Question

Solve the following system. y = (1/2)x^2 + 2x - 1 and 3x - y = 1 The solutions are (____)and (______ ) (remember to include the commas)

Answer 1

The solutions are (0, -1) and (2, 5)

Explanation:

y = (1/2)x^2 + 2x - 1 ------ eqn(I)

3x - y = 1

y = 3x - 1 ------------- eqn(II)

Equate eqn(I) & (II)

(1/2)x^2 + 2x - 1 = 3x - 1

Multiply each term by 2

x^2 + 4x - 2 = 6x - 2

x^2 + 4x - 6x = -2 + 2

x^2 - 2x = 0

x(x - 2) = 0

x = 0, 2

Substitute the values of x in eqn(II)

y = 3x - 1

When x = 0

y = 3(0) - 1 = 0 - 1 = -1

y = 3x - 1

When x = 2

y = 3(2) - 1 = 6 - 1 = 5

The solutions are (0, -1) and (2, 5)

Answer 2

x=0, 2. y=-1, 5.

Explanation:

y=(1/2)x^2+2x-1

3x-y=1

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3x-y=1

y=3x-1

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(1/2)x^2+2x-1=3x-1

(1/2)x^2+2x-3x-1-(-1)=0

(1/2)x^2-x-1+1=0

(1/2)x^2-x=0

factor out the x

x[(1/2)x-1]=0

zero property,

x=0, (1/2)x-1=0

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(1/2)x-1=0

1/2x=0+1

1/2x=1

x=1/(1/2)=(1/1)(2/1)=2/1=2

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x=0, x=2

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y=(1/2)(0)^2+2(0)-1=0+0-1=-1

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y=3(0)-1=0-1=-1

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y=(1/2)(2)^2+2(2)-1=(1/2)(4)+4-1=4/2+4-1=2+4-1=6-1=5

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y=3(2)-1=6-1=5