# How do you solve the system of equations x+y=3, x-y=1 by graphing?

Mar 27, 2018

Point where the lines cross is: $\left(x , y\right) = \left(2 , 1\right)$

#### Explanation:

$\textcolor{b l u e}{\text{Consider: } x + y = 3}$

Subtract $x$ from both sides

$+ x + y = 3$
$\underline{- x \textcolor{w h i t e}{\text{dddddddd}} - x}$
$\textcolor{w h i t e}{\text{d.}} 0 + y = 3 - x$

$\textcolor{w h i t e}{\text{dddddddd}} \textcolor{red}{\underline{\overline{| \textcolor{w h i t e}{\frac{2}{2}} y = - x + 3 \textcolor{w h i t e}{\frac{2}{2}} |}}}$

The line crosses the x-axis at $y = 0 \implies 0 = - x + 3 \implies x = 3$
So we have ${x}_{\text{intercept}} \to \left(x , y\right) = \left(3 , 0\right)$

The line crosses the y-axis at $x = 0 \implies y = 0 + 3$
So we have ${y}_{\text{intercept}} \to \left(x , y\right) = \left(0 , 3\right)$
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$\textcolor{b l u e}{\text{Consider: } x - y = 1}$

Add $y$ to both sides
Subtract 1 from both sides

$x - y \textcolor{w h i t e}{\text{dddd")=color(white)("dddd}} + 1$
$\underline{\textcolor{w h i t e}{x} + y - 1 \textcolor{w h i t e}{\text{dddd}} + y - 1}$
$x + 0 - 1 \textcolor{w h i t e}{\text{d") =color(white)("ddd}} y + 0$

$\textcolor{w h i t e}{\text{dddddddd}} \textcolor{red}{\underline{\overline{| \textcolor{w h i t e}{\frac{2}{2}} y = x - 1 \textcolor{w h i t e}{\frac{2}{2}} |}}}$

The line crosses the x-axis at $y = 0 \implies 0 = x - 1 \implies x = 1$
So we have ${x}_{\text{intercept}} \to \left(x , y\right) = \left(1 , 0\right)$

The line crosses the y-axis at $x = 0 \implies y = 0 - 1$
So we have ${y}_{\text{intercept}} \to \left(x , y\right) = \left(0 , - 1\right)$
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