# 3.03 Solving linear inequalities

Lesson

## 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.

 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.
 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.

But now let's look what happens when we multiply by a negative number:

 $9$9 $<$< $15$15 Given $9\times\left(-3\right)$9×(−3) $\editable{}$ $15\times\left(-3\right)$15×(−3) Multiplying by $-3$−3 $-27$−27 $\editable{}$ $-45$−45 Simplifying

What inequality sign goes in the boxes on lines two and three? Can we just keep the original inequality sign?

If we look at the last line, we can see that $-27$27 is actually greater than $-45$45, as it further to the right on a number line. So we have to reverse the inequality sign whenever we multiply by a negative number. This gives us the following:

 $9$9 $<$< $15$15 Given $9\times\left(-3\right)$9×(−3) $>$> $15\times\left(-3\right)$15×(−3) Multiplying by $-3$−3 and reversing the inequality $-27$−27 $>$> $-45$−45 Simplifying

So we can see that $-27$27 is greater than $-45$45.

To divide by a negative we will need to use the same trick:

 $9$9 $<$< $15$15 Given $\frac{9}{-3}$9−3​ $>$> $\frac{15}{-3}$15−3​ Dividing by $-3$−3 and reversing the inequality $-3$−3 $>$> $-5$−5 Simplifying

So once again, the inequality switches from $<$<to $>$>.

### Summary

Same symbol

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.
Opposite symbol

The following operations reverse the inequality symbol used:

• Multiplying both sides of an inequality by a negative number.
• Dividing both sides of an inequality by a negative 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 $-10$10, $0$0 and $10$10.

Do:

Substituting in $-10$10, we get $-10+3<4$10+3<4 or $-7<4$7<4 which is true, so $-10$10 satisfies the inequality

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 $-10$10 and $0$0 satisfy 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.

##### Question 2

Solve $3-x<4$3x<4 algebraically and show your work.

Think: We can solve this similarly to the equation $3-x=4$3x=4, but we just need to be careful if we are multiplying or dividing by a negative value.

Before we solve this, let's check two values $x=-5$x=5 and $x=5$x=5
Substituting in $x=-5$x=5 gives us $3-\left(-5\right)<4$3(5)<4, which simplifies to $8<4$8<4 which is clearly not true.

Substituting in $x=5$x=5 gives us $3-5<4$35<4, which simplifies to $-2<4$2<4, which is certainly true. So we know $x=5$x=5 is one possible solution.

Do: Now, let's solve the inequality to find all possible solutions

 $3-x$3−x $<$< $4$4 Given $3-x-3$3−x−3 $<$< $4-3$4−3 Subtract $3$3 from both sides $-x$−x $<$< $1$1 Simplify both sides $x$x $>$> $-1$−1 Divide both sides by $-1$−1, remembering to reverse the sign

So our solution is $x>-1$x>1. We can now check that we have the right sign using our initial solution of $x=5$x=5.

As $5$5 is greater than $-1$1, the solution holds, and we know we have solved the inequality correctly. And in fact any number greater than $-1$1 will be a solution to this inequality.

#### Practice questions

##### Question 1

Solve the following inequality: $x-1<15$x1<15

##### question 2

Solve the following inequality: $2-\frac{1}{2}x>5$212x>5

## 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.

Phrases
• $>$> 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 3

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+17.

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 3

Write down the inequality described by "six more than the value of $x$x is at least seven", and solve for $x$x.

##### Question 4

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.

### Outcomes

#### VCMNA336

Solve linear inequalities and graph their solutions on a number line.