Working with inequalities
When we solve equations, we can apply the same operation to both sides and the equation remains true. Consider the following equation:
| $x+7$x+7 | $=$= | $12$12 |
We can subtract $7$7 from both sides of the equation in order to find the value of $x$x. This is because both sides of the equation are identical, so what we do to one side, we should do to the other side.
| $x+7$x+7 | $=$= | $12$12 |
rewriting the equation |
| $x+7-7$x+7−7 | $=$= | $12-7$12−7 |
subtracting $7$7 from both sides |
| $x$x | $=$= | $5$5 |
simplifying both sides |
When working with inequalities, we can follow exactly the same procedure, but with an inequality sign in place of the equals sign. But there are two operations that we need to be careful of.
Exploration
Consider the inequality $9<15$9<15.
If we add or subtract both sides by any number, say $3$3, we can see that the resulting inequality remains true.
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Adding $3$3 to $9$9 and $15$15. $9+3<15+3$9+3<15+3 $12<18$12<18 The inequality stays the same and it is true that $12$12 is less than $18$18. |
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Subtracting $3$3 from $9$9 and $15$15. $9-3<15-3$9−3<15−3 $6<12$6<12 Once again the inequality stays the same and it is true that $6$6 is less than $12$12. |
Adding a negative would have the same effect as subtracting, so we can also add and subtract negative numbers without changing the inequality.
What happens if we multiply or divide both sides?
Let's look at what happens when we multiply by a positive number:
| $9$9 | $<$< | $15$15 |
Given |
| $9\times3$9×3 | $<$< | $15\times3$15×3 |
Multiplying by $3$3 |
| $27$27 | $<$< | $45$45 |
Simplifying |
The inequality stays the same, and it is true that $27$27 is less than $45$45.
Now let's look at what happens when we divide by a positive number:
| $9$9 | $<$< | $15$15 |
Given |
| $\frac{9}{3}$93 | $<$< | $\frac{15}{3}$153 |
Dividing by $3$3 |
| $3$3 | $<$< | $5$5 |
Simplifying |
The inequality stays the same, and it is true that $3$3 is less than $5$5.
Summary
The following operations don't change the inequality symbol used:
- Adding a number to both sides of an inequality.
- Subtracting a number from both sides of an inequality.
- Multiplying both sides of an inequality by a positive number.
- Dividing both sides of an inequality by a positive number.
Solving inequalities
Now that we have seen what happens when we perform addition, subtraction, multiplication and division, we can use this knowledge to solve inequalities.
Before jumping in algebraically, it can be helpful to consider some possible solutions and non-solutions. Then we can look at an algebraic strategy.
Worked examples
Question 1
List at least two values of $x$x which satisfy $x+3<4$x+3<4 and one which does not.
Think: Let's start by picking three values and see if they satisfy the inequality or not. We'll try $0$0 and $10$10.
Do:
Substituting in $0$0, we get $0+3<4$0+3<4 or $3<4$3<4 which is true, so $0$0 satisfies the inequality
Substituting in $10$10, we get $10+3<4$10+3<4 or $13<4$13<4 which is false, so $10$10 does not satisfy the inequality
So $0$0 satisfies the inequality $x+3<4$x+3<4, but $10$10 does not. This means that somewhere between $0$0 and $10$10 there is a point where everything below it satisfies the inequality.
Reflect: How does knowing some true values help us when finding a solution or graphing?
If we know some particular values that are solutions to the equation, we can check that our final answer has the correct inequality sign, in case we have forgotten to reverse the sign at a particular step.
Practice questions
Question 1
Solve the following inequality: $x-1<15$x−1<15
Real life applications
Much as with solving equations from written descriptions, there are certain key words or phrases to look out for. When it comes to inequalities, we now have a few extra key words and phrases to represent the different inequality symbols.
- $>$> greater than, more than
- $\ge$≥ greater than or equal to, at least, no less than
- $<$< less than
- $\le$≤ less than or equal to, at most, no more than
Worked example
Question 2
Construct and solve an inequality for the following situation:
"The sum of $2$2 groups of $x$x and $1$1 is at least $7$7."
Think: "At least" means the same as "greater than or equal to". Also "groups of" means there is a multiplication, and "sum" means there is an addition.
Do: $2$2 groups of $x$x is $2x$2x, and the sum of this and $1$1 is $2x+1$2x+1. So altogether we have that "the sum of $2$2 groups of $x$x and $1$1 is at least $7$7" can be written as $2x+1\ge7$2x+1≥7.
We can now solve the inequality for $x$x:
| $2x+1$2x+1 | $\ge$≥ | $7$7 |
| $2x$2x | $\ge$≥ | $6$6 |
| $x$x | $\ge$≥ | $3$3 |
So the possible values of $x$x are those that are greater than or equal to $3$3.
Practice questions
Question 2
Write down the inequality described by "six more than the value of $x$x is at least seven", and solve for $x$x.
Question 3
Sandy has a budget for school stationery of $\$46$$46, but has already spent $\$18.10$$18.10 on books and folders.
Let $p$p represent the amount that Sandy can spend on other stationery. Write an inequality that shows how much she can spend on other stationery, and solve for $p$p.