Question

Solve the system of equations: y= 3x y= x^2 - 4

Answer 1

Answer:

$x=-1\\y=-3\\\\x=4\\y=12$

Step-by-step explanation:

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

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

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:

$x=-1\\y=-3\\\\x=4\\y=12$