Word Problem And Solving Inequalities JSS2 Mathematics Lesson Note

 SOLVING INEQUALITIES

RULES IN SOLVING INEQUALITIES.

1. An inequality remains true when the same quantity is added to, or subtracted from both sides.
2. An inequality remains true when both sides are multiplied or divided by the same positive quantity.
3. An inequality remains true when both sides are multiplied or divided by a negative quantity provided the inequality sign is reversed.

Example 1: Find the greatest possible value of an X that satisfies the inequality 8 + 2X > 3 + 5X. if X is an integer.

Solution:

8 + 2X > 3 + 5X

Subtract 8 from both sides

2X > 3 – 8 + 5X

2X > -5 + 5X

Subtract 5X from both sides

2X – 5X > -5  = -3x > -5

Divide both sides by -3 and also reverse the inequality

X < 12/3 [rule three]

The greatest integer value of X is 1.

Example2: Find the smallest integer value of X that satisfies the inequality 7X – 2 ≥ 5X – 6

Solution:

7X – 2 ≥ 5X – 6

Add 2 to both sides

7X ≥ 5X – 6 + 2

7X ≥ 5X – 4

Subtract 5X from both sides

7X – 5X ≥ -4 = 2x ≥ -4 , Divide both side by 2

X ≥ -2

WORD PROBLEMS LEADING TO INEQUALITIES

Example 1: David is X years old. In 4 years, his age will still be less than 12 years. (a) Write this information in an inequality in X. (b) Find the maximum age of David to the nearest whole numbers.

Solution:

In 4 years, David age’s will be (X + 4) years.

If at that time his age is less than 12, then

X + 4 < 12

Subtract 4 from both sides

X < 12 – 4

X < 8.

The maximum age of David is 7 years.

Example 2: A man had #X, out of this, he used #1000 to pay his house rent. The amount he had left was not more than 3500. (a) Write this information in an inequality in X (b) Solve for X.

Solution:

The man used #1000 to pay his house rent out of #X, so the amount left is #(x – 1000).

This amount is not more than 3500

X – 1000 ≤ 500

Add 1000 to both sides

X ≤ 500 + 1000 = X ≤ 1500

Hence the man had less than or equal to #1500