# How do you solve the following linear system:  -3x + 2y = 1 , x + 1/2y = 7 ?

Apr 1, 2017

Find the x value using one off the equations.

$x + \frac{1}{2} y = 7$
$x = 7 - \frac{1}{2} y$

Plug the x value into the other equation.

$- 3 \left(7 - \frac{1}{2} y\right) + 2 y = 1$

Simplify

$- 21 + \frac{3}{2} y + 2 y = 1$

$\frac{3}{2} y + 2 y = 22$

$\frac{3}{2} y + \frac{4}{2} y = 22$

$\frac{7}{2} y = 22$

$y = \frac{22 \left(2\right)}{7}$

$y = \frac{44}{7}$

Plug in the y value to find the x value.

$x + \frac{1}{2} \left(\frac{44}{7}\right) = 7$

$x + \frac{22}{7} = 7$

$x = \frac{49}{7} - \frac{22}{7}$

$x = \frac{27}{7}$

ANSWER: $\left(\frac{27}{7} , \frac{44}{7}\right)$

Apr 1, 2017

$\text{answer:" x=27/7" , } y = \frac{44}{7}$

#### Explanation:

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

$x + \frac{1}{2} y = 7 \text{ } \left(2\right)$

$\text{let us multiply both sides of equation (2) by 4}$

$\textcolor{red}{4} \cdot \left(x + \frac{1}{2} y\right) = \textcolor{red}{4} \cdot 7$

$\text{We get ;}$

$4 x + 2 y = 28 \text{ } \left(3\right)$

$\text{the coefficients of term 'y' in both equation (1) and (3) is equal.}$

$\text{let subtract (3) from (1) to eliminate 'y'}$

$- 3 x + 2 y - \left(4 x + 2 y\right) = 1 - 28$

$- 3 x + 2 y - 4 x - 2 y = - 27$

$- 3 x + \cancel{2 y} - 4 x - \cancel{2 y} = - 27$

$- 7 x = - 27$

$x = \frac{27}{7}$

$\text{now ,let us use (1) or (2)}$

$x + \frac{1}{2} y = 7$

$\frac{27}{7} + \frac{y}{2} = 7$

$\frac{y}{2} = 7 - \frac{27}{7}$

$\frac{y}{2} = \frac{49 - 27}{7}$

$\frac{y}{2} = \frac{22}{7}$

$y = \frac{44}{7}$