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# Solving Inequalities

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 Sub Topics The word "inequality" literally means "not equal". An inequality is defined as an expression that instead of having equality sign, contains less than or greater than signs. The objective of solving inequality is to find the possible values of unknown variables. There are basically two types of inequalities:1) Strict Inequalities: An equation with "less than" or "greater than" symbol is known as strict inequalities. For example: a > 5 which means that "a" is strictly greater than 5 and it cannot be less than or equal to 5. Similarly, b < 0 means that "b" is strictly less than 0 and it cannot be greater than or equal to 0.\2) Not-strict Inequalities: An equation with "less than or equal to" or "greater than or equal to" symbol is known as not-strict inequalities. For Example: x $\geq$ 10 means that "x" can at least be 10. i.e. either 10 or greater than 10. Similarly, y $\leq$ 8 means that "y" can at most be 8 i.e. either 8 or less than 8.Few more examples of inequalities are given below:a - 2 < 53s - 7 > 182(x - 1) $\geq$ 3x4x + 1 $\leq$ 2(x - 3)There are some rules to be followed while solving inequalities. These rules are as follows:Rule 1: Flip the inequality sign while switching sidesWhenever we switch the sides of inequality equation, we replace > by < sign and vice versa. Similarly, replace $\geq$ by $\leq$ and vice versa.  This rule can be explained as under:4 < 5 - yIf we change side, then5 - y > 4we changed sign of inequality  from < to >.Rule 2: Addition and Subtraction OperationWhenever addition or subtraction operation is performed on an inequality equation, we must add or subtract same number on both the sides. Let understand this by the help of an example:a - 2 > 5To solve this inequality, we add 2 on both the sides:a - 2 + 2 > 5 + 2a > 7.Rule 3: Multiplication and Division OperationWhenever multiplication or division is performed, it must be done on both the sides. If we multiply or divide by a positive number on both the sides, the inequality sign remains the same. On the other hand, if we multiply or divide by negative number on both the sides, the inequality signs flips.

## Solving Linear Inequalities

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Linear inequalities are similar to linear equations with "less than ($<$)", "greater than ($>$)", "less than or equal to ($\leq$)", or "greater than or equal to ($\geq$)" sign. Linear inequalities are solved in the same manner as linear equations are solved. But, the rules of solving inequalities should be kept in mind and followed while solving linear inequalities.

Let us have a look at one example:

x - 4 > 1

Adding 4 on both the sides, we get

x - 4 + 4 > 1 + 4

x > 5

## Solving Multi-Step Inequalities

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Multi-step inequalities are those which involve various steps. While solving them, we need to follow the same rules of solving inequalities, as well as the order of operation. Let us take an example:

-6(5y - 7) > 5y + 25

First, open the bracket

-30y + 42 > 5y + 25

Subtract 5y from both sides in order to eliminate it from right hand side

-30y + 42 - 5y > 5y + 25 - 5y

-35y + 42 > 25

Subtract 42 on both sides, by which it is eliminated from left hand side

-35y + 42 - 42 > 25 - 42

-35y > -17

Dividing by -35

y < 0.49

## Solving Rational Inequalities

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In order to solve rational inequalities, the following steps are to be adopted.

Step 1: Make sure the inequality is in a form that right side has only zero, while left side has fraction or rational form.
Let us understand this method by the help of an example :

$\frac{t+3}{t^{2}- 5t + 4}$ $\geq 0$

Step
2: Set numerator as well as denominator equal to zero and solve. In this way, we obtain critical values.

t + 3 = 0

t = -3

and

t$^{2}$ - 5t + 4 = 0

(t - 1)(t - 4) = 0

t = 1, 4

Step 3: Now, we have to prepare a sign analysis. In order to do so, create section of these critical values. Pick a number between each section and plug this into polynomial to find the sign of the result.

We have the sections .... -3 ... 1 ... 4...

