Solving Inequalities

Solving inequalities means finding the unknown value of its variable. It is done by keeping the variable on the left and the value on the right side of the inequality sign (‘<,’ ‘>,’ ‘≤,’ and ‘≥’).  

For example, x > 11 means the value of ‘x’ is more than 11, and x < -9 includes all values ‘x’ less than -9. 

Using the Rules

Like solving an equation, solving an inequality involves using inverse operations to keep the variable on one side of the inequality sign and constant on the other. However, the most important part of solving an inequality is to keep an eye on the direction of the inequality. 

By Adding or Subtracting

Adding or subtracting a number on or from both sides of the inequality does not change the direction of the inequality. 

Let us solve the linear inequality x – 7 > 5

Now, on adding 7 to both sides, isolate the variable ‘x’ on the left side.

x – 7 + 7 > 5 + 7

⇒ x > 12

Thus, the solution of the inequality x – 7 > 5 is x > 12

On graphing the solution on the number line, we get

Now, let us solve the linear inequality x + 7 > 5

Here, we subtract 7 from both sides to keep the variable ‘x’ on the left side.

x + 7 – 7 > 5 – 7

⇒ x > -2

Thus, the solution of the inequality x + 7 > 5 is x > -2

On graphing the solution on the number line, we get

By Multiplying or Dividing

Similar to adding and subtracting, multiplying or dividing an inequality by the same positive number also does not change the direction of the inequality. 

Let us find the solution of ${\dfrac{x}{12} <5}$

On multiplying by 12, 

${\dfrac{12x}{12} <12\times 5}$

⇒ x < 60, which is the solution.

On graphing the solution on the number line, we get

Now, if we solve another inequality ${12x\leq 60}$

On dividing both sides of the inequality sign by 12, w

${\dfrac{12x}{12}\leq \dfrac{60}{12}}$

⇒ ${x\leq 5}$

Thus, the solution of ${12x\leq 60}$ is ${x\leq 5}$

On graphing the solution on the number line, we get

Multiplying or Dividing by a Negative Number

When multiplying or dividing by a negative number, the direction of inequality is reversed.

For example, we find the solution of ${\dfrac{-x}{17}\leq 4}$

Here, we multiply both sides of the inequality sign by -17, and this implies

${\dfrac{\left( -x\right) \times \left( -17\right) }{17}\geq 4\times \left( -17\right)}$

⇒ ${x\geq -68}$

On graphing the solution on the number line, we get

Now, let us find the solution of -24x < 72

Here, we divide both sides by the negative quantity -24, and the inequality symbol will be reversed.

We get, ${\dfrac{-24x}{-24} >\dfrac{72}{-24}}$ 

⇒ x > -3, the solution.

On graphing the solution on the number line, we get

Multiplying or Dividing by a Variable

Sometimes, to solve the inequality, we must multiply or divide both sides of the inequality sign by a variable. 

For example,

${\dfrac{3}{x} >1}$

Here, by multiplying both sides with the variable ‘x,’ 

${x\cdot \dfrac{3}{x} >1\cdot x}$

⇒ 3 > x

⇒ x < 3, which is the solution.

On graphing the solution on the number line, 

Now, let us solve the polynomial inequality xy > 7x

Here, we divide both sides of the inequality sign by the variable ’x.’

If x > 0, the inequality sign does not change, and the solution yields

${\dfrac{xy}{x} >\dfrac{7x}{x}}$

⇒ ${y >7}$, and its graph on the number line is plotted below.

If x < 0, the inequality sign gets reversed, and the solution yields

${\dfrac{xy}{x} <\dfrac{7x}{x}}$

⇒ ${y <7}$, and its graph is as shown.

Thus, we get two solutions depending on the nature of the variable.

These inequalities typically involve a single variable and can be solved by isolating the variable using a single mathematical operation (known as one-step inequalities).

However, not all inequalities can be solved using single-step or multi-step inequalities, where one or more than one step is required to solve. 

Let us solve the inequality 5x + 3 < 13

On subtracting 3 from both sides of the inequality, 

5x + 3 – 3 < 13 – 3

⇒ 5x < 10

On dividing by 5, 

${\dfrac{5x}{5} <\dfrac{10}{5}}$

⇒ x < 2, which is solved using two steps.

Even some inequalities require more than 2 steps to get to the solution.

Compound Inequalities

It includes solving inequalities with two or more inequality symbols.

Let us solve the compound inequality ${-3 <\dfrac{8x-12}{2} <6}$

On multiplying by 2, 

${-6 <8x-12 <12}$

On adding by 12, 

${6 <8x <24}$

On dividing by 8, 

${\dfrac{6}{8} <x <\dfrac{24}{8}}$

⇒ ${\dfrac{3}{4} <x <3}$

Thus, the solution is ${\dfrac{3}{4} <x <3}$

On graphing the solution on the number line, we get

Absolute Value Inequalities

It involves solving inequalities with the modulus or the absolute value.

Let us solve the inequality |x + 2| > 4

By splitting the inequality into separate inequalities, 

-(x + 2) > 4 or (x + 2) > 4

Now, by reversing the negative sign in the first inequality, 

(x + 2) < -4 or (x + 2) > 4

On subtracting 2 from both sides of the inequalities, 

x + 2 – 2 < -4 – 2 or x + 2 – 2 > 4 – 2

⇒ x < -6 or x > 2

Thus, the solution is x < -6 or x > 2

On graphing the solution on the number line, we get

Quadratic Inequalities

Simplifying a second-order inequality involves the same steps as solving a quadratic equation where the middle term is factored to find the two probable values of the variable.  

Let us simplify the quadratic inequality x² – 11x + 28 < 0

Factoring the left side, we get

x² – 4x – 7x + 28 < 0

⇒ x(x – 4) – 7(x – 4) < 0

⇒ (x – 4)(x – 7) < 0

Now, determining the sign of the expression for distinct intervals, we get

If (x – 4) < 0, (x – 7) < 0 ⇒ x < 4, x > 7

At x < 4, the original inequality gives a positive value.

At x > 7, the original inequality gives a positive value.

Now, if (x – 4) > 0, (x – 7) < 0 ⇒ x > 4, x < 7

At x > 4, the original inequality holds true.

At x < 7, the original inequality holds true.

Thus, the solution is 4 < x < 7

On graphing the solution on the number line, we get

Rational Inequalities

Let us solve the inequality with a fraction (also known as rational inequality) ${\dfrac{x+3}{x-5} >0}$

Equating the numerator and denominator to zero, we get the critical points as

x + 3 = 0 and x – 5 = 0 ⇒ x = -3 and x = 5

Now, at x < -3, both the numerator and denominator are negative; thus, the expression yields a positive value.

At -3 < x < 5, the numerator is positive, while the denominator is negative; thus, the expression gives a negative value.

At x > 5, the numerator and denominator are both positive, which means the expression yields a positive value.

Hence, the solution is x < -3 or x > 5

On graphing the solution on the number line, we get