© William James Calhoun, 2001 7-1: Solving Inequalities by Using Addition and Subtraction OBJECTIVE: You will be able to solve inequalities by using addition

© William James Calhoun, 2001

7-1: Solving Inequalities by Using Addition and Subtraction

OBJECTIVE:You will be able to solve inequalities by using addition and subtraction.

This section is essentially the same as solving equations.

The major difference is that instead of finding one single value which makes an equation true, the answers to inequalities is a set of answers.

You need to recall the work done in Chapter 2 involving inequalities.Specifically, the significance of which way they point and how their number-line graphs appear.

If you understood how to solve equations in Chapter 3, these inequalities will be easy.

If you did not understand Chapter 3, you need to go back and review it.



The symbols and their meanings again:

In solving inequalities by using addition and subtraction, the rule is the same as solving equations:

YOU CAN ADD OR SUBTRACT ANYTHING AS LONG AS YOU ADD OR SUBTRACT THE SAME THING ON BOTH SIDES.

With that, we will start our first example.

•less than•fewer than

•greater than•more than

•at most•no more than•less than or equal to

•at least•no less than•greater than or equal to

< > Inequalities



The questions are the same.What is the letter?

gWhat is on the same side as the g?

positive 29How are the 29 and the g combined?

additionGet rid of the 29 by…

subtracting 29 from both sides.Do it.

29 + g 68-29 -29

g 39

This is the answer.The book gives its answers in set-builder notation.In set-builder notation, the answer would look like this:

{g | g 39}

If we were to graph this solution set, it would appear like this:

Notice the full circle at the point and the arrow showing the solution set goes on forever.

34 35 36 37 38 39 40 41 42 43 44 45 46 47

This is read, “g such that g is less than or equal to 39.”

EXAMPLE 1: Solve 29 + g 68.



Example 2: Solve 13 + 2z < 3z - 39. Then graph the solution.

What is the letter?z

What is on the same side as z?which z?

We must get all the z’s on the same side.For now, make sure the resulting z’s are positive.

We will talk about negative variables and their implications on inequalities next section.

Because of this, we need to move the 2z by…subtracting 2z from both sides.Do it.

13 + 2z < 3z - 39 - 2z -2z 13 < z - 39

Now, what is on the same side as the z?negative 39

Get rid of it by…adding 39 to both sides.

+39 +3952 < z{z | z > 52}

Now graph it.

49 50 51 52 53 54



EXAMPLE 3: Alvaro, Chip, and Solomon have earned $500 to buy equipment for their band. They have already spent $275 on a used guitar and a drum set. They are now considering buying a $125 amplifier. What is the most they can spend on promotional materials and T-shirts for the band if they buy the amplifier?You cannot spend more than what you have, so the amount the group spends MUST be less than or equal to the amount they have.

what they spend what they haveWhat did they start with?$500Let “p” be the amount of promo materials.What will be the amount they spend?$275 + $125 + p

275 + 125 + p 500

Combine like terms on the left-hand side.400 + p 500

What is the letter?p

What is on the same side as p?400

How to get rid of it?subtract 400 from both sides

-400 -400p 100

If the band purchases the amplifier, they will have at most $100 to spend on promo materials.



EXAMPLE 4: Write an inequality for the sentence below. Then solve the inequality and check the solution.

Three times a number is more than the difference of twice that number and three.

3 * x > -2* x 3

3x > 2x - 3Rewrite it.Solve it. -2x -2x

x > -3

To check it, plug in any number larger than -3 and see if it makes a true statement.

0 > -33(0) > 2(0) - 3

0 > -3True

10 > -33(10) > 2(10) - 3

30 > 17True

100 > -33(100) > 2(100) - 3

300 > 197True

-2 > -33(-2) > 2(-2) - 3

-6 > -7True



HOMEWORK

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© William James Calhoun, 2001 7-1: Solving Inequalities by Using Addition and Subtraction OBJECTIVE: You will be able to solve inequalities by using addition