Simulatenous Equation PLEASE BEGGING HELP!

1) y=x^3 and y=x^2 +6x

The answer is x=0, y=0 and x=-2, y=-8 and x=3, y=27

How do u solve?

you start by making both values for y equal to each other, because 1y is x^3, but also x^2 + 6

so you get: x^3 = x^2 + 6x
then you bring all the x's to one side: x^3 - x^2 - 6x = 0
simplify: x(x^2 - x - 6) = 0 -> you end up with a quadratic, so you need to factorise
x[(x - 3)(x + 2)] = 0
x = 0 , x = 3 or x = -2

then you find the values for y by using the equation that would be the easiest to solve, and by replacing x by the values you found.

if x = 0
y = 0^3 = 0

if x = -2
y = -2^3 = -8

if x = 3
y = 3^3 = 27

so your answers are : x = 0 , y = o
x = 2 , y = -8
x = 3 , y = 27

hope this was helpful enough x

y = x^3
y = x^2 + 6x

substitute y = x^3 into the second equation.

x^3 = x^2 + 6x

-> x^3 - x^2 - 6x

-> x^3 + 2x^2 - 3x^2 - 6x

-> x^2 (x + 2) - 3x (x + 2)

-> (x^2 - 3x) (x + 2)

if the equation is equivalent to 0
then

(x^2 - 3x) (x + 2) = 0 (divide by (x + 2))
x^2 - 3x = 0 (divide by x)
x - 3 = 0
x = 3

again.

(x^2 - 3x) (x + 2) = 0 (divide by x^2 - 3x)
x + 2 = 0
x = -2

substitute x = 3 in y = x^3. y = 3^3 = 27
" x = -2 in y = x^3. y = (-2)^3 = -8

when x = 0, y = 0 in both equations.

there you go.

x = -2, 0, 3
y = -8, 0, 27

y=x^3 and y=x^2 +6x

plug in y=x^3 to y=x^2 + 6x with respect to y

x^3 = x^2 + 6x
x^3 - x^2 - 6x = 0
x(x^2 - x - 6) = 0
x = 0 (x+2)(x-3)=0
.........x+2=0 x-3=0
........x=-2 x=3

now solving for y

y=x^3

when x=0
y= 0^3
y=0

when x=-2
y = (-2)^3
y = -8

when x=3

y=(3)^3
y=27

there fore the points are (0,0),(-2,-8),(3,27).. answer//

x^3=x^2+6x
x^2-x^2-6x=0
x(x-3)(x+2)=0, x=0,y=0
x=3,y=27
x=-2,y=-8
God bless you.