PREC11
Section 8.1 Extra Practice (optional)
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)
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
b) y  (x  2)(x  7)
yx7
c) 0  x  2y  10
y  1(x  3)2  4
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
b)
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?
8.1 Key
1. Point (1, 3):
LS  x2  4x  y
RS  0
  (1)2  4(1)  (3)
 0
LS  RS
LS  x  y  4
RS  0
  1  (3)  4
 0
LS  RS
Therefore, point (1, 3) is a
solution.
Point (4, 0):
LS  x2  4x  y
RS  0
  (4)2  4(4)  (0)
 0
LS  RS
LS  x  y  4
RS  0
  4  (0)  4
 0
LS  RS
Therefore, point (4, 0) is a
solution.
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)
c) no solution
d) (1, 8) and (0, 1)
c) (0.50, 11.25) and (1.67,
2.22)
d) no solution
5. 78 items, or $195
4. a) (2.50, 1.75)
6. a) x  Max’s age: x  y  35
y  father’s age: x2  5  y
b)
b) (1.00, 2.00) and (9.00,
154.00)
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.
PREC11
Section 8.2 Extra Practice (optional)
1. Verify that (1, 11) and (2, 5) are solutions
to the following system of equations.
2x  y  9
2x2  4x  y  5
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.
a) y  2x  1
y  x2  5x  13
b) 3x  y  4  0
2x2  4x  y  2  0
c) y  x2  3x  14
y  3x2  5x  18
d) 4x  y  5  x2
x2  5x  2y
4. Solve each system of equations by
elimination.
a) 3x2  x  3y  8
x  3y  9
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.
y  x2  2x  3
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 x-intercepts 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.
9. Consider the given rectangle.
b) y  2x2  x  1
2y  2x2  x  1
c) x  6y  12
1 2 5
x  x y2
2
3
d) x2  y  4x  5
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
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.
7. a) k   4 b) k   4 c) k   4
8.1 Key
1. Point (1, 11):
LS  2x  y
  2(1)  11
 9
LS  RS
LS  2x2  4x  y
  2(1)2  4(1)  11
  5
LS  RS
RS  9
RS  5
Therefore, (1, 11) is a solution.
Point (2, 5):
LS  2x  y
RS  9
  2(2)  5
 9
LS  RS
LS  2x2  4x  y
RS  5
  2(2)2  4(2)  5
  5
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)
3 17
b)  ,  and (2, 2)
 2 2
c) (4, 10) and (2, 4)
d) (5, 0) and (2, 7)
 3 9  b) no solution
c) (0, 2) and (3, 1.5) d)  1 , 63  and (5, 0)
4 16


4. a) 1, 10 and 1 , 26
3
5. a) (3, 18)
b) (1.62, 0.21) and (0.62, 0.54)
6. a) k  7 b) (0, 7)
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