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Math
A |
Solving
a Linear Quadratic System Algebraically |
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We
will now look at how to solve a linear quadratic system of
equations algebraically |
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We'll
use the following example.
Solve
algebraically:
y = x2 - x - 6
y = 2x - 2 |
First
we solve for one
of the variables in the
linear
equation. |
y
= 2x - 2
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Since
this is already done for us in this example, we can go to the
next step. |
Next
we substitute for
that variable
in the
quadratic
equation,
and
solve the resulting
equation. |
y
= x2 - x - 6
2x-2 = x2 - x - 6
2x = x2 - x - 4
0 = x2
- 3x - 4
0
=(x-4)(x+1)
x-4=0
x+1=0
x = 4 x = -1 |
Subtract
2 from both sides.
Subtract 2x from both sides.
Factor.
Set
each factor =0 and solve.
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| We
now have two values for x, but we
still need to find the corresponding values
for y. |
We
find the y-values by
substituting each value
of x into the linear
equation.
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| y
= 2x - 2 |
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y
= 2(4)
- 2
y = 8 - 2
y = 6 |
(4,6) |
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| y
= 2x - 2 |
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y
= 2(-1)
- 2
y = -2 - 2
y = -4 |
(-1,-4) |
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| Now
we have 2
possible solutions for the system: (4,6) and (-1,-4).
We need to check each solution in each equation.
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| Check#1:
(4,6) |
| y
= x2 - x - 6
6
= (4)2 - 4 - 6
6 = 16 - 4 - 6
6 = 6 it
checks !
y
= 2x - 2
6
= 2(4) - 2
6 = 8 - 2
6 = 6
it also checks ! |
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| Check#2:
(-1,-4) |
| y
= x2 - x - 6
-4
= (-1)2 - (-1) - 6
-4 = 1 + 1 - 6
-4 = -4 it
checks !
y
= 2x - 2
-4
= 2(-1) - 2
-4 = -2 - 2
-4 = -4
it also checks ! |
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| We
finally have our solution set for this linear quadratic system. |
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