4.07 One-step inequalities

Working with inequalities

When we manipulate with 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, this is not necessarily always the case.

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

Subtracting $3$3 from $9$9 and $15$15. $9-3<15-3$9−3<15−3 $6<12$6<12

 

Note that 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?

Multiplying by a positive

$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

Dividing by a positive

$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

Multiplying by a negative

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

The inequality switches from $<$< to $>$>!

Dividing by a negative

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

The inequality switches from $<$<to $>$>!

 

Summary

 

Practice questions

Question 1

question 2

question 3

 

Solving one-step 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 4

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?

 

Question 5

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

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

Do: 

$x+3$x+3	$<$<	$4$4	Given
$x+3-3$x+3−3	$<$<	$4-3$4−3	Subtract $3$3 from both sides
$x$x	$<$<	$1$1	Simplify both sides

 

Reflect: How could the two numberlines below help us to visualize this problem?

 

Question 6

Solve $-\frac{x}{2}\ge5$−x2​≥5 algebraically and show your work.

Think: We can solve this similarly to the equation $-\frac{x}{2}=5$−x2​=5, but we just need to be careful when we are multiplying or dividing by a negative value.

Do: 

$-\frac{x}{2}$−x2​	$\ge$≥	$5$5	Given
$-\frac{x}{2}\times\left(-2\right)$−x2​×(−2)	$\le$≤	$5\times\left(-2\right)$5×(−2)	Multiply both sides by $-2$−2, we must flip the sign
$x$x	$\le$≤	$-10$−10	Simplify both sides

 

Practice questions

QUESTION 7

Question 8

Question 9

 

Graphing solutions to one-step inequalities

As we have previously seen, we can plot inequalities by using number lines.

We can first solve an inequality and then graph it. This can also help to check our answer.

Worked example

Question 10 

Solve and graph the solution to $x+3>5$x+3>5.

Think: We do not need to multiply or divide by a negative, so we end up with a numberline with a hollow (open) endpoint and the ray point to the right.

Do: 

$x+3$x+3	$>$>	$5$5	Given
$x+3-3$x+3−3	$>$>	$5-3$5−3	Subtract $3$3 from both sides
$x$x	$>$>	$2$2	Simplify both sides

 

So the plot will show "all numbers greater than $2$2" on the number line, which looks like this:

 

 

Practice Questions

Question 11

Question 12

Question 13