# Linear Inequalities in Two Variables

## Key Questions

• The solution set of a single linear inequality is always a half-plane, so there are infinitely many solutions.

I hope that this was helpful.

The direction of the inequality will tell you this information.

#### Explanation:

Example: $y \ge 2 x + 3$

You would draw the line $y = 2 x + 3$ and shade above the line, since $y$ is also greater than $2 x + 3$.

graph{y>=2x+3 [-10, 10, -5, 5]}

Example: $y < \frac{1}{2} x - 2$

You would draw the line $y = \frac{1}{2} x - 2$ as a dashed line, then shade below the line since $y$ is less than $\frac{1}{2} x - 2$.

graph{y<1/2x-2 [-10, 10, -5, 5]}

• The best way (when possible!) is to express the inequality bringing all the terms involing a variable on the left, and all the terms involving the other variable on the right. In the end, you'll have an inequality of the form $y \setminus \le f \left(x\right)$ (or $y \setminus \ge f \left(x\right)$), and this is easy to graph, because if you can draw the graph of $f \left(x\right)$, then you'll have that $y \setminus \le f \left(x\right)$ represents all the area under the function $f$, and $y \setminus \ge f \left(x\right)$, of course, the area over the function.

For example, consider the inequality
$y {x}^{2} < - y + 3 x$
In this form, it would be very hard to say which points satisfy the inequality, but with some manipulations we obtain
$y {x}^{2} + y < 3 x$
$y \left({x}^{2} + 1\right) < 3 x$
$y < \setminus \frac{3 x}{{x}^{2} + 1}$

Now, the graph of $\setminus \frac{3 x}{{x}^{2} + 1}$ is easy to draw, and the inequality is solved considering all the area below the graph, as showed:

graph{y<3x/{x^2+1} [-10, 10, -5, 5]}

Note that if you have $y < f \left(x\right)$ the graph of the function $f$ is not included, since it represents the points for which $y = f \left(x\right)$.