# How do you solve the system y = 2x - 3 and y = -x + 3?

Apr 3, 2017

You may set the two $y$'s equal to each other:

#### Explanation:

$2 x - 3 = - x + 3$. Now add $3$ to both sides:

$2 x - \cancel{3} + \cancel{3} = - x + 3 + 3$. Now add $x$:

$2 x + x = \cancel{x} - \cancel{x} + 6 \to 3 x = 6 \to x = 2$

Plug this into both equations (one for check!):
$y = 2 \cdot 2 - 3 = 1$
$y = - 2 + 3 = 1$
(they agree).

Apr 3, 2017

$\left(2 , 1\right)$

#### Explanation:

Label the equations.

$\textcolor{red}{y} = 2 x - 3 \to \left(1\right)$

$\textcolor{red}{y} = - x + 3 \to \left(2\right)$

Since both equations give $\textcolor{red}{y}$ expressed in terms of x we can equate the right sides.

$\Rightarrow 2 x - 3 = - x + 3$

$2 x + x - 3 = \cancel{- x} \cancel{+ x} + 3$

$\Rightarrow 3 x - 3 = 3$

$3 x \cancel{- 3} \cancel{+ 3} = 3 + 3$

$\Rightarrow 3 x = 6$

divide both sides by 3

$\frac{\cancel{3} x}{\cancel{3}} = \frac{6}{3}$

$\Rightarrow x = 2$

Substitute this value into either ( 1 ) or ( 2 ) and evaluate for y

$\text{Substituting in } \left(1\right)$

$x = 2 \to y = \left(2 \times 2\right) - 3 = 4 - 3 = 1$

$\textcolor{b l u e}{\text{As a check}}$

$\text{Substituting in } \left(2\right)$

$x = 2 \to y = - 2 + 3 = 1 \to \text{ true}$

$\Rightarrow \text{ the point of intersection } = \left(2 , 1\right)$
graph{(y-2x+3)(y+x-3)=0 [-10, 10, -5, 5]}