LINEAR INEQUALITIES
You will remember that a number line is a line running from left to right (horizontal) having numbers shown on it as shown below:
-5 -4 -3 -2 -1 0 1 2 3 4 5 x
Example 1 : Represent the inequality x > 3 on the number line
-3 -2 -1 0 1 2 3 4 5 6 7 x
X > 3 means all numbers greater than 3 I e 4,5,6,7 etc
3 is excluded from these numbers to show this, we have an empty circle around the number 3
Example 2 : Represent x ≥ 3 on the number line
-3 -2 -1 0 1 2 3 4 5 6 7 x
X > 3 means all numbers greater than 3. I e 4,5,6,7 etc
3 is included in these numbers to show this, we have a shaded circle around the number 3.
Example 3: Solve the inequality x + 4 < 6
Solution
X +4 < 6
Subtract 4 from both sides
X + 4 – 4< 6 – 4
X < 2
Example 4: Solve the inequality 6 ≤ 2x – 1 and sketch the graph of the solution.
Solution
Add 1 to both sides
6 + 1 ≤ 2x – 1 + 1
7 ≤ 2x
Divide both sides by 2
≤
3≤ x, then x ≤ 3
-3 -2 -1 0 1 2 3 3 4 5 6 7 x
Multiplication and Division by negative numbers
As a rule, if both sides of an inequality are multiplied or divided by a negative number, the inequality must be reversed. I.e.> ≥ become < ≤.
Example 5:Solve 19≥ 4 – 5x
Solution
19 ≥ 4 – 5x
Subtract 4 from both sides
19 – 4 ≥ -5x
15 ≥ -5x
Divide both sides by -5 and reverse the inequality sign
≤
-3 ≤ x
Example 6: Solve 5 – x > 3
Solution
Subtract 5 from both sides
-x > -2
Multiply both sides by -1 and reverse the inequality
(-1) x (-x) < (-1) x(-2)
X <2
Word problems involving inequalities
Example 7: if 9 is added to a number x the result is greater than 17. Find x?
Solution
Let the number be x.
x + 9 > 17
Subtract 9 from both sides
x + 9 – 9 > 17 – 9
X> 8
Example 8: A triangle has sides of xcm, (x+7) cm and 10cm, where x is a whole number. If the perimeter of the triangle is less than 31, find the possible value of x.
Solution
Perimeter of a triangle ={ x + (x+7) +10}cm
Thus, { x + x+7 +10} < 31x+x=2
2x + 17 <31
Subtract 17 from both sides
2x + 17 – 17 = 31 – 17
2x < 14
Divide both sides by 2
<
X < 7
Also , in any triangle, the sum of the length of any two sides must be greater than the length of the third sides.
Thus, x + (x+7) > 10
2x + 7 > 10
2x > 10 – 7
2x > 3
X > 1
Thus x < 7and x > 1. But x must be a whole number. Thus the possible values of x are 2, 3 4, 5, 6.
Assignment