9.7 Solving Quadratic
Systems
Algebra II
Solving Systems
(Linear or Quadratic)
• Graphing
• Substitution
• Linear combination
# of possible solutions
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No Solution
1 solution
2 solutions
3 solutions
4 solutions
Ex. 1 Find the points of
intersection
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x2 + y2 = 13 & y = x + 1
We will use….. Substitution.
x2 + (x + 1)2 = 13
Now plug these values into the
x2 + (x2 + 2x + 1) = 13 Equation to get y!!
2x2 + 2x – 12 = 0
(-3,-2) and (2,3) are the points
2(x2 + x – 6) = 0
where the two graphs intersect.
2(x + 3)(x – 2) = 0
Check it on your calculator!
x = -3 & x = 2
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Ex. 2 Solve by substitution:
x2 + 4y2 – 4 = 0
-2y2 + x + 2 = 0
The second equation has no x2 term so solve for x →
x = 2y2 – 2 and substitute it into the first equation.
(2y2 – 2)2 + 4y2 - 4 = 0
4y4 – 8y2 + 4 + 4y2 – 4 = 0
Now plug these x values into
4
2
4y – 4y = 0
The revised equation
2
2
4y (y – 1) = 0
4y2 (y-1)(y+1) = 0
Which gives you :
(-2,0) (0,1) (0,-1)
y = 0, y = 1, y = -1
Ex. 3 Linear combination
• x2 + y2 – 16x + 39 = 0
• x2 – y2 – 9 = 0
• If you add these two equations together, the
y’s will cancel
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x2 + y2 – 16x + 39 = 0
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x2 – y2
- 9 =0
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x2 + y2 – 16x + 39 = 0
x2 – y2
- 9 =0
2x2
– 16x + 30 = 0
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2(x2 – 8x + 15) = 0
2 (x-3) ( x-5) = 0
x = 3 or x = 5
• Plugging these into one of the ORIGINAL
equations to get: (3,0) (5,4) ( 5,-4)
Your turn!
• Find the points of intersection of:
• x2 + y2 = 5 & y = -x + 3
• (1,2) & (2,1)
Solving Quadratic Systems
by Graphing
• You may use a graphing calculator to check
your work though you need to solve each
equation for y first.
Assignment