# Solving Linear Inequalities

Solving linear inequalities is almost like solving linear equations. Here’s how it works:

Solve *x* + 3 < 0

If the problem was *x* + 3 = 0, we’d have subtracted 3 from both sides and we can do the same here:

x+ 3 <

x+ 3 − 3 <

x<0

0 − 3

− 3

Because the solution is written as an inequality, this format is referred to as inequality notation. Three other commonly taught formats, covered in Set Notation, are included as a comparison but, for the purposes of this lesson, we’ll use inequality notation:

Notation | Format | Spoken as |
---|---|---|

inequality | x < −3 |
x is less than negative three |

number line | the interval of all numbers less than negative three | |

interval | (−, −3) | the interval from minus infinity to negative three |

set | {x | x < −3 } |
The set of all real numbers x such that x is less than negative three |

Solve *x* − 4 ≥ 0

Again, if they’d given us *x* − 4 = 0, we would have added 4 to each side and, again, we do the same here:

x− 4 ≥

x− 4 + 4 ≥

x≥0

0 + 4

− 4

Solve 2*x* ≤ 9

If the problem had been 2*x* = 9, you would have divided each side by 2 and you do the same thing here:

2x≤

2x÷ 2 ≤

x≤9

9 ÷ 2

4.5

Solve ^{x}/₄ > ½

With ^{x}/₄ = ½, you would have multiplied both sides by 4 and again, the same thing applies here:

Solve −2*x* < 5

Up to now, solving linear inequalities has followed the same procedures as solving linear equations, but here is the one case where it’s different. To explain, consider 3 > 2. With a linear equation, we could multiply both sides by −1 without changing the value but doing so here would give us −3 > −2, but −3 is NOT greater than −2.

When solving inequalities, if you multiply or divide through by a negative, you must reverse the inequality sign.

To solve −2*x* < 5, we need to divide by a negative, so we must reverse the inequality:

−2x<

−2x÷ 2 <

x>5

5 ÷ 2

2.5

Solve

Solve 10 ≤ 3*x* + 4 ≤ 19.

This is what is called a compound inequality. It works just like regular inequalities, except that it has three elements. So to solve, we subtract the 4 from all three elements then divide all three elements by 3.

10 ≤

6 ≤

2 ≤3x+ 4 ≤

3x≤

x≤19

15

5

Solve 5*x* + 7 < 3(*x* + 1)

First we’ll multiply through the parentheses then solve as usual:

5x+ 7 <

5x+ 7 <

2x+ 7 <

2x<

x<3(x+ 1)

3x+ 3

3

−4

−2

Solve 3(*x* − 2) + 4 ≥ 2(2*x* − 3)

Again, we multiply through the parentheses, simplify, then solve:

3(x− 2) + 4 ≥

3x− 6 + 4 ≥

3x− 2 ≥

−x−2 ≥

−x≥

x≤2(2x− 3)

4x− 6

4x− 6

− 6

−4

4

The velocity of an object fired directly upward is given as V = 80 − 32*t*, where *t* is in seconds. When will the velocity be between 32 and 64 feet per second?

We’re looking for the situation where the velocity (80 − 32*t*) is greater than 32 and less than 64 so we set up the compound inequality and solve for *t*:

32 <

32 (− 80) <

−48 <

>

1.5 >80 − 32t<

80 (− 80) − 32t<

−32t<

>

t>64

64 (− 80)

−16

0.5

Note that, since we divided through by a negative, we reversed the inequality signs and the solution was 1.5 > *t* > 0.5 however, it would more easily understood if written the other way around:

0.5 < *t* < 1.5

Looking back at the original question, it did not ask for the value of the variable *t*, but asked for the times when the velocity was between certain values. So the answer would be written as:

The velocity will be between 32 and 64 feet per second between 0.5 and 1.5 seconds after launch.

Always remember when doing word problems, that you need to go back and re-read the problem to make sure that you’re answering the actual question.

You want to invest $30,000. Part of this will be deposited in a stable 5% simple interest rate account. The remainder will be invested in your father’s business, and he says that he’ll pay you back with 7% interest. What is the least you can invest and still get at least $1900 in interest?

First, we have to set up equations for this. The interest formula for simple interest is *I = Prt, *where *I* is the interest, *P* is the beginning principal, *r* is the interest rate expressed as a decimal, and *t* is the time in years. Since no time-frame is specified for this problem, we’ll assume that *t* = 1. We’ll let *x* be the amount that you’re going to invest. Then there will be 30000 − *x* left to deposit in the savings account. The interest on the business investment will be:

(*x*)(0.07)(1) = 0.07*x*

The interest on the savings account will be: (30000 − *x*)(0.05)(1) = 1500 − 0.05*x*

Then the total interest is: 0.07*x* + (1500 − 0.05*x*) = 0.02*x* + 1500

You need to get at least $1900, so:

0.02x+ 1500 ≥

0.02x≥

x≥1900

400

20000

An alloy needs to contain between 44% and 48% copper. Find the least and greatest amounts of a 60% copper alloy that should be mixed with a 40% copper alloy in order to end up with thirty pounds of an alloy containing an allowable percentage of copper.

This is similar to a mixture word problem, except that it will involve inequalities but we’ll set it up the same way:

pounds % copper pounds copper 60% x0.6 0.6 x40% 30 − x0.4 0.4(30 − x) = 12 − 0.4xmix 30 between 0.44 and 0.48 between 13.2 and 14.4

Note: To get those values in the bottom right cell, we multiply the total number of pounds in the mixture (30) by the minimum and maximum percentages (44% and 48%, respectively). The total amount of copper in the mixture will be the sum of the copper from the two alloys put into the mixture, so we add the expressions for the amount of copper from the alloys and place the total between the minimum and the maximum allowable amounts of copper:

13.2 ≤

13.2 ≤

1.2 ≤

6 ≤0.6x+ (12 − 0.4x) ≤

0.2x+ 12 ≤

0.2x≤

x≤14.4

14.4

2.4

12

So, you’ll need to use between 6 and 12 pounds of the 60% alloy.