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Copyright
Copyright
© 2011
© 2011
Pearson
Pearson
Education,
Education,
Inc. Publishing
Inc. Publishing
as Prentice
as Prentice
Hall Hall
Chapter 4
Systems of
Equations
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
4.3
Systems of Linear
Equations and Problem
Solving
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
Example
Hilton University Drama Club sold 311 tickets for a
play. Student tickets cost $0.50 each; non-student tickets
cost $1.50 each. If total receipts were $385.50. How
many of each ticket were sold?
Solution
1. UNDERSTAND Read and reread the problem.
Suppose 200 student tickets were sold.
(311 – 200) = 111 non-student tickets sold
Receipts: 200(0.50) + 111(1.50) = $266.50
Proposed solution is incorrect.
Let s = student tickets
Let n = non-student tickets
continue
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
Hilton University Drama Club sold 311 tickets for a play. Student tickets cost
$0.50 each; non-student tickets cost $1.50 each. If total receipts were $385.50.
How many of each ticket were sold?
2. TRANSLATE.
student tickets + non-student tickets = total tickets
s
+
n
= 311
price of student + price of non-students = total receipts
$0.50s + $1.50n
= $385.50
3. SOLVE.  s  n  311

0.50s  1.5n  385.50
Multiply the second equation by 2.
continue
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
Hilton University Drama Club sold 311 tickets for a play. Student tickets cost
$0.50 each; non-student tickets cost $1.50 each. If total receipts were $385.50.
How many of each ticket were sold?
 s  n  311

0.50s  1.5n  385.50
2(0.50s  1.5n)  2(385.50)   s  3n  771
s  n  311
 s  3n  771
 2n  460
n  230
Substitute n = 230 into the first equation to determine s.
s  n  311
s  230  311
s  81
continue
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
Hilton University Drama Club sold 311 tickets for a play. Student tickets cost
$0.50 each; non-student tickets cost $1.50 each. If total receipts were $385.50.
How many of each ticket were sold?
 s  n  311

0.50s  1.5n  385.50
4. INTERPRET.
Check: Substitute s = 81 and n = 230 into both
equations.
s  n  311
81  230  311
311  81
0.5s  1.5n  385.50
0.5(81)  1.5(230)  385.50
40.50  345  385.50
385.50  385.50
State: There were 81 student tickets and 230 nonstudent tickets sold for the play.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
Example 2
One number is 4 more than twice the second number.
Their total is 25. Find the numbers.
Solution
1. UNDERSTAND Read and reread the problem. If
the second number is 5. The first number is 4 more than
twice the second number, (4 + 2•5) = 14.
Is the total 25? No: 14 + 5 = 19. Our proposed
solution is incorrect, but we have a better
understanding of the problem.
Since we are looking for two numbers, we let
x = first number
y = second number
continue
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
2. TRANSLATE. Translate the given facts.
Words: One number is 4 more than twice the second
Translate:
x
= 4 + 2y
Second statement:
Words:
Their total is 25.
Translate:
x + y = 25
3. SOLVE. Solve the system.  x  4  2 y

 x  y  25
Since the first equation is solved for x, we will use
substitution.
continue
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
x  4  2 y

 x  y  25
x  y  25
4  2 y  y  25
3 y  4  25
3 y  21
y7
Replace y with 7 in the equation x = 4 + 2y and
solve for x. x  4  2 y
x  4  2(7)
x  18
The ordered pair solution of the system is (18, 7).
continue
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
4. INTERPRET Since the solution of the system is
(18, 7), then the first number we are looking for is
18 and the second number is 7.
Check: 18 is 4 more than twice 7, the sum of the
two number is 25.
State: The numbers are 18 and 7.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
Example 3
Terry Watkins can row about 10.6 kilometers in 1 hour
downstream and 6.8 kilometers upstream in 1 hour. Find
how fast he can row in still water, and find the speed of
the current.
Solution
1. INTERPRET. Read and reread the problem. Use the
formula d = rt (distance = rate  time). However, we
have the effect of the water current in this problem. The
rate when traveling downstream would be r + w and the
rate upstream would be r – w, where r is the speed of
the rower in still water, and w is the speed of the
current.
continue
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
Terry Watkins can row about 10.6 kilometers in 1 hour downstream and 6.8
kilometers upstream in 1 hour. Find how fast he can row in still water, and find
the speed of the current.
1. INTERPRET (continued)
Suppose Terry can row 9km/hr in still water, and the
water current is 2 km/hr. Since he rows for 1 hour in
each direction, downstream would be (r + w)t = d
or (9 + 2)(1) = 11 km
Upstream would be (r – w)t = d or (9 – 2)(1) = 7 km
Proposed solution is incorrect.
We are looking for two rates:
Let r = rate of the rower in still water
Let w = rate of the water current
continue
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
Terry Watkins can row about 10.6 kilometers in 1 hour downstream and 6.8
kilometers upstream in 1 hour. Find how fast he can row in still water, and find
the speed of the current.
2. TRANSLATE.
Rate
Time (hr)
Distance
Upstream
r+w
1
10.6
Downstream
r w
1
6.8
3. SOLVE. r  w  10.6

