# How do solve the following linear system?:  x+2y=1 , 6 x+3y=11 ?

May 18, 2018

See a solution process below:

#### Explanation:

Step 1) Solve the first equation for $x$:

$x + 2 y = 1$

$x + 2 y - \textcolor{red}{2 y} = 1 - \textcolor{red}{2 y}$

$x + 0 = 1 - 2 y$

$x = 1 - 2 y$

Step 2) Substitute $\left(1 - 2 y\right)$ for $x$ in the second equation and solve for $y$:

$6 x + 3 y = 11$ becomes:

$6 \left(1 - 2 y\right) + 3 y = 11$

$\left(6 \cdot 1\right) - \left(6 \cdot 2 y\right) + 3 y = 11$

$6 - 12 y + 3 y = 11$

$6 + \left(- 12 + 3\right) y = 11$

$6 + \left(- 9\right) y = 11$

$6 - 9 y = 11$

$6 - \textcolor{red}{6} - 9 y = 11 - \textcolor{red}{6}$

$0 - 9 y = 5$

$- 9 y = 5$

$\frac{- 9 y}{\textcolor{red}{- 9}} = \frac{5}{\textcolor{red}{- 9}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 9}}} y}{\cancel{\textcolor{red}{- 9}}} = - \frac{5}{9}$

$y = - \frac{5}{9}$

Step 3) Substitute $- \frac{5}{9}$ for $y$ in the solution to the first equation at the end of Step 1 and calculate $x$:

$x = 1 - 2 y$ becomes:

$x = 1 - \left(2 \cdot - \frac{5}{9}\right)$

$x = 1 - \left(- \frac{10}{9}\right)$

$x = 1 + \frac{10}{9}$

$x = \frac{9}{9} + \frac{10}{9}$

$x = \frac{19}{9}$

The Solution Is:

$x = \frac{19}{9}$ and $y = - \frac{5}{9}$

Or

$\left(\frac{19}{9} , - \frac{5}{9}\right)$