Question

solve the system of equations y = 4x + 1 y = x^2 + 2x - 2 A. (-3,-13) and (1,3) B. (-3,13) and (1,-3) c. (1,3) and (-3,13) D. (-1,-3) and (3,13)

Answer 1

D. (-1,-3) and (3,13)

Explanation:

The given equation is;

`y=4x+1`

and

`y=x^2+2x-2`

Equate both equations to get;

`x^2+2x-2=4x+1`

`x^2+2x-4x-2-1=0`

`x^2-2x-3=0`

Split the middle term to obtain;

`x^2-3x+x-3=0`

Factor by grouping;

`x(x-3)+1(x-3)=0`

`(x-3)(x+1)=0`

x=-1,x=3

When x=-1, y=4(-1)+1=-3

(-1,-3)

When x=3, y=4(3)+1=13

(3,13)

Answer 2

option D

(-1,-3) and (3,13)

Explanation:

Given in the question two equations

y = 4x + 1

y = x² + 2x - 2

Equate both functions

4x + 1 = x² + 2x - 2

rearrange the x term and constant

-x² + 4x - 2x + 2 + 1 = 0

-x² + 2x + 3 = 0

factors

-x * 3x = -3x²

-x + 3x = 2x

-x² -x + 3x + 3 = 0

-x(x+1) +3(x+1) = 0

solve

(x+1)(3-x) = 0

x = -1

and

x = 3

Plug these values in equation to find y

x = -1

y = 4(-1)+ 1

y = -3

x = 3

y = 4(3)+ 1

y = 13