r  w  6.8
Use the elimination method. r  w  10.6
r  w  6.8
2r  17.4
r
 8.7
continue
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
Terry Watkins can row about 10.6 kilometers in 1 hour downstream and 6.8
kilometers upstream in 1 hour. Find how fast he can row in still water, and find
the speed of the current.
3. SOLVE. Since r = 8.7 substitute into equation 1 and
determine w.
r  w  10.6

r  w  6.8
8.7  w  10.6
w  1.9
4. INTERPRET.
Check: Check that r + w = 10.6 and r – w = 6.8, both
equations check.
State: Terry’s rate in still water is 8.7 km/hr and the
rate of the water current is 1.9 km/hr.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
Example 4
The Candy Barrel shop manager mixes candy worth $2
per pound with trail mix worth $1.50 per pound. Find
how many pounds of each she should use to make 50
pounds of a party mix worth $1.80 per pound.
Solution
1. INTERPRET. Read and reread the problem.
To find the cost of any quantity of items we need to use
the formula
price per unit  number of units = price of all units
Suppose the manager decides to mix 20 pounds of
candy. Since the total mixture will be 50 pounds, we
need 30 pounds of trail mix.
continue
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
The Candy Barrel shop manager mixes candy worth $2 per pound with trail
mix worth $1.50 per pound. Find how many pounds of each she should use to
make 50 pounds of a party mix worth $1.80 per pound.
candy
$2

20 pounds =
$40
trail mix
$1.50 
30 pounds =
$45
mixture
$2.80 
50 pounds =
$90
Since $40 + $45 ≠ $90 our proposed solution is
incorrect.
We are looking for two quantities:
Let x = the amount of candy
Let y = the amount of trail mix
continue
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
The Candy Barrel shop manager mixes candy worth $2 per pound with trail
mix worth $1.50 per pound. Find how many pounds of each she should use to
make 50 pounds of a party mix worth $1.80 per pound.
Use a table to organize the given data.
 x  y  50

2 x  1.5 y  90
Number of
pounds
Cost per
pound
Total cost
candy
x
$2
2x
trail mix
y
$1.50
1.5y
total
50
$1.80
$90
2. Translate.
Words: amount of
candy
x
Words: price of candy
2x
+
+
+
+
amount of = 50
trail mix
y
= 50
price of mix = (1.8)(50)
1.5y
= 90
continue
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
The Candy Barrel shop manager mixes candy worth $2 per pound with trail
mix worth $1.50 per pound. Find how many pounds of each she should use to
make 50 pounds of a party mix worth $1.80 per pound.
3. SOLVE.  x  y  50
2 x  1.5 y  90
Use elimination. Multiply equation (1) 3 x  3 y  150
4 x  3 y  180
by 3 and equation (2) by 2.
 x  30
x  30
Substitute x = 30 into equation (1).
x  y  50
30  y  50
y  20
continue
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
The Candy Barrel shop manager mixes candy worth $2 per pound with trail
mix worth $1.50 per pound. Find how many pounds of each she should use to
make 50 pounds of a party mix worth $1.80 per pound.
4. INTERPRET.  x  y  50

2 x  1.5 y  90
Check:
x  y  50
30  20  50
50  50
2 x  1.5 y  90
2(30)  1.5(20)  90
60  30  90
90  90
State: The store manager needs to mix 30 pounds of
candy with 20 pounds of trail mix to create 50 pounds
of party mix worth $1.80 per pound.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
Example 6
The measure of the largest angle of a triangle is 90
more than the measure of the smallest angle, and the
measure of the remaining angle is 30 more than the
measure of the smallest angle. Find the measure of each
angle.
Solution
1. INTERPRET. Read and reread the problem. The
sum of the measures of the angles of a triangle is 180.
Guess a solution.
Smallest angle = 30, largest angle = 90 + 30 or 120
remaining angle = 30 + 30 = 60
The sum of the three angles = 30 + 120 + 60 = 210, too
large.
continue
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
The measure of the largest angle of a triangle is 90 more than the measure of
the smallest angle, and the measure of the remaining angle is 30 more than
the measure of the smallest angle. Find the measure of each angle.
Let x = smallest angle
Let y = remaining angle
Let z = largest angle
2. TRANSLATE.
x + y + z = 180 (sum of angles in triangle)
z = 90 + x (largest angle is 90 more than smallest)
y = 30 + x (remaining angle is 30 more than smallest angle)
3. SOLVE.
 x  y  z  180

 z  90  x
 y  30  x

Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
continue
The measure of the largest angle of a triangle is 90 more than the measure of
the smallest angle, and the measure of the remaining angle is 30 more than
the measure of the smallest angle. Find the measure of each angle.
3. SOLVE. Substitute the last two equations into the
first equation.
 x  y  z  180
x  y  z  180
x  (30  x)  (90  x)  180
3x  120  180
3x  60
x  20
y  30  x
y  30  20
z  90  x
z  90  20
y  50
z  110

 z  90  x
 y  30  x

The ordered triple
is (20, 50, 110).
continue
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
The measure of the largest angle of a triangle is 90 more than the measure of
the smallest angle, and the measure of the remaining angle is 30 more than
the measure of the smallest angle. Find the measure of each angle.
4. INTERPRET.
Check: 20 + 50 + 110 = 180 The measure of the
largest angle is 90 more than the smallest.
The measure of the remaining angle is 30 more than the
smallest.
State: The measure of the smallest angle is 20, the
measure of the largest angle is 110 and the measure of
the remaining angle is 50.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall