Chapter Twenty-Three — Word Problems
The Humongous Book of Algebra Problems
516
Determining Unknown Values
Integer and age problems
23.1 What two integers have a sum of 64, if one of them is three times the other?
Let x be the smaller of the two integers. The other integer is three times as
large, so it is equal to 3x.
x + 3x = 64
Solve the equation for x.
Recall that x represents the smaller of the two integers. The second integer is
3x = 3(16) = 48. The two integers described by the problem are 16 and 48.
23.2 What two consecutive integers have a sum of 85?
Let x be one of the integers and x + 1 be the other. Create an equation stating
that the sum of the integers is 85.
x + (x + 1) = 85
Solve the equation for x.
The integers are x = 42 and x + 1 = 42 + 1 = 43.
23.3 What two consecutive odd counting numbers have a product of 1,443?
Let x be one of the unidentified numbers and let x + 2 be the other. Create an
equation stating that the product of the numbers is 1,443.
x(x + 2) = 1,443
Distribute x.
x
2
+ 2x = 1,443
Subtract 1,443 from both sides of the quadratic equation.
x
2
+ 2x – 1,443 = 0
The problem
says that the sum
of the numbers is 64,
so add the numbers (x
and 3x) together and
set the sum equal
to 64.
Consecutive
integers are right
next to each other
on the number line,
like 4 and 5 or –12
and –11.
The next
consecutive
integer after x is
x + 1. For example, if
x = 9, then the next
consecutive integer
is 9 + 1 = 10.
The counting
numbers are 1, 2,
3, 4, 5, ..., basically
the positive integers.
Negative integers are
excluded, as is zero.
To go from
one odd number to
the next, you have to
skip over the even number
between them, so add 2
to x instead of 1.
Chapter Twenty-Three — Word Problems
The Humongous Book of Algebra Problems
517
Apply the quadratic formula to solve for x.
Evaluate both values of x.
Discard the solution x = –39 because it is not a counting number. The odd
counting numbers are x = 37 and x + 2 = 37 + 2 = 39.
23.4 The product of x = 8 and an unknown integer y is 29 less than the sum of x and
y. Calculate y.
The product of x and y is xy, and the sum of x and y is x + y. Construct an
equation stating that the product is 29 less than the sum.
xy = x + y – 29
Substitute x = 8 into the equation and solve for y.
The problem
states that the
numbers you’re looking
for are counting
numbers, which
means they’ve got
to be positive.
Chapter Twenty-Three — Word Problems
The Humongous Book of Algebra Problems
518
23.5 What two positive integers have a sum of 25 and a product of 154?
Let x be one of the positive integers and y be the other. The sum of the numbers
is 25, so x + y = 25. The product of the numbers is 154, so xy = 154. Thus, the
solution to the problem is the solution to the system of equations.
To solve the system by substitution, solve the first equation for x.
Substitute x = 25 – y into the second equation of the system.
Distribute y.
25y – y
2
= 154
Set one side of the equation equal to zero.
y
2
– 25y + 154 = 0
Apply the quadratic formula to solve the equation.
Evaluate both values of x.
The positive integers are 11 and 14.
You’ve
got two
different
equations, and
unlike Problems
23.1–23.4, both
of the equations
contain two
variables. To solve
the problem, you
need to nd an
x and a y that
make both
of the
equations
true.
Multiply
25y – y
2
= 154 by
–1 and add 154 to
both sides to get
this equation.
Chapter Twenty-Three — Word Problems
The Humongous Book of Algebra Problems
519
23.6 Six years ago, Nick was three times as old as Erin. Two years later, Nick was
twice as old as Erin. How old is Erin now?
Let x be Nick’s current age and let y be Erin’s current age. Six years ago, Nick’s
age was x – 6 and Erin’s age was y – 6. At that time, Nick was three times as old
as Erin.
Two years later (four years ago), Nick’s age was x – 4 and Erin’s age was y – 4. At
that time, Nick was twice as old as Erin.
The solution to this problem is the solution to the system of equations
describing Nick and Erin’s ages four and six years ago.
Solve the system by elimination.
Erin is currently eight years old.
This reads
“x – 6” (Nick’s age
six years ago) “=” (is)
“3(y – 6)” (three times
Erin’s age six years
ago). Distribute the 3
and then put the linear
equation in standard
form (x and y left of
the equal sign, constant
on the right side, and a
positive x-term).
Multiply
the
equation
x – 2y = –4 by
–1 and then add
the equations
together to get
–y = –8. To solve
for y, multiply
both sides
by –1.
Chapter Twenty-Three — Word Problems
The Humongous Book of Algebra Problems
520
23.7 Sara is six years older than Ted. Thirteen years ago, Sara was four times as old
as Ted. How old is Ted now?
Let x be Ted’s current age. Sara is six years older than Ted, so Sara’s age is x + 6.
Thirteen years ago, Ted’s age was x – 13 and Sara’s age was (x + 6) – 13 = x – 7.
At that time, Sara was four times as old as Ted.
Ted is currently 15 years old.
23.8 In ten years, Mel’s age will be five less than twice Alice’s age. If their current
ages total 65, how much older is Mel?
Let x be Mel’s current age and y be Alice’s current age. In ten years, Mel’s age
will be x + 10 and Alice’s will be y + 10. At that time, Mel’s age will be five less
than twice Alice’s age.
Construct an equation stating that the sum of Mel’s and Alice’s current ages
is 65.
x + y = 65
The solution to this problem is the solution to the system of equations.
Use as
few variables
as possible.
Instead of saying
that Sara’s age is y,
you can say her age is
x + 6 (six years older
than Ted). You couldn’t
do that in Problem 23.6
because Nick and
Erin’s CURRENT
ages were never
compared.
This says “x – 7”
(Sara’s age 13
years ago) “=”
(is) “4(x – 13)” (four
times Ted’s age 13
years ago). That 7 is
there because you’re
subtracting 13 from
Sara’s age (just like you
did with Ted’s) and
her age started
out as x + 6.
This reads
“x + 10” (Mel’s
age in ten years)
“=” (is) “2(y + 10) – 5”
(ve less than two
times Alice’s age in
ten years). Distribute
the 2 and put the
linear equation in
standard form.
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