Solving Inequalities by using different Operations using Multiplication and Division
Inequalities in mathematics are used mainly to make a comparison of numbers on a number line. There are certain symbols that are used to represent an inequality. Greater than, >, less than, <, greater than and equal to, ≥, and, lesser than and equal to, ≤, are the symbols used to depict an inequality. We will look at the multiplication and the division property of inequalities and then observe how both these operations affect both sides of the inequality.
Remember that multiplying or dividing both sides of an inequality with a negative integer, will result in the inequality sign being altered.
Multiplication Property of Inequality
Let’s take three numbers, x, y and z, where z is a positive number, so z > 0.
If x < y, and z > 0 then x × z < y × z
Example:
Suppose 2 < 5
Multiplying both sides by 10
2 × 10 < 5 × 10 ( Notice that z = 10 and 10 > 0)
20 < 50
Let’s take three numbers again x, y and z, where z is a negative number so z < 0.
If x < y and z < 0 then x × z > y × z
Example:
Suppose 2 < 5
Multiplying both sides by -4
2 × -4 > 5 × -4
-8 > -20 (z = -4 and -4 < 0)
Division Property of Inequality
Let’s take three numbers x, y and z, where z > 0
If x < y, and z > 0 then x ÷ z < y ÷ z
Example:
Suppose 2 < 4
Dividing both sides by 2
then 2 ÷ 2 < 4 ÷ 2
1 < 2
Let’s take three numbers x, y and z, where z is a negative number so z < 0.
If x < y, and z < 0 then x ÷ z > y ÷ z
Example:
Suppose 4 < 8
Dividing both sides by -2
then 4 ÷ -2 > 8 ÷ -2
-2 > -4
Let’s consider the following inequality,
\( 8x\geq 16 \)
To simplify the inequality, both sides should be divided by 8.
\( \frac{8x}{8}\geq \frac{16}{8} \)
\( x\geq 2 \)
The following set of representations shows us how this inequality can be represented on a number line.
In case the equation was \( x>2 \) instead of \( x\geq 2 \).
When the number is included in the inequality the graph is represented by a closed circle on that number, on the other hand, when the number is not included then it is represented by an open circle.
Solved Examples
Example 1:
A man digs the ground at the rate of 39 meters per hour with the help of a driller in search of a hidden treasure. According to the information he received, the treasure is greater than 3900 meters below the surface. How long would the man take to drill a hole deep enough to reach the treasure?
Solution:
Let t be the time taken, in hours, for the driller to reach the treasure.
To represent the descent and drilling below ground level we take the numbers as negative.
We know that in case of negative numbers, when the absolute value is greater than the absolute value of the other number, then the number is smaller than the other number.
For instance, \( -4< -2 \) but \( \left | -4 \right |=4,~\left | -2 \right |=2 \) and \( 4> 2 \).
Similarly, the distance to be covered has to be greater than 3900 meters.
\( -39.t< -3900 \)
Applying the division property of inequality, we get
\( \frac{-39.t}{-39}> \frac{-3900}{-39} \)
\( t> 100 \)
Therefore the driller will take more than 100 hours to reach the treasure.
Example 2:
Show how 5x < -20 is different from -5x < 20
Solution:
Let’s solve both the inequalities first.
5x < -20
We use the division property of inequalities and divide both sides by 5.
\( \frac{5x}{5}< -\frac{20}{5} \)
This gives us
\( x< -4 \)
This is the result we get from this inequality.
Let’s take a look at the next one.
\( -5x< 20 \)
We will divide both sides by 5 and this gives us the following.
\( -\frac{5x}{5}< -\frac{20}{5} \)
\( x> -4 \)
If we check the final form of both the inequalities, this is what we get:
\( x< -4 \)
\( x> -4 \)
It can be observed that in the first inequality the value of “x” is less than -4 and in the other case the value of “x” is greater than -4.
Example 3:
Solve the following inequality and plot a graph for the same
9r > 81
Solution:
Dividing 9 from both sides we get
\( \frac{9r}{9}=\frac{81}{9} \)
\( r> 9 \)
The graph for this expression is as follows.
Example 4:
Solve the following inequality and plot a graph for the same
65 – 965g
Solution:
Dividing by – 965 on both sides,
\( -\frac{65}{965}\geq \frac{965}{965}g \)
\( -\frac{13}{193}\geq g \)
\( g\leq -\frac{13}{193} \)
\( g\leq -0.067 \)
Plotting this on a graph, we get the following.
Example 5:
The boundary for a triangular field needs to be constructed having an area greater than 80 square feet. The base length of the triangle is 40 meters. What is the probable height of the triangle?
Solution:
The formula for the area of a triangle is \( \frac{1}{2}bh \). The given area is greater than 80, based on the circumstances given in the question, the inequality is as shown below.
\( 80> \frac{1}{2}\times 40\times h \)
Simplifying the equation further we get,
\( 80> 20h \)
Dividing both sides by 20 we get,
\( h< 40 \)
So, the length of the side cannot be greater than 40.
Example 6:
Working as a part-time tutor, Teresa earns $8.40 per hour. She wants to buy a tab that costs at least $672. What is the minimum number of hours Teresa has to work to earn the amount shown here. Also plot a graph representing the resulting inequality.
Solution:
Let’s assume the minimum time required for Teresa is t.
The equation representing the situation is as follows:
\( 8.4\times t\geq 672 \)
Dividing both sides by 6.4 we get the following equation.
\( \frac{8.4}{8.4}\times t\geq \frac{672}{8.4} \)
\( t\geq 80 \)
The graph for the equality is as follows
So, Teresa has to work for a minimum of 80 hours to attain this amount.
The multiplication property of inequality states that when the same positive number is multiplied on both sides of the inequality the inequality remains the same. But when multiplied by a negative number the inequality reverses.
If one side of the inequality is divided by a positive number, the other side of the inequality can be divided by the same positive number and the inequality remains the same. But when divided by a negative number the inequality reverses.
Let’s understand this by taking 3 numbers, x, y and z.
If x > y and y > z, then x > z
Example: Suppose 10 > 5 and 5 > 2, then 10 > 2
x < y and y < z, then x < z
5 < 10 and 10 < 20, then 5 < 20