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)

option D

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

Step-by-step 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

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

Step-by-step explanation:

The given equation is;

and

Equate both equations to get;

Split the middle term to obtain;

Factor by grouping;

x=-1,x=3

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

(-1,-3)

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

(3,13)

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?