Plug t = -4 (any value between $- \infty$ and -3) in $\frac{t+3}{t^{2}- 5t + 4}$

$\frac{-4+3}{(-4)^{2}- 5(-4) + 4}$ = $\frac{-1}{16+20 + 4}$ = -ve

Similarly by substituting t = -1, (value between -3 and 1), we get

$\frac{-1+3}{(-1)^{2}- 5(-1) + 4}$ = $\frac{2}{1+6 + 4}$ = +ve

Plugging t = 2, we get

$\frac{2+3}{2^{2}- 5(2) + 4}$ = $\frac{5}{4-10 + 4}$ = -ve

Putting t = 5, we obtain

$\frac{5+3}{5^{2}- 5(5) + 4}$ = $\frac{8}{25-25 + 4}$ = +ve

Step 4: Choose the intervals that satisfy given inequality in the question.

Here, inequality is $\geq$, so we shall choose the intervals which give +ve values.

t = -3 to 1 and t = 4 to infinity

Also, 1 and 4 cannot be included since from question, it is implied that t $\neq$ 1, 4.

Thus, the answer will the following interval -

$[-3, 1) \cup (4, \infty)$

## Solving Absolute Value Inequalities

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An equation of the form $\left | x \right | = k$ is called an absolute value equation. An absolute value inequality has inequality sign in place of "equals" sign.

There are two types of patterns in solving absolute value inequalities.

Absolute value inequality with $< or \leq$ sign: An absolute value inequality with $< or \leq$ sign is to solved by using the following pattern:

For k > 0

Absolute value inequality with $> or \geq$ sign: An absolute value inequality with $> or \geq$ sign is to solved by using the following pattern:

For k > 0

## Examples

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Few examples based on inequalities are discussed below:

### Solved Examples

Question 1: Solve$x^{2} + 2x - 3 \geq 0$
Solution:
Step 1: Find the roots of quadratic equation by equating it to zero.
$x^{2} + 2x - 3 = 0$
$x^{2} + 3x - x - 3 = 0$
$(x - 1)(x + 3) = 0$
$x = 1, x = -3$

Step 2: Now, the number line is divided into three intervals based on the roots as shown below:
$[-\infty ,-3], [-3,1], [1,\infty]$

Step 3:
Choose any point among these intervals and determine whether the result is greater than zero or less than zero. This process is called sign analysis.
Here, we substitute x = - 4, x = 0 and x = 2 in $f(x) = x^{2} + 2x - 3$.
$f(-4) = (-4)^{2} + 2(-4) - 3$
$= 5 > 0$
$f(0) = (0)^{2} + 2(0) - 3$
$= -3 < 0$
$f(2) = (2)^{2} + 2(2) - 3$
$= 5 > 0$

Step 4:
Now, find the intervals which satisfies the given inequality.
In example, inequality is "greater than or equal to" which is satisfied by the intervals $(-\infty ,-3)$ and $(1,\infty )$
The following graph shows above interval on number line:

We may write this in interval notation as $[-\infty ,-3] \bigcup [1,\infty ]$.

Question 2: Solve $x + 3 \geq 0$
Solve the given inequality and determine the value of x.
Solution:
$x + 3 \geq 0$
$x + 3 - 3 \geq 0 - 3$
$x \geq - 3$
It means that x is either -3 or greater than -3. Therefore, it represents the following interval:
$[-3,\infty ]$ which is left closed interval.
To graph this interval, we need to draw number line and then, draw a closed circle at -3. In this way, we get the following graph:

Question 3: Solve $3x + 8 > 2x - 7$
Solution:
$3x + 8 > 2x - 7$
$3x + 8 - 2x > 2x - 7 - 2x$
$x + 8 > -7$
$x + 8 - 8 > -7 - 8$
$x > -15$

Question 4: Solve $7t - 5 \leq 8$
Solution:
$7t - 5 \leq 8$
$7t - 5 + 5 \leq 8 + 5$
$7t \leq 13$
$t \leq$$\frac{13}{7}$