SOLVING INEQUALITIES BY MULTIPLYING OR DIVIDING

Remember, solving inequalities is similar to solving equations. To solve an inequality that contains multiplication or division, undo the operation by dividing or multiplying both sides of the inequality by the same number.

The rules below show the properties of inequality for multiplying or dividing by a positive number.

Multiplication by Positive Numbers

Words : 

Both sides of an inequality can be multiplied by the same positive number, and the statement will still be true.

Numbers : 

5 < 8

5(3) < 8(3)

15 < 24

Algebra : 

If x < y and z > 0, then, 

xz < yz

Division by Positive Numbers

Words : 

Both sides of an inequality can be multiplied by the same positive number, and the statement will still be true.

Numbers : 

20 < 45

20/5 < 45/5

4 < 9

Algebra : 

If x < y and z > 0, then, 

x/z < y/z

Note : 

The above explained properties are also true for inequalities that use the symbols >, ≥, and ≤.

Multiplying or Dividing by a Positive Number

Example 1 : 

Solve the following inequality and graph the solution. 

3x < 18

Solution : 

3x < 18

Because x is multiplied by 3, divide each side by 3 to undo the multiplication.

3x/3 < 18/3

x < 6

Example 2 : 

Solve the following inequality and graph the solution. 

y/2 > -1.5

Solution : 

y/2 > -1.5

Because y is divided by 2, multiply each side by 2 to undo division.

2(y/2) > 2(-1.5)

y > -3

Multiplication by Negative Numbers

Words : 

If each side of an inequality is multiplied by the same negative number, you have to reverse the inequality symbol for the statement to still be true.

Numbers : 

8 > 5

8(-3) < 5(-3)

-24 < -15

Algebra : 

If x > y and z < 0, then, 

xz < yz

Division by Negative Numbers

Words : 

If each side of an inequality is divided by the same negative number, you have to reverse the inequality symbol for the statement to still be true.

Numbers : 

45 > 20

45/(-5) < 20/(-5)

-9 < -4

Algebra : 

If x > y and z < 0, then, 

x/z < y/z

Note : 

The above explained properties are also true for inequalities that use the symbols >, ≥, and ≤.

Multiplying or Dividing by a Negative Number

Example 3 : 

Solve the following inequality and graph the solution. 

-10a > 15

Solution : 

-10a > 15

Because a is multiplied by -10, divide each side by -10 and change > to <. 

-10a/(-10) < 15/(-10)

a < -1.5

Example 4 : 

Solve the following inequality and graph the solution. 

-b/10 ≥ -0.11

Solution : 

-b/10 ≥ -0.11

Because -b is divided by 10, multiply each side by -10 and change ≥ to ≤. 

(-b/10)(-10) ≤ -0.11(-10)

10b/10 ≤ 1.1

b ≤ 1.1

Consumer Application

Example 5 :

James has a $16 gift card for a health store where a smoothie costs $2.50 with tax. What are the possible numbers of smoothies that James can buy?

Solution : 

Let y represent the number of smoothies James can buy.

2.50y ≤ 16

Because y is multiplied by 2.50, divide each side by 2.50. The symbol does not change.

y ≤ 6.4 

James can buy only a whole number of smoothies.

So, he can buy 0, 1, 2, 3, 4, 5 or 6 smoothies.

Example 5 :

Solve the inequality and graph the solution

24 + 6k < -6(-4 - k)

Solution :

24 + 6k < -6(-4 - k)

Distributing -6, we get

24 + 6k < 24 + 6k

Subtracting 24 and subtracting 6k, we get

24 - 24 < 6k - 6k

0 < 0

So, there is no solution.

Example 6 :

Solve the inequality and graph the solution

-4(4 + 7x) + x ≥ -6x + 5

Solution :

-4(4 + 7x) + x ≥ -6x + 5

Distributing -4, we get

-16 - 28x + x ≥ -6x + 5

-16 - 27x ≥ -6x + 5

Adding 6x on both sides

-27x + 6x - 16 ≥ 5

Adding 16 on both sides

-21x ≥ 5 + 16

-21x ≥ 21

When we divide by negative sign on both sides of the inequality, we get

x ≤ -1

Example 7 :

Tom is deciding whether or not he should become a member gym to use their basketball courts. The membership cost is $135. Members pay $2 to rent out the basketball courts. Non-members can rent the court also, but they have to pay $11 each time. how many times would Tom need to rent the court in order for it be cheaper to be a member than a non member?

Solution :

Membership cost = $135

Rent per hour for members = $2

Rent for non members = $11

Let x be the number of hours renting.

Rent paid by members for renting x hours = 135 + 2x

Rent paid by non members for renting x hours = 11x

135 + 2x < 11x

To solve for x, we have to subtract 2x on both sides.

135 < 11x - 2x

135 < 9x

Dividing by 9 on both sides

135/9 < x

15 < x

At least 15 times.

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