Applications of the Subtraction and Addition Properties

Learning Outcomes

Translate a sentence into an algebraic equation
Solve an equation that has been translated and verify the solution is correct
Solve an application problem by writing and solving an algebraic equation

In previous chapters, we translated word sentences into equations. This skill will help you when you solve word problems. Previously, you translated phrases into expressions, and now we will translate phrases into mathematical equations so we can solve them.

The first step in translating phrases into equations is to look for the word (or words) that translate(s) to the equal sign. The table below reminds us of some of the words that translate to the equal sign.

Equals (=)
is	is equal to	is the same as	the result is	gives	was	will be

Let’s review the steps we used to translate a sentence into an equation.

Translate a word sentence to an algebraic equation.

Locate the “equals” word(s). Translate to an equal sign.
Translate the words to the left of the “equals” word(s) into an algebraic expression.
Translate the words to the right of the “equals” word(s) into an algebraic expression.

In our first example we will translate and solve a one-step equation.

Example

Translate and solve: five more than [latex]x[/latex] is equal to [latex]26[/latex].

Solution:

Translate.	Five more than [latex]x[/latex] [latex]\Rightarrow\quad{x+5}[/latex] is equal to [latex]\Rightarrow\quad{=}[/latex] [latex]26[/latex] [latex]\Rightarrow\quad{26}[/latex] [latex]x+5=26[/latex]
Subtract 5 from both sides.	[latex]x+5\color{red}{-5}=26\color{red}{-5}[/latex]
Simplify.	[latex]x=21[/latex]
Check: Is [latex]26[/latex] five more than [latex]21[/latex] ? [latex]21+5\stackrel{\text{?}}{=}26[/latex] [latex]26=26\quad\checkmark[/latex] The solution checks.

TRY IT

Example

Translate and solve: The difference of [latex]5p[/latex] and [latex]4p[/latex] is [latex]23[/latex].

Show Solution

TRY it

Watch this video for more examples of how to translate a phrase into an equation, then solve it.

Translate and Solve Applications

In most of the application problems we solved earlier, we were able to find the quantity we were looking for by simplifying an algebraic expression. Now we will be using equations to solve application problems. We’ll start by restating the problem in just one sentence, then we’ll assign a variable, and then we’ll translate the sentence into an equation to solve. When assigning a variable, choose a letter that reminds you of what you are looking for.

Example

The Robles family has two dogs, Buster and Chandler. Together, they weigh [latex]71[/latex] pounds.
Chandler weighs [latex]28[/latex] pounds. How much does Buster weigh?

Show Solution

try it

Devise a problem-solving strategy.

Read the problem. Make sure you understand all the words and ideas.
Identify what you are looking for.
Name what you are looking for. Choose a variable to represent that quantity.
Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebra equation.
Solve the equation using good algebra techniques.
Check the answer in the problem and make sure it makes sense.
Answer the question with a complete sentence.

Let’s take a look at the problem-solving strategy in action.

example

Shayla paid $[latex]24,575[/latex] for her new car. This was $[latex]875[/latex] less than the sticker price. What was the sticker price of the car?

Show Solution

Now you can try translating an equation from a statement that represents subtraction.

try it

In the following video you will see another example of how to translate a phrase into an equation and solve.

Read the problem carefully.
Identify what you are asked to find, and choose a variable to represent it.	How much does Buster weigh? Let [latex]b=[/latex] Buster’s weight
Write a sentence that gives the information to find it.	Buster’s weight plus Chandler’s weight equals 71 pounds.
We will restate the problem, and then include the given information.	Buster’s weight plus 28 equals 71.
Translate the sentence into an equation, using the variable [latex]b[/latex].	[latex]b+28=71[/latex]
Solve the equation using good algebraic techniques.	[latex]b+28-28=71-28[/latex] [latex]b=43[/latex]
Check the answer in the problem and make sure it makes sense.
Is 43 pounds a reasonable weight for a dog?	Yes.
Does Buster’s weight plus Chandler’s weight equal 71 pounds?	[latex]43+28\stackrel{?}{=}71[/latex]
	[latex]71=71\quad\checkmark[/latex]
Write a complete sentence that answers the question, “How much does Buster weigh?”	Buster weighs [latex]43[/latex] pounds.

What are you asked to find?	“What was the sticker price of the car?”
Assign a variable.	Let [latex]s=[/latex] the sticker price of the car.
Write a sentence that gives the information to find it.	$24,575 is $875 less than the sticker price. $24,575 is $875 less than [latex]s[/latex].
Translate into an equation.	[latex]24,575=s-875[/latex]
Solve.	[latex]24,575+875=s-875+875[/latex] [latex]24,575=s[/latex]
Check:	Is $875 less than $25,450 equal to $24,575? [latex]25,450 - 875\stackrel{?}{=}24,575[/latex] [latex]24,575=24,575\quad\checkmark[/latex]
Write a sentence that answers the question.	The sticker price was $[latex]25,450[/latex].

Translate.	The difference of [latex]5p[/latex] and [latex]4p[/latex] [latex]\Rightarrow\quad{5p-4p}[/latex] is [latex]\Rightarrow\quad{=}[/latex] [latex]23[/latex] [latex]\Rightarrow\quad{23}[/latex] [latex]5p-4p=23[/latex]
Simplify.	[latex]p=23[/latex]
Check: [latex]5p-4p=23[/latex] [latex]5(23)-4(23)\stackrel{?}{=}23[/latex] [latex]115-92\stackrel{?}{=}23[/latex] [latex]23=23\quad\checkmark[/latex]
The solution checks.

Module 5: Multi-Step Linear Equations