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Zigmanuir [339]
2 years ago
8

How to solve the following system y=(1/2)x^2+2x-1 and 3x-y=1

Mathematics
1 answer:
ruslelena [56]2 years ago
7 0

Answer:

(0,-1),(2,5)

Step-by-step explanation:

3x-y=1

0.5x^{2}+2x-1=y

Substitute the equation of y into the other equation

3x-(0.5x^{2}+2x-1)=1

3x-0.5x^{2}-2x+1=1

collect like terms

-0.5x^{2}+x=0

Solve for x

x(-0.5x+1)=0

0=-0.5x+1

-1=-0.5x

x=2 x=0

Back substitute into the other original equation and solve for y

3(2)-y=1

6-y=1

y=5

(2,5)

3(0)-y=1

-1(-y)=-1(1)

y=-1

(0,-1)

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Answer:

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Step-by-step explanation:

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