Question

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

Answer 1

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 + 3y > 8`.

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

`x + 3y > 8`

Subtract `x` from both sides:

`x + 3y - x > 8 - x`

`3y > -x + 8`

Then divide by `3`:

`(3y)/3 > (-x + 8)/3`

`y > -1/3 x + 8/3`

Let's graph the (dashed!) line `y = -1/3 x + 8/3`:

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

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