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Sub Topics
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Inequality literally means "not equal". An inequality is defined as the 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:
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Rule 1: Flip the inequality sign while switching sides
Whenever 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:
$4 < 5 - y$
If we change side, then
$5 - y > 4$
We changed sign of inequality $<$ to $>$.
Rule 2: Addition and Subtraction OperationIf we change side, then
$5 - y > 4$
We changed sign of inequality $<$ to $>$.
Whenever addition or subtraction operation is performed on an inequality equation, we must add or subtract same number on both the sides.
We can explain it by an example:
$a - 2 > 5$
To solve this inequality, we add 2 on both the sides:
$a - 2 + 2 > 5 + 2$
$a > 7$.
Rule 3: Multiplication and Division Operation
Whenever 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.
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.Whenever 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.
Let us assume one example:
$x - 4 > 1$
Adding 4 on both the sides, we get
$x - 4 + 4 > 1 + 4$
$x > 5$
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, we solve the bracket:
$-30y + 42 > 5y + 25$
Subtract 5y on both sides:
$-30y + 42 - 5y > 5y + 25 - 5y$
$-35y + 42 > 25$
Subtract 42 on both sides:
$-35y + 42 - 42 > 25 - 42$
$-35y > -17$
Dividing by -35
$y < 0.49$
Compound inequality is a combination of two inequalities which are joined together by logical operators and or or.
Compound Inequality with "and":
It represents "intersection" i.e. for the compound inequality with "and" to be true, both the inequalities must be true separately.
Compound Inequality with "or":
It represents "union" i.e. for the compound inequality with "or" to be true, at least one of the two inequalities must be true.
Solved Examples
Question 1: Solve $3x - 5 > 10 \text{or} -2x - 1 > 9$
Solution:
Solution:
We can solve both the equations simultaneously with "or" operator in between.
$3x - 5 + 5 > 10 + 5 \text{or} -2x - 1 + 1 > 9 + 1$
$3x > 15 \text{or} - 2x > 10$
$x > 5 \text{or} x < -5$
This compound inequality looks at number line as below:
This can also be written in interval notation as:
$(-\infty ,-5)\bigcup (5,\infty )$
$3x - 5 + 5 > 10 + 5 \text{or} -2x - 1 + 1 > 9 + 1$
$3x > 15 \text{or} - 2x > 10$
$x > 5 \text{or} x < -5$
This compound inequality looks at number line as below:
This can also be written in interval notation as:
$(-\infty ,-5)\bigcup (5,\infty )$
Question 2: Solve $x + 3 \geq 5$ and $2x - 1 \leq 7$.
Solution:
Solution:
$x + 3 - 3 \geq 5 - 3$ and $2x - 1 + 1 \leq 7 + 1$
$x \geq 2$ and $2x \leq 8$
$x \geq 2$ and $x \leq 4$.
This inequality is shown on number line as below:
This can be written in interval notation as:
$\left [2,4 \right ]$
$x \geq 2$ and $2x \leq 8$
$x \geq 2$ and $x \leq 4$.
This inequality is shown on number line as below:
This can be written in interval notation as:
$\left [2,4 \right ]$
Step 1: Equate polynomial to zero and find all the zeros of it.
Step 2: Now, divide the number line into intervals according to the roots.
Step 3: The values of a polynomial can only be either positive or negative between two roots. So, choose any point among these intervals and evaluate whether the result is greater than zero or less than zero. This process is called sign analysis.
Step 4: Now, determine the intervals which satisfy given inequality.
Step 5: Draw those intervals on number line.
An equation of the form $ax^{2} + bx + c = 0$ is known as quadratic equation. A quadratic equation with "inequality" sign in place of "equals" sign, is referred as quadratic inequality.
The process of solving quadratic inequalities is the same as that of solving polynomial inequalities.
Solved Example
Question: Solve$x^{2} + 2x - 3 \geq 0$
Solution:
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 $[-\infty ,-3] \bigcup [1,\infty ]$.
$x^{2} + 2x - 3 = 0$
$x^{2} + 3x - x - 3 = 0$
$(x - 1)(x + 3) = 0$
$x = 1, x = -3$
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 $[-\infty ,-3] \bigcup [1,\infty ]$.
Solved Examples
Question 1: Solve $x + 3 \geq 0$
Solve the given inequality and determine the value of x.
Solution:
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:
$x + 3 - 3 \geq 0 - 3$
$x \geq - 3$
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 2: $x - 2 < 0$
Solve the given inequity and determine the value of x.
Solution:
Solve the given inequity and determine the value of x.
Solution:
$x - 2 + 2 < 0 + 2$
$x < 2$
It means that, x is strictly less than 2. Therefore, it represents the interval $(-\infty,2)$ which is an open interval.
To graph this interval, we need to draw number line and then, draw an open circle at 2. In this way, we get the following graph:
$x < 2$
It means that, x is strictly less than 2. Therefore, it represents the interval $(-\infty,2)$ which is an open interval.
To graph this interval, we need to draw number line and then, draw an open circle at 2. In this way, we get the following graph:
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
For k > 0
Solved Example
Question: Solve $\sqrt{7x + 5} \geq 2$
Solution:
Solution:
$\left \{\sqrt{7x + 5} \right \}^{2} \geq (2)^{2}$ (squaring both the sides)
$7x + 5 \geq 4$
$7x + 5 - 5 \geq 4 - 5$
$7x \geq -1$
$x \geq - $$\frac{1}{7}$
$7x + 5 \geq 4$
$7x + 5 - 5 \geq 4 - 5$
$7x \geq -1$
$x \geq - $$\frac{1}{7}$
Solved Examples
Question 1: Solve $3x + 8 > 2x - 7$
Solution:
Solution:
$3x + 8 > 2x - 7$
$3x + 8 - 2x > 2x - 7 - 2x$
$x + 8 > -7$
$x + 8 - 8 > -7 - 8$
$x > -15$
$3x + 8 - 2x > 2x - 7 - 2x$
$x + 8 > -7$
$x + 8 - 8 > -7 - 8$
$x > -15$
Question 2: Solve $7t - 5 \leq 8$
Solution:
Solution:
$7t - 5 \leq 8$
$7t - 5 + 5 \leq 8 + 5$
$7t \leq 13$
$t \leq $$\frac{13}{7}$
$7t - 5 + 5 \leq 8 + 5$
$7t \leq 13$
$t \leq $$\frac{13}{7}$