# How do you solve this set of linear equations: 36x + 19y = 35; 10x + 12y = 19?

Jun 3, 2017

$x = \frac{59}{242}$ and $y = \frac{167}{121}$

#### Explanation:

Strategy: There are several different options to solving these equations. This one may be susceptible to picking one of them and solving for $x$ in terms of $y$. Then plug that equation with $y$s into the other equation. Then solve for $y$. Finally, with $y$ equal to some value, plug it in to the first equation and solve for $x$.

Step 1. Pick one equation and solve for $x$.

Pick one of the two equations. It doesn't matter which.
$36 x + 19 y = 35$

Subtract $19 y$ from both sides.
$36 x = - 19 y + 35$

Divide both sides by $36$.
$\textcolor{red}{x = - \frac{19}{36} y + \frac{35}{36}}$

Step 2. Plug that equation into the other, $10 x + 12 y = 19$.

$10 \textcolor{red}{x} + 12 y = 19$

$10 \left(\textcolor{red}{- \frac{19}{36} y + \frac{35}{36}}\right) + 12 y = 19$

Step 3. Solve for $y$.

Multiply $10$ through.
$- \frac{190}{36} y + \frac{350}{36} + 12 y = 19$

Subtract $350 / 36$ from both sides.
$- \frac{190}{36} y + 12 y = 19 - \frac{350}{36}$

$\frac{121}{18} y = \frac{167}{18}$

Multiply both sides by $18 / 121$.
$y = \frac{167}{18} \times \frac{18}{121}$

$\textcolor{b l u e}{y = \frac{167}{121}}$

Step 3. Plug this into your $x$ equation of Step 1.

$x = - \frac{19}{36} \textcolor{b l u e}{y} + \frac{35}{36}$

$x = - \frac{19}{36} \left(\textcolor{b l u e}{\frac{167}{121}}\right) + \frac{35}{36}$

$x = - \frac{19}{36} \left(\textcolor{b l u e}{\frac{167}{121}}\right) + \frac{35}{36}$

$x = \frac{59}{242}$

ANSWER: $x = \frac{59}{242}$ and $y = \frac{167}{121}$