December 8th
December 8th
Inequalities
Rules for Inequalities
The section following this one will deal with the solution of
inequalities. As with the solution of equations, there are
certain rules that may be used. In the case of inequalities these
are
Rule 1 An equal quantity may be added to, (or
subtracted from) both sides of an inequality without changing the
inequality.
Rule 2 An equal positive quantity may
multiply (or divide) both sides of an inequality without changing
the inequality.
Rule 3 If both sides of an inequality are
multiplied (or divided) by a negative quantity then the
inequality is reversed .
N.B. It is very important to be careful with
the last rule.
Before looking at the solution of inequalities it is useful to
see why the above rules hold. This is done in the following
example.
Example 1
Given the (true) inequality 4 >  1, verify each of the
rules
(a) by adding 3 to both sides,
(b) by subtracting 3 from both sides,
(c) by multiplying both sides by 3,
(d) by multiplying both sides by  3.
Solution
(a) Adding 3 to both sides gives 7 > 2, which is also true.
(b) Subtracting 3 from both sides gives 4  3 >  1  3 or
1 >  4, which is also true.
(c) Multiplying both sides by 3 gives 12 > 3, which is
also true.
(d) Multiplying the inequality 4 > 1 by 3 gives.
according to rule 3,
4 × (3) < (1) ×(3), or 12 < 3, which is correct.
Solving Inequalities
Example 2
Solve the following inequalities.
(a) x  3 > 5, (b) 2x  1 > 7, (c) 3  2x > 5.
Solution
(a) x  3 > 5, add 3 to both sides
x  3 + 3 > 5 + 3,
x > 8 .
(b) 2x  1 > 7 , add 1 to both sides
2x > 8 , divide both sides by 2,
x > 4.
(c) 3  2x > 5, subtract 3 from both sides
2x > 8 , divide both sides by  2
x < 4, rule 3 has reversed the inequality!
Here are some exercises on solving inequalities.
Exercise
Solve each of the following inequalities using the rules given
in section 2.
Solutions
(a)
3x  4 < 5
3x  4 + 4 < 5 + 4
3x < 9
x < 3.
(b)
x + 1 < 0
x + 1  1 < 0  1
x <  1
(c)
2x  6 10
2x  6 + 6 10 + 6
2x 16
x 8
(d)
2x x  3 substract x from both sides
2x  x x  3  x
x  3
(e)
3x + 1 < 2x + 5
3x + 1  1 < 2x + 5  1
3x < 2x + 4
3x  2x < 2x + 4  2x
x < 4
(d) In this case the brackets must first be
removed using the standard rules.
3(x  1) > 2(1  x)
3x  3 > 2  2x
3x  3 + 3 > 2  2x + 3
3x > 5 + 2x
3x  2x > 5 + 2x  2x
x > 5
Now try this short quiz.
Quiz
Which of the following is the solution to the inequality 15 
x > 7 + x ?
(a) x > 4, (b) x > 11, (c) x
Solution
The solution is as follows.
15  x > 7 + x
15  x + x > 7 + x + x
15 > 7 + 2 x
15  7 > 7 + 2 x  7
8 > 2 x
4 > x or equivalently
x < 4
The first step in this solution was adding x to both sides so
that (the positive) 2x appears on the right. This meant that the
subsequent division was by the positive number 2. Division by
positive numbers is always preferable as it generally leads to
fewer mistakes.
