Solving Systems of Linear and
Quadratic Equations
2x  y  8
x  y  10
Systems of Linear Equations
We had 3 methods to solve them.
Method 1 - Graphing
Solve for y.
y  2 x  8
y  x  10
(6, – 4)
2x  y  8
x  y  10
Systems of Linear Equations
Method 2 - Substitution
y  2 x  8
x x  y  10  10
x  2 x  8  10
3 x  18
x6
y y 2(
2 x ) 8
8
y  12  8
y  4
(6, – 4)
2x  y  8
x  y  10
Systems of Linear Equations
Method 3 - Elimination

2x  y  8
x  y  10
3x  18
x 6
( x)  y  10
y  4
y  4
(6, – 4)
Solve the System Algebraically
Use Substitution
y  x2 1
y  x 1
( xy2x1
) 1 x  1
x x0
x  x  1  0
x  0 or x  1  0
x  0 or x  1
2
x0
y  0 1
y 1
2
(0, 1)
x 1
y  1 1
2
(1, 2)
y2
Answer: (0,1) (1,2)
Solve the System Algebraically
Use Substitution
y  x2  2x  2
y  2 x  2
y  2x  2
2
2
( 2 xy 
 2x) x2 x2 x2 2
0  x  4x  4
0   x  2 x  2
2
x2
y  2  2   2
(2, 2)
y  4  2
y2
0  x2
x2
Answer: (2,2)
Solve the System Algebraically
Use Substitution
x 2  y 2  26
x y 6
y  x6
x   x  6   26
2
2
x 2  x 2  12 x  36  26
2 x 2  12 x  10  0
x2  6x  5  0
 x  5 x 1  0
x  5 or x  1
x5
5 y  6
y 1
y  1
x 1
1 y  6
y  5
y  5
(5, –1)
(1, – 5)
Answer: (5, –1) (1, – 5)
Solve the System Algebraically
Use Substitution
3
y x
4
2
3 
2
x   x   25
4 
9 2
2
x  x  25
16
16 x  9 x  400
25 x 2  400
2
2
x 2  16
x  4
x 2  y 2  25
4 y  3x
x4
4 y  3 4
4 y  12
y3
x  4
4 y  3  4 
(4, 3)
(– 4, – 3)
4 y  12
y  3
Answer: (4, 3) (–4, – 3)
Now let’s look at the Graphs of
these Systems!
Quadratic
What
Parabola
does
theequation
graph
Classify
each
of
each
look like?
as
linear/quadratic.
Linear
Line
Point of Intersection
(-2, 0)
What is the solution
to the system?
Point of Intersection
(1, -3 )
We have solved the following algebraically
Now use your calculator
to check it graphically.
y  x2 1
y  x 1
Answer:
(0,1) (1,2)
y  x2  2x  2
y  2 x  2
Answer:
(2,2)
Equations must be solved for y!!
x 2  y 2  26
x y 6
x 2  y 2  26
y 2  26  x 2
The first equation
can
entered as two separate entries:
Zoom
6:5:be
ZStandard
Zoom
ZSquare
y   26  x 2
and
y  x6
or a "list" notation may be used:
Answer: (5, –1) (1, – 5)