| Lesson
Page |
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Math
A |
Solving
a Linear Quadratic System Algebraically |
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| Example
2 |
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Solve
algebraically:
x2 + y2 = 26
x - y = 6 |
| First
we solve for x in the linear
equation. |
x
- y = 6
x = y + 6
|
Add y to
both sides |
Now
substitute this value
of x
into the quadratic
equation
replacing the
x.
Solve the
resulting equation. |
x2
+ y2 = 26
(y+6)2 + y2 = 26
y2+12y+36+y2=26
2y2+12y+36 = 26
2y2+12y+10=0
y2
+ 6y + 5 = 0
(y+5)(y+1)=0
y+5=0 y+1=0
y= -5 y= -1 |
Expand (y+6)2
Combine
similar terms.
Divide
each term by 2.
Factor.
Set
each factor = 0.
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| Next,
find the values of y by substituting in the linear
equation.
|
| x
- y = 6 |
|
x
- (-5) = 6
x + 5 = 6
x = 1* |
(1,-5) |
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| x
- y = 6 |
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x
- (-1) = 6
x + 1 = 6
x = 5* |
(5,-1) |
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*
Remember to write the x-values first
in the ordered pairs. |
|
Solution
Set
{(1,-5),(5,-1)} |
| CHECK: x2 + y2 = 26
x - y = 6
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|
Check:
(1,-5) - plug into the two equations |
(1)2 +
(-5)2 = 26
1 + 25 = 26
26 = 26 Check |
(1) - (-5) = 6
1 + 5 = 6
6 = 6 Check |
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| Check:
(5, -1) - plug into the two equations |
(5)2 +
(-1)2 = 26
25 + 1 = 26
26 = 26 Check |
(5) -
(-1) = 6
5 + 1 = 6
6 = 6 Check |
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