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Stage 5.1-2

4.06 Solving linear inequalities

Lesson

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=12

We can subtract 7 from both sides of the equation in order to find the value of 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.

\displaystyle x+7\displaystyle =\displaystyle 12Rewrite the equation
\displaystyle x+7-7\displaystyle =\displaystyle 12-7Subtract 7 from both sides
\displaystyle x\displaystyle =\displaystyle 6Evaluate

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.

Consider the inequality 9<15.

If we add or subtract both sides by any number, say 3, we can see that the resulting inequality remains true.

A horizontal number line. An arrow from 9 points right to 9 plus 3 and another arrow from 15 points right to 15 plus 3.

\text{Adding }3 \text{ to }9 \text{ and } 15 \\ 9+3 < 15 +3\\12<18

The inequality stays the same and it is true that 12 is less than 18.

A horizontal number line. An arrow from 9 points left to 9 minus 3 and another arrow from 15 points left to 15 minus 3.

\text{Subtracting }3 \text{ from }9 \text{ and } 15 \\ 9-3 < 15 -3\\6<12

Once again the inequality stays the same and it is true that 6 is less than 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:

\displaystyle 9\displaystyle <\displaystyle 15Given
\displaystyle 9\times3\displaystyle <\displaystyle 15\times3Multiply both sides by 3
\displaystyle 27\displaystyle <\displaystyle 45Evaluate

The inequality stays the same, and it is true that 27 is less than 45.

Now let's look at what happens when we divide by a positive number:

\displaystyle 9\displaystyle <\displaystyle 15Given
\displaystyle \dfrac{9}{3}\displaystyle <\displaystyle \dfrac{15}{3}Divide both sides by 3
\displaystyle 3\displaystyle <\displaystyle 5Evaluate

The inequality stays the same, and it is true that 3 is less than 5.

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

\begin{aligned} 9& &< &&&15 &&\text{Given} \\ 9 \times \left(-3\right)& &⬚ &&&15 \times \left(-3\right) &&\text{Multiply both sides by} -3 \\ -27& &⬚ &&&-45 &&\text{Evaluate} \end{aligned}

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 is actually greater than -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:

\displaystyle 9\displaystyle <\displaystyle 15Given
\displaystyle 9\times(-3)\displaystyle >\displaystyle 15\times(-3)Multiply both sides by 3, reverse inequality
\displaystyle -27\displaystyle >\displaystyle -45Evaluate

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

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

\displaystyle 9\displaystyle <\displaystyle 15Given
\displaystyle \dfrac{9}{-3}\displaystyle >\displaystyle \dfrac{15}{-3}Divide both sides by -3, reverse inequality
\displaystyle -3\displaystyle >\displaystyle -5Evaluate

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

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.

Examples

Example 1

Solve the following inequality: x-10<13

Worked Solution
Create a strategy

Use the inverse operations to find the range of values of x.

Apply the idea
\displaystyle x-10+10\displaystyle <\displaystyle 13+10Add 10 to both sides
\displaystyle x\displaystyle <\displaystyle 23Evaluate

Example 2

Solve the following inequality: \dfrac{x}{-9} \geq 4

Worked Solution
Create a strategy

Use the inverse operations to find the range of values of x.

Apply the idea
\displaystyle \dfrac{x}{-9} \displaystyle \geq\displaystyle 4Write the inequality
\displaystyle \dfrac{x}{-9} \times -9\displaystyle \leq\displaystyle 4 \times -9Multiply both sides by -9, reverse inequality
\displaystyle x\displaystyle \leq\displaystyle -36Evaluate
Idea 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.

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.

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 that can be used to represent an inequality symbol:

  • > greater than, more than
  • \geq greater than or equal to, at least, no less than
  • < less than
  • \leq less than or equal to, at most, no more than

Examples

Example 3

Neville is saving up to buy a plasma TV that is selling for \$950. He has \$650 in his bank account and expects a nice sum of money for his birthday next month.

If the amount he is to receive for his birthday is represented by x, which of the following inequalities models the situation where he is able to afford the TV?

A
x+650\leq 950
B
x+650\geq 950
C
x-650\geq 950
D
x-650\leq 950
Worked Solution
Create a strategy

Find the total money he will have which needs to be at least as much as the price.

Apply the idea

To afford the TV, the money he will have must be greater or equal to the price of the TV.

The money he will have is made up of his birthday money, x, plus his saving, \$650. So we add these amounts and make them greater or equal to the price of the TV: x+650\geq 950 The answer is option B.

Idea summary

Phrases that can be used to represent an inequality symbol:

  • > greater than, more than
  • \geq greater than or equal to, at least, no less than
  • < less than
  • \leq less than or equal to, at most, no more than

Outcomes

MA5.2-8NA

solves linear and simple quadratic equations, linear inequalities and linear simultaneous equations, using analytical and graphical techniques

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