The solution to a linear inequality is an interval, which is also true or systems of inequalities, but the solution is the interval which is common to both inequalities. The procedure is the following: solve both inequalities and then searh for their intersection, which is the solution to the system of inequalities.

Example 1. Solve the system of linear inequalities.

$ 2x + 1 \ge 3$

$ 5x – 2 > 6x – 6$

Solve inequalities individually.

$2x + 1 \ge 3$

$ 2x \ge 2$

$ x \ge 1$

The solution of this inequality is the interval $ [1, +\infty>$, which is represented graphically:

The solution of the second inequality:

$ 5x – 2 > 6x – 6$

$ -x > -4$

$ x < 4$

$<- \infty, 4>$

The solution to the system of the two given inequalities is the intersection of $ [1, +\infty>$ and $<- \infty, 4>$, which is $[1, 4>$.

Example 2. What would happen if we altered this system just a little bit? If we got a system:

$ 2x + 1 \ge 3$

$ 5x – 2 > 6x – 2$

The first inequality is the same, so is its solution: $ [1, + \infty>$.

But the second inequality did change, let’s find its solution.

$ 5x – 2 > 6x – 2$

$ – x > 0$

$ x < 0$

We got interval $<- \infty, 0>$. Let’s draw these two intervals.

From the picture we can conclude that these two intervals have no intersection which means that this systems of inequalities has no solutions.

Example 3. Solve the inequality:

$\frac{4 – x}{x + 2} > 4$.

If this were an equality, we would multiply the expression with $(x + 2)$, however we can’t do this here. When the inequality is multiplied with a negative number the inequality sign is changed. Since $(x + 2)$ is positive for all $x$ strictly greater then 2, and negative for all $x$ strictly less then 2. This is why this problem is considere as the system of inequalities. To solve this inequality, we must solve both of these:

$4 – x > 4(x + 2)$

$4 – x < 4(x + 2)$

1. For $x+2>0$,

$\frac{4 – x}{x + 2} > 4$

$ 4 – x > 4(x + 2)$

$ 4 – x > 4x + 8$

$ – 5x > 4$

$ x < -\frac{4}{5}$.

The condition is $x + 2 > 0$ or $x>-2$. These two inequalities make a systems of inequalities. Let’s draw these solutions on a number line:

The solution of this case is the interval $<-2, -\frac{4}{4}>$.

2. For $ x + 2 < 0$,

$\frac{4 – x}{x + 2} > 4$

$ 4 – x < 4(x + 2)$

$ 4 – x < 4x + 8$

$ – 5x < 4$

$ x > -\frac{4}{5}$.

The condition is $x + 2 < 0$ or $x<-2$.

The solution of this case is an empty set or $\emptyset$.

The solution of our starting inequality is the union of these two intervals:

$<- 2, -\frac{4}{5}>$ $ \cup$ $\emptyset$ $= <- 2, -\frac{4}{5}>$.

## Systems of inequalities worksheets

**Standard form** (5.7 MiB, 421 hits)

**Slope-intercept form** (5.7 MiB, 400 hits)