Transcript Algebra for College Students, 6e

§ 1.5
Problem Solving and Using Formulas
Solving Word Problems
Strategy for Solving Word Problems
STEP 1
Read the problem carefully. Attempt to state the problem in your own
words and state what the problem is looking for. Let x (or any variable)
represent one of the quantities in the problem.
STEP 2
If necessary, write expressions for any other unknown quantities in the
problem in terms of x.
STEP 3
Write an equation in terms of x that describes the verbal conditions of the
problem. (translate from English to “Math”)
STEP 4
STEP 5
Solve the equation and answer the problem’s question.
Check the solution in the original wording of the problem, not in the
equation obtained from the words.
Blitzer, Algebra for College Students, 6e – Slide #2 Section 1.5
Solving Word Problems
Study Tip
When solving word problems, particularly problems
involving geometric figures, drawing a picture of the situation
is often helpful. Label x on your drawing and where
appropriate, label other parts of the drawing in terms of x.
For some people, just hearing “word problem” is about as unsettling as hearing
words like “cancer” or “terror”. But remember – real world problems come in
words and if you are to use your algebra – you must learn to make word
problems your friends. You can do it! Read the problem slowly, draw a picture,
think of what you are looking for and call it “x”, and you will be well on your way
to solving that problem that was written in words.
Blitzer, Algebra for College Students, 6e – Slide #3 Section 1.5
Solving Word Problems
EXAMPLE
A rectangular soccer field is twice as long as it is wide. If the
perimeter of the soccer field is 300 yards, what are its
dimensions?
SOLUTION
STEP 1: Let x represent one of the quantities.
x = the width of the soccer field.
STEP 2: Represent other unknown quantities in terms of x.
Since the field is twice as long as it is wide, then
2x = the length of the soccer field.
Blitzer, Algebra for College Students, 6e – Slide #4 Section 1.5
Solving Word Problems
CONTINUED
STEP 3: Write an equation in x that describes the conditions.
The soccer field is in the shape of a rectangle and
therefore has a perimeter equal to twice the length plus
twice the width. This can be expressed as follows:
2(2x) + 2(x) = 300
Blitzer, Algebra for College Students, 6e – Slide #5 Section 1.5
Solving Word Problems
CONTINUED
STEP 4: Solve the equation and answer the question.
2(2x) + 2(x) = 300
4x + 2x = 300
6x = 300
x = 50
Multiply
Add like terms
Divide both sides by 6
Therefore the width of the soccer field is 50 yards. Also, I
therefore know that 2x = 2(50) = 100. Therefore, the length of
the soccer field is 100 yards.
Blitzer, Algebra for College Students, 6e – Slide #6 Section 1.5
Solving Word Problems
CONTINUED
STEP 5: Check the proposed solution in the original wording
of the problem.
The problem states that the perimeter of the soccer field is
300 yards. Let’s use this information to verify our answer.
The formula for the perimeter of a rectangle is repeated as
follows:
2(2x) + 2(x) = 300
2(2(50)) + 2(50) = 300
Replace x with 50
200 + 100 = 300
Multiply
300 = 300
Add
So, the dimensions of the soccer field are 50 yards by 100 yards.
Blitzer, Algebra for College Students, 6e – Slide #7 Section 1.5
Solving Word Problems
EXAMPLE – You try!
The length of an American football field is 200 feet more than
the width. If the perimeter of the field is 1040 feet, what are
its dimensions?
SOLUTION
STEP 1: Let x represent one of the quantities.
x=
STEP 2: Represent other unknown quantities in terms of x.
Blitzer, Algebra for College Students, 6e – Slide #8 Section 1.5
Solving Word Problems
CONTINUED
STEP 3: Write an equation in x that describes the conditions.
STEP 4: Solve the equation and answer the question.
Blitzer, Algebra for College Students, 6e – Slide #9 Section 1.5
Solving Word Problems
EXAMPLE
A coupon book for a bridge costs $30 per month. The toll for
the bridge is normally $5, but it is reduced to $3.50 for people
who have purchased the coupon book. Determine the number
of times in a month the bridge must be crossed so that the
total monthly cost without the coupon book is the same as the
total monthly cost with the coupon book.
SOLUTION
STEP 1: Let x represent one of the quantities.
x = number of times someone crosses the bridge in a month
Blitzer, Algebra for College Students, 6e – Slide #10 Section 1.5
Solving Word Problems
CONTINUED
STEP 2: Represent other unknown quantities in terms of x.
There are no other unknown quantities, so we can skip this
step.
STEP 3: Write an equation in x that describes the conditions.
The amount of money spent crossing the bridge without the
coupon book is 5x.
The amount of money spent crossing the bridge with the
coupon book is 30 + 3.50x (30 is not multiplied by x since it
is a fixed cost).
Blitzer, Algebra for College Students, 6e – Slide #11 Section 1.5
Solving Word Problems
CONTINUED
Since we want to know when the it begins to be profitable to
use the coupon book, we want to know when the two
methods of paying the bridge toll are equal. Therefore the
equation to be used is 5x = 30 + 3.50x.
STEP 4: Solve the equation and answer the question.
5x = 30 + 3.50x
1.5x = 30
x = 20
Subtract 3.5x from both sides
Divide both sides by 1.5
Therefore, after using the coupon book 20 times, it becomes
profitable.
Blitzer, Algebra for College Students, 6e – Slide #12 Section 1.5
Solving Word Problems
CONTINUED
STEP 5: Check the proposed solution in the original wording
of the problem.
Because the only unknown quantity is the variable
for which we are solving, it is not necessary to
check our answer. However, it is always a very
good idea to verify our calculations.
Verify the calculations on your own right now
and then say the following ten times: “I can
solve word problems!”
Blitzer, Algebra for College Students, 6e – Slide #13 Section 1.5
Solving a Formula for a Variable
• We know that solving an equation is the process of finding the
number or number that makes the equation a true statement.
Formulas contain two or more letters, representing two or
more variables. The formula for the perimeter P of a
rectangle is 2l +2w = P where l is the length and w is the
width of the rectangle. We say that the formula is solved for
P, since P is alone on one side and the other side does not
contain a P.
Blitzer, Algebra for College Students, 6e – Slide #14 Section 1.5
Solving a Formula for a Variable
To solve a formula for one of its variables, treat that variable as if it were
the only variable in the equation. Think of the other variables as if they
were just numbers.
Use the addition property of equality to isolate all terms with the specified
variable on one side.
Then use the multiplication property of equality to get the specified
variable alone. The next example shows how to do this.
Blitzer, Algebra for College Students, 6e – Slide #15 Section 1.5
Solving a Formula for a Variable
EXAMPLE
Solve the formula 2l 2w  P for l.
SOLUTION

2l + 2w = P
-2w -2w
Think of l saying, “I really want to be alone.”
Subtract 2w from both sides.
2l = P-2w
2l = P-2w
2 2
Divide both sides by 2 to isolate l.
l = P-2w
2
Blitzer, Algebra for College Students, 6e – Slide #16 Section 1.5
Solving a Formula for a Variable
EXAMPLE – You Try!
Solve the formula
SOLUTION
V  r 2 h
For h.

Blitzer, Algebra for College Students, 6e – Slide #17 Section 1.5
1.5 Assignment
p. 59 (2,4,6,8,27,28,29,47)
(50-56 even, 66,70)