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:

$$ax+b<c$$

$$ax+b>c$$

$$ax+b \leqslant c$$

$$ax+b \geqslant c$$

First step in solving two – step inequalities is the same as the one in one step inequalities. In other words, we first need to isolate the variable on one side of the inequality.

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

Obviously, we have to divide our whole inequality by number $a$. Here is where you have to be careful. Remember that an inequality sign depends on whether number $a$ is negative or positive.

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 see a few examples.

Example 1:

$5x + 4 < 14$

Solution:

$5x + 4 < 14 /+(-4)$
$5x +4-4< 14 + (-4)$
$5x < 10$

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

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

On the number line (again, since we have a ”strictly less than” sign, number $2$ isn’t included in the solution set): Example 2:

$– 7x + 8 ≤ 1$
Solution:

$– 7x + 8 ≤ 1/+(-8)$

$- 7x +8 + (-8)≤ 1 +(-8)$

$- 7x ≤ – 7$

In this case, $x$ is multiplied by $-7$, so we should divide the whole inequality by $-7$. But, $-7$ is a negative number, so the inequality sign will change from $≤$ to $≥$.

$- 7x ≤ -7 /:(-7)$

$x ≥ 1$,

Here we have that $x$ is greater than or equal to $1$. That means that $1$ will be included in the solution set. $$x \in \left[1, +\infty\right>$$

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 sign changes from $≥ to ≤$ )
Since we have ”less than or equal to” sign, the number $1$ is included in the solution set. The solution set on the number line looks like this: In the interval form: $x \in \left< -\infty, 1\right]$.

Example 3:

$7x + 9 ≤ 1$

Solution:

$7x ≤ 1 – 9$

$7x ≤ – 8 / : 7$

$x ≤ -\frac{8}{7}$

$-\frac{8}{7}= -1 \frac{1}{7}$ $x \in \left<-∞ ,-\frac{8}{7}\right]$

Example 4:

$$\frac{1}{5}x + 2 ≥ \frac{11}{5}$$

Solution:

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

$x + 10 ≥ 11$

$x ≥ 11 – 10$

$x ≥ 1$ $$x \in \left[1, \infty\right>$$

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