

$4 x + 5 y = 2$ Eqn (1)

$y + 5 = 3 x$

$y = 3 x - 5$ Eqn (2)

Substituting value of y term in Eqn (1) in terms of x,

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

$4 x + 15 x - 25 = 2$

Rearranging variables on L H S, constants on R H S,

$4 x + 15 x = 27$

$19 x = 27$ or $x = \frac{27}{19}$

Substituting value of x in Eqn (2),

$y = 3 x - 5 = \left(3 \cdot \left(\frac{27}{19}\right)\right) - 5 = \frac{81 - 95}{19} = - \frac{14}{19}$