Chapter 8 Review Section 8.1
1. Verify that (1, 3) and (4, 0) are solutions to the
following system of equations.
x2  4x  y  0
xy40
2. Use the graph to solve the system of equations. Then,
write the system of equations represented in each graph
a)
5. When the cost to produce n items is equal to the
revenue from selling n items, this is called the
breakeven point. If the cost is $100 plus a variable cost,
the function is C(n)  100  (2  0.01n)n. The selling
price is $2.50 per unit. The revenue function is R(n) 
2.50n. Determine the breakeven point graphically, to
the nearest whole number of units.
6. The ages of Max and his father add up to 35 years.
Max’s father’s age is the same as five more than the
square of Max’s age.
a) Write a system of equations to represent this
situation. Define your variables.
b) Solve the system graphically. Are all possible
solutions meaningful? Explain.
c) How old are Max and his father?
Section 8.2
1.Verify that (1, 11) and (2, 5) are solutions to the
following system of equations.
2x  y  9
2x2  4x  y  5
b)
2. Verify that (1,  4) is a solution to the following
system of equations.
y  x2  2x  3
y  x2  2x  5
3. Solve each system of equations by substitution. Verify
your solutions.
a) y  2x  1
y  x2  5x  13
c) y  x2  3x  14
y  3x2  5x  18
b) 3x  y  4  0
2x2  4x  y  2  0
d) 4x  y  5  x2
x2  5x  2y
4. Solve each system of equations by elimination.
3. Solve each system of equations by graphing. Express
answers to the nearest whole units. Verify your
solutions.
a) x2  4x  3y  5
x2
c) 0  x  2y  10
y  1(x  3)2  4
b) y  (x  2)(x  7)
yx7
d) 2x2  5x  y  1
7x  y  1
4. Solve each system of equations by graphing. Express
answers to the nearest hundredth.
a) x2  8x  y  12
x2  y  8
b) y  2x2  x  1
y  x2  9x  8
c) y  5x2  10x  5
y  x2  3x  10
d) y  3(x  4)2  2
y  2(x  3)2  2
a) 3x2  x  3y  8
x  3y  9
b) y  2x2  x  1
2y  2x2  x  1
c) x  6y  12
d) x2  y  4x  5
1 2 5
x  x y2
2
3
5x 
1
y  x2
3
5. Solve each system of equations algebraically. Round
answers to the nearest hundredth.
1
3
a) y  x 2 
2
x
3
3y  2x2  3x  1
b) x2  5x  y  6
2x2  x  y  3
6. Consider the following system of equations.
x2  6x  y  k  0
3x  y  k  0
a) Determine the value of k if a solution is (3, 2).
b) Determine the second solution.
7. Consider the following system of equations.
c) no solution
d) (1, 8) and (0, 1)
y  x  2x  3
2
yk
Determine the value of k, if the system has
a) two solutions b) one solution
c) no solution
8. A parabola’s vertex is at (4, 4) and one of its xintercepts is at (6, 0). A second parabola’s vertex is at
(1, 9) and its y-intercept is at (0, 8).
a) Determine the equations of the parabolas.
b) Solve the system of equations to determine the
point(s) of intersection.
4. a) (2.50, 1.75)
b) (1.00, 2.00) and (9.00, 154.00)
9. Consider the given rectangle.
c) (0.50, 11.25) and (1.67, 2.22)
d) no solution
The perimeter is equal to y, and the area is equal to 3y.
a) Determine equations to represent the perimeter and
area.
b) Solve the system of equations algebraically.
c) Are both solutions possible? Explain.
d) State the value of x, the perimeter, and the area.
Answers
5. 78 items, or $195
Section 8.1
6. a) x  Max’s age: x  y  35
y  father’s age: x2  5  y
1. Point (1, 3):
LS  x2  4x  y RS  0
LS  x  y  4
RS  0
  (1)2  4(1)  (3)
 1  (3)  4
 0
0
LS  RS
LS  RS
Therefore, point (1, 3) is a solution.
Point (4, 0):
LS  x2  4x  y RS  0
LS  x  y  4
  (4)2  4(4)  (0)
 4  (0)  4
 0
0
LS  RS
LS  RS
Therefore, point (4, 0) is a solution.
RS  0
2. a) (2, 4) and (0, 0);
y  x2  4x
y  x2
b) (1, 2) and (4, 8);
y  2x2  8x  8
y  2x
3. a) (2, 3)
b) (3, 4) and (7, 0)
6b. The two solutions to the system are (6, 41) and (5, 30).
(6, 41) is not meaningful because Max cannot be 6 years old.
c) Max is 5 and his father is 30.
Section 8.2
1.Point (1, 11):
LS  2x  y
RS  9
LS  2x2  4x  y
RS  5
  2(1)  11
 2(1)2  4(1)  11
 9
 5
LS  RS
LS  RS
Therefore, (1, 11) is a solution.
Point (2, 5):
LS  2x  y
RS  9
LS  2x2  4x  y
RS  5
  2(2)  5
 2(2)2  4(2)  5
 9
 5
LS  RS
LS  RS
Therefore, (2, 5) is a solution.
2. Point (1, 4):
LS  y
RS  x2  2x  3
  4
 (1)2  2(1)  3

