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# Algebra/Quick Algebra Question(12.7-2)

Question
y = x^2 + 5x + 4
y = 8x + 8

y = x^2 + 5x + 4
and
y = 8x + 8

so

x^2 + 5x + 4 = 8x + 8

x^2 - 3x - 4 = 0

(x-4)(x+1) = 0

x = 4 or x = -1

if x = 4  , substitute 4 for x in y = 8x + 8

y = 40

if x = -1 , substitute -1 for x in y = 8x + 8

y = 0

so the solutions to the system are

x=4 and y=40  , x=-1 and y=0

Algebra

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#### Socrates

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