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matrenka [14]
3 years ago
10

Solve the following linear quadratic system of equations algebraically. y=^2+3x-2 y+3=5x

Mathematics
1 answer:
USPshnik [31]3 years ago
3 0

Answer:

x=1

y=2

Step-by-step explanation:

y=x^2+3x-2 , y+3=5x

Replace all occurrences of y in y+3=5x with x^2+3x-2.

(x^2+3x-2)+3=5x

y=x^2+3x-2

Add -2 and 3.

x^2+3x+1=5x

y=x^2+3x-2

Subtract 5x from both sides of the <u>equation.</u>

x^2+3x+1-5x=0

y=x^2+3x-2

Subtract 5x from 3x.

x^2-2x+1=0

y=x^2+3x-2

Rewrite 1 as 1^2.

x^2-2x+1^2=0

y=x^2+3x-2

Check that the middle <u>term</u> is two times the <u>product</u> of the numbers being squared in the first <u>term</u> and third <u>term</u>.

2x=2 · x · 1

y=x^2+3x-2

Rewrite the <u>polynomial</u>.

x^2-2 · x · 1 + 1^2=0

y=x^2+3x-2

Factor using the <u>perfect square</u>

<u>trinomial</u> rule a^2-2ab+b^2=(a-b)^2,

where a = x and b = 1.

(x-1)^2=0

y=x^2+3x-2

<u>Set</u> the x-1 equal to 0.

x-1=0

y=x^2+3x-2

Add 1 to both sides of the <u>equation</u>.

x=1

y=x^2+3x-2

Replace all occurrences of x in

y=x^2+3x-2 with 1.

y=(1)^2+3(1)-2

x=1

y=2

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Answer:

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Step-by-step explanation:

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