# How do you graph the system of linear inequalities -x<y and x+3y>8?

##### 1 Answer
Dec 29, 2017

Graph the inequalities as if they are equations (but let the lines be dashed if the inequality is merely "less than" or "more than"), then shade accordingly.

#### Explanation:

Our first inequality is $- x < y$, or $y > - x$.

To start, graph $y = - x$, with a dashed line:

I chose to draw a dashed line here because the inequality only says "$>$", not "≥", meaning that the equality itself is not contained; the dashed line indicates that line is not part of the later shaded area.

Then what would $y > - x$ be? Well, that inequality tells us that $y$ must be more than whatever the value of $- x$ happens to be. Since $y = - x$ has been graphed as a line, $y > - x$ would be anything above the line:

Notice how I left the line dashed.

The second inequality is $x + 3 y > 8$.

Let's rearrange that to put it in the slope-intercept form of linear equations, $y = m x + b$.

$x + 3 y > 8$

Subtract $x$ from both sides:

$x + 3 y - x > 8 - x$

$3 y > - x + 8$

Then divide by $3$:

$\frac{3 y}{3} > \frac{- x + 8}{3}$

$y > - \frac{1}{3} x + \frac{8}{3}$

Let's graph the (dashed!) line $y = - \frac{1}{3} x + \frac{8}{3}$:

Again, the inequality asks us to shade everything above the line.

The purple area is the shaded area that both inequalities share.