3. Equations & Inequalities

Pennsylvania Algebra 1 - 2020 Edition

3.08 Solving multi-step inequalities

Lesson

We have seen that it is easiest to plot an inequality on a number line by first solving the inequality. We have also looked at solving simple inequalities. We are now going to combine these ideas together - let's recap through an example.

Suppose we want to plot the solutions to the inequality $2\left(3+x\right)<8$2(3+`x`)<8 on a number line. That is, we want to plot the values of $x$`x` which can be added to $3$3 and then doubled to result in a number less than $8$8.

To solve this inequality, we want to undo these operations in **reverse order**. That is, we can solve this inequality by first dividing both sides by $2$2, then subtracting $3$3 from both sides:

$2\left(3+x\right)$2(3+x) |
$<$< | $8$8 | ||

$3+x$3+x |
$<$< | $4$4 | Dividing both sides by $2$2 | |

$x$x |
$<$< | $1$1 | Subtracting $3$3 from both sides |

In this case, we arrive at the result $x<1$`x`<1. We can test some values in the original inequality to see if this is the right solution set - let's say $x=0$`x`=0 and $x=2$`x`=2.

- When $x=0$
`x`=0, we have $2\left(3+x\right)=2\left(3+0\right)=6$2(3+`x`)=2(3+0)=6, which**is**less than $8$8. - When $x=2$
`x`=2, we have $2\left(3+x\right)=2\left(3+2\right)=10$2(3+`x`)=2(3+2)=10, which**is not**less than $8$8.

So our result of $x<1$`x`<1 seems to be correct.

Remember

When solving any inequality:

- Multiplying or dividing both sides of an inequality by a negative number reverses the inequality symbol.
- Writing an inequality in reverse order also reverses the inequality symbol.

When solving an inequality with two (or more) operations:

- It is generally easiest to undo one operation at a time, using the properties of inequality.

We can now plot the solutions on a number line as follows, using a hollow circle for the endpoint (since $x=1$`x`=1 is **not** included in the solutions):

Remember

When plotting an inequality:

- The symbols $<$< and $>$>
**don't**include the end point, which we show with a**hollow**circle. - The symbols $\ge$≥ and $\le$≤
**do**include the endpoint, which we show with a**filled**circle.

Solve the following inequality: $\frac{a}{5}+3>3$`a`5+3>3

Consider the inequality $3x+1>4$3`x`+1>4.

Solve the inequality.

Now plot the solutions to the inequality $3x+1>4$3

`x`+1>4 on the number line below.

Consider the inequality $7-x>13$7−`x`>13.

Solve the inequality.

Now plot the solutions to the inequality $7-x>13$7−

`x`>13 on the number line below.

We have now looked at solving inequalities that involve one or two steps to solve. We're now going to take a look at how we can use inequalities to solve problems given a written description.

Much as with solving equations from written descriptions, there are certain keywords or phrases to look out for. When it comes to inequalities, we now have a few extra keywords 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.

Construct and solve an inequality for the following situation:

"The sum of $2$2 lots of $x$`x` and $1$1 is at least $7$7."

**Think**: "At least" means the same as "greater than or equal to". Also "lots of" means there is a multiplication, and "sum" means there is an addition.

**Do**: $2$2 lots of $x$`x` is $2x$2`x`, and the sum of this and $1$1 is $2x+1$2`x`+1. So altogether we have that "the sum of $2$2 lots of $x$`x` and $1$1 is at least $7$7" can be written as $2x+1\ge7$2`x`+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.

Consider the following situation:

"$2$2 less than $4$4 groups of $p$`p` is no more than $18$18".

Construct and solve the inequality described above.

What is the largest value of $p$

`p`that satisfies this condition?$p=5$

`p`=5A$p=-5$

`p`=−5BThere is no largest value.

C$p=4$

`p`=4D$p=5$

`p`=5A$p=-5$

`p`=−5BThere is no largest value.

C$p=4$

`p`=4D

Lachlan is planning on going on vacation. He has saved $\$2118.40$$2118.40, and spends $\$488.30$$488.30 on his airplane ticket.

Let $x$

`x`represent the amount of money Lachlan spends on the rest of his holiday.Write an inequality to represent the situation, and then solve for $x$

`x`.What is the most that Lachlan could spend on the rest of his holiday?

At a sport clubhouse the coach wants to rope off a rectangular area that is adjacent to the building. He uses the length of the building as one side of the area, which measures $26$26 meters. He has at most $42$42 meters of rope available to use.

If the width of the roped area is $W$

`W`, form an inequality and solve for the range of possible widths.

So far we have looked at how to solve inequalities for all of the possible solutions, and we found that the process was very similar to how we solve equations.

We don't always need to solve for all of the possible solutions, however. Sometimes we just want to know if a particular value will satisfy an inequality or not.

Think about the inequality given by $p-7<12$`p`−7<12. There is an interval of values that $p$`p` can take to make the inequality true.

For example, we can test whether $p=20$`p`=20 satisfies $p-7<12$`p`−7<12 by simply substituting that value into it and see what happens.

So, if $p=20$`p`=20, the expression becomes $20-7<12$20−7<12 or, in other words, $13<12$13<12 which is clearly not true. This means that $p=20$`p`=20 is not in the solution set. It doesn't satisfy the inequality.

On the other hand, a number like $p=10$`p`=10 is in the solution set because $10-7<12$10−7<12.

**Solve:** Does $t=6$`t`=6 satisfy the equation $3t-5\ge12$3`t`−5≥12?

**Think:** We can simply substitute $t=6$`t`=6 to find out.

**Do:** So when $t=6$`t`=6 we have:

$3\times6-5$3×6−5 | $\ge$≥ | $12$12 |

$18-5$18−5 | $\ge$≥ | $12$12 |

$13$13 | $\ge$≥ | $12$12 |

So the answer is yes, $t=6$`t`=6 does satisfy the inequality.

It's also possible to apply this method to a practical situation. See the practice question below for an example.

Neville is saving up to buy a plasma TV that is selling for $\$950$$950. He has $\$650$$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 $x$

`x`, which of the following inequalities models the situation where he is able to afford the plasma TV?$x+650\le950$

`x`+650≤950A$x+650\ge950$

`x`+650≥950B$x-650\ge950$

`x`−650≥950C$x-650\le950$

`x`−650≤950D$x+650\le950$

`x`+650≤950A$x+650\ge950$

`x`+650≥950B$x-650\ge950$

`x`−650≥950C$x-650\le950$

`x`−650≤950DHow much money would he have in total if his parents were to give him $\$310$$310 for his birthday?

Would he have enough to buy the plasma TV if his parents were to give him $\$310$$310 for his birthday?

Yes

ANo

BYes

ANo

B

Represent, solve and interpret equations/inequalities and systems of equations/inequalities algebraically and graphically.

Identify or graph the solution set to a linear inequality on a number line.

Interpret solutions to problems in the context of the problem situation. Note: Linear inequalities only.