Math 20-1 Chapter 8 Systems of Equations
8.2 Solving Systems Algebraically
Teacher Notes
8.2 Solve a Linear-Quadratic System of Equations Algebraically
Solve the following system of equations.
y  5  x2
y  x  15
Substitution Method
Substitute x – 15 for y in the quadratic equation and simplify
x  15  5  x 2
x 2  x  20  0
( x  5)( x  4)  0
x  5 or x  4
Substitute to determine the corresponding values of y.
y  x  15
For x = –5
For x = 4
y  (–5)  15
y  –20
y  (4)  15
y  –11
The two solutions are (–5, –20) and (4, –11).
8.2.1
Solve a Linear-Quadratic System of Equations Algebraically
Verify the solutions (–5, –20)
and (4, –11).
y  5  x2
y  x  15
For (–5, –20)
For (4, –11)
y  5  x2
y  5  x2
20  5  (–5) 2
11  5  (4) 2
11  5  16
20  5  25
20  20
11  11
y  x  15
20  (5)  15
y  x  15
11  (4)  15
20  20
11  11
Both solutions are correct.
The two solutions are (–5, –20) and (4, –11).
8.2.2
Solve a Linear-Quadratic System of Equations Algebraically
Solve the following system of equations.
y  x 2  3x
x  y  3
Elimination Method
• Align the terms in the same degree.
• Since the quadratic term is in the
variable x, eliminate the y-term.
• Solve the resulting equation.
y  x 2  3x
– (y 
x  3)
Subtract to
eliminate the
y-term
0  x 2  2x  3
0  x 2  2x  3
0  ( x  3)( x  1)
• Determine the corresponding
values of y.
x  3
y  x 3
y  3  3
y 0
x 1
y  x 3
y  1 3
y 4
x  3 or x  1
The two solutions are
(–3, 0) and (1, 4).
8.2.3
Solve a Linear-Quadratic System of Equations Algebraically
Verify the solutions (–3, 0) and
(1, 4).
For (–3, 0)
y  x 2  3x
x  y  3
For (1, 4)
y  x 2  3x
y  x 2  3x
0  (–3) 2  3(3)
4  (1) 2  3(1)
0  99
00
4  1 3
44
x  y  3
3  0  3
x  y  3
1  4  3
3  3
3   3
Both solutions are correct.
The two solutions are (–3, 0) and (1, 4).
8.2.4
Your Turn
Solve the system:
y  x 2  3x  4
y   2x  4
0  x 2  5x
0  x 2  5x
0  x ( x  5)
y  x 2  3x  4
2x  y  4
x0
2(0)  y  4
y  4
y  4
or
x 5
2(5)  y  4
10  y  4
y 6
x  0 or x  5
The two solutions are
(0, –4) and (5, 6).
8.2.5
Solve a Linear-Quadratic System of Equations Algebraically
A main support cable of a suspension bridge hangs in the shape of a parabola
modeled by the equation y = .25x2 -10x + 100, where x represents the number of
feet from its left-most support and where y represents the number of feet the cable
is above the road deck for any given x value. A surveyor’s line of sight is shown in
the diagram below.
(a) Write an equation for the line of sight in y = mx + b form. (Hint – The line of sight
goes through the origin and (40,100) .)
(b) Find the coordinates of the point where the line of sight first intersects the cable,
point P, by solving the system of equations consisting of y = .25x2 -10x + 100 and
your linear equation from part (a).
8.2.6
Solve a Linear-Quadratic System of Equations Algebraically
a) Slope of the line of sight is
100  0
40  0
5

2
 2.5
m
b) Solve the system
y-intercept is 0
The equation of the line of sight is
y = 2.5x
y  .25x 2  10 x  100
y  2.5x
y  .25x 2  10 x  100
y
2.5 x
0  .25 x 2  12.5 x  100
0  .25x 2  12.5x  100
0  ( x  10)( x  40)
x  10 or x  40
0  x 2  50 x  400
x  10 y  25
x  40 y  100
The line of sight intersects the cable at the point (10, 25).
8.2.7
Solve a Quadratic-Quadratic System of Equations Algebraically
Solve the following system of equations.
y  2x 2  2x  3
y  x 2  5x  7
y  2x 2  2x  3
y  x 2  5x  7
0  x 2  7 x  10
0  x 2  7 x  10
x 5
0  ( x  5)( x  2)
y  2x 2  2x  3
y  2x 2  2x  3
x  5 or x  2
y  2(5) 2  2(5)  3
y  2( 2) 2  2( 2)  3
y  50  10  3
y 843
y  43
y 7
or
x2
The two solutions are (5,43) and (2, 7).
8.2.8
Solve a Quadratic-Quadratic System of Equations Algebraically
Verify the solutions (5, 43) and
(2, 7).
y  2x 2  2x  3
y  x 2  5x  7
For (5, 43)
For (2, 7)
y  2x 2  2x  3
y  2x 2  2x  3
43  2(5) 2  2(5)  3
43  50  10  3
43  43
7  2(2) 2  2(2)  3
7  843
77
y  x 2  5x  7
y  x 2  5x  7
7  (2) 2  5(2)  7
7  4  10  7
77
7  (2) 2  5(2)  7
7  4  10  7
77
Both solutions are correct.
The two solutions are (5, 43) and (2, 7).
8.2.9
Solve a Quadratic-Quadratic System of Equations Algebraically
6 x 2  x  y  1
Your Turn
4 x 2  4 x  y  6
Solve the system:
6 x 2  x  y  1
4 x 2  4 x  y  6
2 x 2  3x
5
x
2 x 2  3x  5
2 x 2  3x  5  0
(2 x  5)( x  1)  0
2 x  5  0 or x  1  0
5
x   or x  1
2
5
2
x 1
2
 5  5
6         y  1
 2  2
 25  5
6     y  1
 4  2
40  y  1
6 1  1  y  1
2
6  1  y  1
5  y  1
y 6
y  41
 5


,
41
The two solutions are (1, 6) and 
.
 2

8.2.10
Problem Solving:
Solve the following system of equations.
4 x 2  y  2 x  5
3x 2  4 y  46 x  37  0
Problem Solving:
Solve the system.
1 2
y  x  4 x  12
2
2
y  2 x  12 x  23
Problem Solving: Determine the values of m and n if (3, 4) is a solution
to the following system of equations.
mx 2  y  32
mx 2  5 y  n
Suggested Questions:
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