solving systems of equations?

Question

how do I solve this with substitution? I cant figure it out. The two equations are 3x + y = 2 and x^3 -2 + y = 0 any help is appreciated...

Answer 1

3x + y = 2
y = 2 - 3x ---(1)

x^3 - 2 + y = 0
from (1)
x^3 - 2 + (2 - 3x) = 0
x^3 - 3x = 0
x(x^2 -3) = 0
x (x + sqrt(3)) (x - sqrt(3)) = 0
x = 0, sqrt(3), -sqrt(3)

x = 0, y = 2
x = sqrt(3), y = 2 - 3sqrt(3)
x = -sqrt(3), y = 2 + 3sqrt(3)

Answer 2

(1) 3x + y = 2

(2) (x^3) - 2 + y = 0


solve (1) for x or y, lets try solving for y, then,

3x + y = 2 becomes y = 2 - 3x


substituting this equation into (2) we get

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

(x^3) - 3x = 0

Answer 3

3x + y = 2, y = - 2x + 3
x^3 -2 + y = 0
x^3 -2 - 2x + 3 = 0
x^3 - 2x + 1 = 0
(x - 1)(x^2 + x - 1) = 0
x = 1, (- 1 + √5)/2, (- 1 - √5)/2
x = 1, 0.61803, -3.2361

Answer 4

so i would do this:
switch around the equation of x^3-2+y=0 to be y=-x^3+2
then do this:
3x+(-x^3+2)=2
then:
3x-x^3+2=2
then:
0=x^3-3x-2+2
then:
0=x^3-3x
then factor out an x:
0=x(x^2-3)
so then you just complete the factoring. hope this helped!

Answer 5

3x + y = 2 (-)
x³ + y = 2

-3x - y = -2
x³ + y = 2
--------------
-3x + x³ = 0
x(-3 + x²) = 0
x = 0
x² = 3
x = \/3; x = -\/3; x = 0
For x = 0; y = 2 ==> (0, 2)
For x = \/3; y = 3 ==> (\/3, 3)
For x = -\/3; y = -3 ==> (-\/3, 3)
::==::

Answer 6

3x + y =2
y = 2 - 3x
x^3 - 2 + y = 0
x^3 - 2 + (2 - 3x) = 0
x^3 -3x = 0
x ( x^2 - 3) = 0
x= 0 or
x^2 =3 x =+ (3)^0.5 or x = - (3)^0.5 cant make square root:)