 4
LS  RS
LS  y
RS  x2  2x  5
  4
 (1)2  2(1)  5

 4
LS  RS
Therefore, (1, 4) is a solution.
3. a) (3, 7) and (4, 9)
Verify:
Point (3, 7)
LS  y
RS  2(3)  1
 7
7
LS  RS
LS  y
RS  (3)2  5(3)  13
 7
7
LS  RS
Therefore, (3, 7) is a solution.
Point (4, 9)
LS  y
RS  2(4)  1
 9
9
LS  RS
LS  y
RS  (4)2  5(4)  13
 9
9
LS  RS
Therefore, (4, 9) is a solution.
3 17
b)  ,  and (2, 2)
 2 2
Verify:
3 17
Point  , 
 2 2
2
LS  (3) 3  17  4

2
2

 3 9  b) no solution
1 63
c) (0, 2) and (3, 1.5) d)  ,  and (5, 0)
4 16


10
1 26
4. a) 1,
and ,
RS  0
18
  24  34  8
4
4
4
4
0
LS  RS
3 17
Therefore,  ,  is a solution.
 2 2
Point (2, 2):
LS  3(2)  (2)  4
RS  0
0
LS  RS
LS  2(2)2  4(2)  (2)  2
RS  0
0
LS  RS
Therefore, (2, 2) is a solution.



RS  0
0
LS  RS
 2   (4)  23   172   2
d) (5, 0) and (2, 7)
Verify:
Point (5, 0)
LS  4(5)  0  5
RS  52
  25
 25
LS  RS
LS  52
RS  5(5)  2(0)
  25
 25
LS  RS
Therefore, (5, 0) is a solution.
Point (2, 7)
LS  4(2)  7  5 RS  (2)2
 4
4
LS  RS
LS  (2)2
RS  5(2)  2(7)
 4
4
LS  RS
Therefore, (2, 7) is a solution.

9 17
 4
2
2
LS  (2) 3
c) (4, 10) and (2, 4)
Verify:
Point (4, 10)
LS  y
RS  (4)2  3(4)  14
  10
 10
LS  RS
LS  y
RS  3(4)2  5(4)  18
  10
 10
LS  RS
Therefore, (4, 10) is a solution.
Point (2, 4)
LS  y
RS  (2)2  3(2)  14
 4
4
LS  RS
LS  y
RS  3(2)2  5(2)  18
 4
4
LS  RS
Therefore, (2, 4) is a solution.
3
5. a) (3, 18) b) (1.62, 0.21) and (0.62, 0.54)
6. a) k  7 b) (0, 7)
7. a) k   4 b) k   4 c) k   4
8. a) y  1(x  4)2  4 and y  (x  1)2  9
b) (2, 0) and (1, 5)
9. a) perimeter: 2(3x)  2(x  5)  y; area: (3x)(x  5)  3y
b) (5, 50) and (2, 6)
c) The only possible solution is (5, 50). You cannot have a
negative perimeter or area.
d) x  5; perimeter  50; area  150 units2