# Two-step inequalities

Two-step inequalities are called like that because we need to perform two steps to find the solution. They come in form:

$x+b<c$, $x+b \geqslant c$,  $x+b>c$,  $x+b \leqslant c$

First step in solving them is the same as the one in one step inequalities.

Since the variable $x$ is the value we are looking for, there is still something in the way. a multiplies $x$, so we have to deal with that before we can have our $x$.

Here is where you have to be careful. We have to divide our whole inequality with that a, but an inequality sign depends on whether it is negative or positive.

As we already said, a single division was required to solve this inequality, but this example required us to remember a very important information: when the variable changes signs, the inequality sign changes to its opposite as well!

for $a>0$

$ax > c-b /:a$

$x>\frac{c-b}{a}$

______________

for $a<0$

$ax > c-b /:(-a)$

$x<\frac{c-b}{-a}$

$x<\frac{b-c}{a}$

Let’s try it on a few examples:

$5x + 4 < – 6 /+(-4)$
$5x +4-4< – 6 – 4$
$5x < – 6 – 4$

$5x < -10 /:5$ ($5 > 0$, so the inequality sign remains the same)

$x < -2$ ,
$x \in <-\infty, -2>$

On the number line (again, since it is strictly lesser, number two isn’t included in the solution set):

#### Example 2:

$– 7x + 8 ≤ 1$

$- 7x ≤ 1 – 8$

$- 7x ≤ – 7$

In this case x is multiplied by (-7) so we should divide it by (-7), but it is a negative number, so the inequality will change from $≤$ to $≥$.

$- 7x ≤ -7 /:(-7)$

$x ≥ 1$,

$x \in [1, +\infty>$

Here we have that $x$ is greater or equal than $1$. That means that $x$ will be included in the solution set.

It would have been similar if the task was set like this:

$– 7x + 8 ≥ 1$

$– 7x ≥ 1 – 8$

$– 7x ≥ -7 /:(-7)$ (we divide by a negative number
$x ≤ 1$ so the inequality changes from $≥ to ≤$ )
Since it is less than or equal to, the number $1$ is included in the solution set, which makes the solution on the number line look like this:

And in the form of an interval: $x \in < -\infty, 1 ]$.

Example 3:

$7x + 9 ≤ 1$

$7x ≤ 1 – 9$

$7x ≤ – 8 / : 7$

$x ≤ -\frac{8}{7}$

$-\frac{8}{7}= -1 \frac{1}{7}$

$x \in <-∞ ,-\frac{8}{7}]$

Example 4:

$0.2 = \frac{2}{10} = \frac{1}{5}$ -> $0.2x + 2 ≥ 11 / * 5$

$\frac{1}{5}x + 2 ≥ \frac{11}{5} / \cdot 5$

$x + 10 ≥ 11$

$x ≥ 11 – 10$

$x ≥ 1$

## Two-step inequalities worksheets

Solve integer inequalities (409.0 KiB, 1,001 hits)

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