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

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

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