Common Algebra Word Problems Made Easy!
(Consecutive Integer, Triangles, Mixed
Interest, D=RT, Mixtures, etc.)
Mr. Martin, 8th Grade Algebra (©
2004-2007 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 AÕs rate was 40
mph. Train BÕs rate was x + 10 =
50 mph.