# How do you solve the system y= 1/2x + 1 and y= -2x + 6?

Mar 3, 2017

$x = 3$,$y = 0$
$S o {l}^{n} = \left(3 , 0\right)$

#### Explanation:

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

$2 y = x + 2$----------(i)

$y = - 2 x + 6$---------(ii)

Substitute eq ii in eq i

$2 \left(- 2 x + 6\right) = - 2 x + 6$

$- 4 x + 12 = - 2 x + 6$

$2 x = 6$

$\underline{x = 3}$

Put $x = 3$ in eq ii

$y = - 6 + 6$

$\underline{y = 0}$

$S o {l}^{n} = \left(3 , 0\right)$

Mar 3, 2017

2,2)

#### Explanation:

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

$\textcolor{red}{y} = - 2 x + 6 \to \left(2\right)$

Since both equations are expressed with $\textcolor{red}{y}$ as the subject, we can equate the right sides.

$\Rightarrow \frac{1}{2} x + 1 = - 2 x + 6$

To eliminate the fraction, multiply ALL terms on both sides by 2, the denominator of the fraction.

$\left({\cancel{2}}^{1} \times \frac{x}{\cancel{2}} ^ 1\right) + \left(2 \times 1\right) = \left(2 \times - 2 x\right) + \left(2 \times 6\right)$

$\Rightarrow x + 2 = - 4 x + 12$

$x + 4 x + 2 = \cancel{- 4 x} \cancel{+ 4 x} + 12$

$\Rightarrow 5 x + 2 = 12$

subtract 2 from both sides.

$5 x \cancel{+ 2} \cancel{- 2} = 12 - 2$

$\Rightarrow 5 x = 10$

divide both sides by 5

$\frac{\cancel{5} x}{\cancel{5}} = \frac{10}{5}$

$\Rightarrow x = 2$

To find the corresponding value of y, substitute x = 2 into either (1) or (2). I've chosen equation (2)

$x = 2 \to y = \left(- 2 \times 2\right) + 6 = - 4 + 6 = 2$

$\text{solution is } \left(2 , 2\right)$