Common Algebra Word Problems Made Easy!
(Consecutive Integer, Triangles, Mixed Interest, D=RT, Mixtures, etc.)
Mr. Martin, 8^{th} Grade Algebra (© 20042007 Mark D. Martin)
Some Examples You Already Know How to Do

Name three consecutive integers that equal 39.

Second angle of a triangle is 20° more than the first. The measure of the third angle is twice the measure of the first angle. Find all three angles.

I invest $30,000 for one year. Part is invested at 2% interest per annum and the rest is invested at 3% per annum. I earn $800 after one year. How much did I invest at 2% and how much did I invest at 3%?
General Method to Solve
First, read the problem carefully. Decide what information you are given and what is being asked. Draw a diagram if applicable. Remember, there is more than one piece of information missing in all these problems. After you have done this, do the following:

Define only one variable. Let n = (describe in English what n equals)

Using the same variable, define the other unknowns.

Using the information given, write an equation.

Solve the equation for the variable.

Find the other answers by going back to the definitions you made.
Solutions to Examples
Problem 1: Three consecutive integers equal 39.
Step 1: Define variable
Let n = the first integer.
Step 2: Using the same variable, define the other unknowns.
n + 1 = the second integer.
n + 2 = the third integer.
Step 3: Write equation that sum of integers equals 39.
n + (n + 1) + (n + 2) = 39
Step 4: Solve equation
n + (n + 1) + (n + 2) = 39
3n + 3 = 39
3n = 36
n = 12
Step 5: Find the other answers by going back to the definitions you made.
n + 1 = the second integer. 12 + 1 = 13
n + 2 = the third integer. 12 + 2 = 14
Answer is 12, 13, 14
Problem 2 – Triangle –second angle 20° more than first, third angle twice the first
Step 1: Let n = the first angle.
Step 2: Let n + 20 = the second angle
Let 2n = the third angle
Step 3: n + (n + 20) + 2n = 180
Step 4: 4n + 20 = 180
4n = 160
n = 40
Step 5: second angle = n + 20 = 40 + 20 = 60
third angle = 2n = 2(40) = 80
Answer: 40°, 60°, 80°
Problem 3 – invest $30,000 for one year, part at 2%, part at 3%, earn $800
Step 1: Let x = amount invested at 2%
Step 2: $30,000 – x = amount invested at 3%
Step 3: Remember I = prt. Here, interest (I) ($800) will equal interest on amount (x) at 2% (.02) and interest on amount ($30,000 –x ) at 3% (.03). t is one year. Hence, x(.02)(1) + (30,000 – x)(.03)(1) = 800
Step 4:^{1}
x(.02)(1) + (30,000 – x)(.03)(1) = 800
.02x + 900  .03x = 800 (applying distributive property)
.01x + 900 = 800 (combining like terms)
.01x = 100 (subtracting 800 from both sides of equation)
x = $10,000 (dividing both sides by .01)
Step 5:
x = $10,000 = amount invested at 2%
$30,000 – x = $30,000  $10,000 = $20,000 = amount invested at 3%
New Types of Problems
Solve other types of problems the same way.
Distance = Rate x Time Problems
Example 1: Two trains leave Los Angeles at the same time. Train A travels north. Train B travels south. At the end of two hours they are 180 miles apart. Find the rate of both trains if Train A is traveling 10 miles per hour slower than Train B.
Preliminary Steps: On these types of problems it is helpful to draw a diagram and a chart. Also, remember that distance = rate x time. Rate is the same as speed. The other steps are the same.
Drawing and Chart

Rate

Time

Distance

Train A

x

2 hours

2x

Train B

x + 10

2 hours

2(x +10)

To San Francisco
Los Angeles
To San Diego
Step 1: Let x = the rate of train A
Step 2: x + 10 = the rate of train B
Step 3: The total distance, 180 miles, equals the distance train A went plus the distance train B went. Distance = rate multiplied by time or D = rt. t = 2 hours
2x + 2(x + 10) = 180
Step 4:
2x + 2(x + 10) = 180
2x + 2x + 20 = 180 (distributive property)
4x + 20 = 180 (combine like terms)
4x = 160 (subtract 20 from both sides)
x = 40 (Divide both sides by 4)
Step 5: Train A’s rate was 40 mph. Train B’s rate was x + 10 = 50 mph.
