– ALGEBRA I – Unit 1 – Section 2 Translating Problems into Equations and Solutions Solving word problems can be one of the trickiest processes in Algebra

– ALGEBRA I – Unit 1 – Section 2

Translating Problems into Equations and Solutions

Solving word problems can be one of the trickiest processes

in Algebra.

The key lies in having a solid approach and plan to

attacking word problems.

– ALGEBRA I – Unit 1 – Section 2

Translating Problems into Equations and Solutions

We will be using a FIVE-STEP approach for solving word

problems in this class…

1 2 3 4 5

FIVE-STEP Problem Solving Plan

This step involves

several things:

STEP #1 – GET ORGANIZED1. Figure out what you know.2. Figure out what you DON’T

know.3. Draw a sketch (if necessary).

Getting organized is the key to success. Take the time to sort through the information and

develop a game plan.


Choose a letter that makes sense to you.

STEP #2 – PICK A VARIABLE

Pick something that reminds you of what you are solving for. For

instance, “t” may stand for time.

Write any “unknowns” in terms of your variable.


STEP #3 – WRITE AN EQUATION

Remember that if you write one equation and want to be able to solve it, you can only have

one variable in it.

GOOD EXAMPLE: w + w + 6 = 20BAD EXAMPLE: w + l = 20

FIVE-STEP Problem Solving PlanSTEP #4 – SOLVE THE

EQUATION

I think that this step is pretty self explanatory.

Until we officially start solving equations, you will be given possible

answers to guess and test with.


STEP #5 – CHECK YOUR ANSWER

There are several

things to do:

1. Check your math and labels.

2. Make sure that you are answering the question.

3. Make sure your answers make sense in the context of the problem.


Let’s recap. The five

steps are:

1. ORGANIZE

2. VARIABLE

3. EQUATION

4. SOLVE

5. CHECK

Example ProblemA rectangle is 5 inches longer than it is wide.

What are the dimensions of the rectangle if the perimeter is 38 inches? (Choices for w: 5, 7, or 8)

ORGANIZEWhat do

you know?What don’t you know?

Draw a sketch

1. Perimeter = 38 in2. Length is 5 inches

more that the width3. Add all sides to get

the perimeter

1. Length2. Width

length

wid

th

p = 38 in



VARIABLEw + 5

w p = 38 in

Since the two unknowns are length and width, make one of them your variable and write the other one in terms of that variable. Edit your sketch as you go…

Width = w Length = w + 5



EQUATIONw + 5

w p = 38 in

Write the equation that will allow you to solve for the variable.

w + w + 5 + w + w + 5 = 38



SOLVEw + 5

w p = 38 in

If you can solve the equation algebraically, the go ahead. Otherwise, guess and test with the possible choices.

w = 7 inches



CHECKw + 5

w p = 38 in

The question is asking for the dimensions. Thus, we need to know the width AND length.

w = 7 inches l = 12 inches



CHECKw + 5

w p = 38 in

Do the answers make sense? Do the sides of the rectangle add up to 38 inches? If yes, then we are done!

w = 7 inches l = 12 inches

Try This Problem…Use the problem solving plan to solve the following problem. Be

sure to show your work. Possible answers for the shorter piece have

been provided.

1. Suppose that a 2'×4' piece of lumber is originally 16 feet long, but it is cut into two pieces. The longer piece is three times as long as the shorter piece. How long is each piece? (Possible answers for short piece: 4, 5, or 6)

**The answers can be found at the end of the PowerPoint.

ALGEBRA IS FUN

AND EASY!**Answers: 1) 4 feet and 12 feet

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– ALGEBRA I – Unit 1 – Section 2 Translating Problems into Equations and Solutions Solving word problems can be one of the trickiest processes in Algebra