Question

Solve the system of equations: y= 3x y= x^2 - 4 A. (-1, -3) and (4, 12) B.(-2, 0) and (2,0) C.(-1,3) and (4,5) D. (-2,-6) and (2,6)

Answer 1

Answer:

A. (-1,-3) and (4,12)

Step-by-step explanation:

Make both equations equal to each other and solve for the variable x:

$y=y\\3x=x^2-4\\x^2-3x-4=0$

Factor the expression. Find two numbers that when you add them you get -3 and the product of both numbers be -4. These numbers are 1 and -4. Then:

$(x+1)(x-4)=0\\\\x+1=0\\x=-1\\\\x-4=0\\x=4$

Substitute these values into one the original equations and solve for y:

$y=3(-1)\\y=-3\\\\y=3(4)\\y=12$

Therefore, you get that the solution is:

(-1,-3) and (4,12)

Answer 2

Answer:

A. (-1,-3) and (4,12)

Step-by-step explanation:

Make both equations equal to each other and solve for the variable x:

Factor the expression. Find two numbers that when you add them you get -3 and the product of both numbers be -4. These numbers are 1 and -4. Then:

Substitute these values into one the original equations and solve for y:

Therefore, you get that the solution is:

(-1,-3) and (4,12)