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2 years ago

What are the solutions to the following system of equations? Select the correct answer below.

2 answers:

5 0

(a). 8x + y = 20, where (b). y = x2 - 2x + 4
So where y is in equation (a), substitute in the value of y in (b).
8x + (x2 - 2x + 4) = 20
8x + (4) = 20
Get rid of the 4 on both sides (since what you do to one side, you must do to the other)
8x + 4 - 4 = 20 - 4
8x = 16
Get rid of the 8 to leave x on its own
8x ÷ 8 = 16 ÷ 8
x = 2
So now that you have x, find y but substituting in the value you just found (2) into one of the starting equations.
8x + y = 20 where x is 2.
8(2) + y = 20
16 + y = 20
Minus 16 from both sides
y = 20 - 16 = 4
So x is 2 and y is 4

MariettaO [177]2 years ago

4 0

Answer:

Solution of the equation is (2,4) and (-8,84) .

Step-by-step explanation:

As given the equation

y = x² - 2x + 4

8x + y = 40

Putting y = x² - 2x + 4 in 8x + y = 40 .

8x + x² - 2x + 4 = 40

x² + 6x + 4 = 20

x² + 6x + 4 -20 = 0

x² + 6x - 16 = 0

x² + 8x - 2x - 16 = 0

x(x+8)-2(x+8) = 0

(x+8)(x-2) = 0

x = -8 , x = 2

Put x = -8 in the equation 8x + y = 20 .

8 × -8 + y = 20

-64 + y = 20

y = 20 + 64

y = 84

Put x = 2 in the equation 8x + y = 20 .

8 × 2 + y = 20

16 + y = 20

y = 20 - 16

y = 4

Therefore solution of the equation is (2,4) and (-8,84) .

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Quotient rule:

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