Answer
(0, 2) and ($\sqrt 3$, 2 - 3$\sqrt 3$) and (-$\sqrt 3$, 2 + 3$\sqrt 3$)
Work Step by Step
For this problem,
Equation 1: 3x + y = 2
Equation 2: $x^{3}$ - 2 + y = 0
For substitution we need to solve for a variable. In this problem, we will solve for y in Equation 1 to get Equation 3:
y = 2 - 3x (Equation 3)
We then substitute that into Equation 2 and solve for x:
$x^{3}$ - 2 + (2 - 3x) = 0
$x^{3}$ - 2 + 2 - 3x = 0
$x^{3}$ - 3x = 0
x ($x^{2}$ - 3) = 0
x = 0, $\sqrt 3$, -$\sqrt 3$
We then use Equation 3:
y = 2 - 3(0) = 2
y = 2 - 3($\sqrt 3$)
y = 2 - 3(-$\sqrt 3$)
Therefore (0, 2) and ($\sqrt 3$, 2 - 3$\sqrt 3$) and (-$\sqrt 3$, 2 + 3$\sqrt 3$) are solutions. Furthermore, the two graphs cross at (0, 2), ($\sqrt 3$, 2 - 3$\sqrt 3$) and (-$\sqrt 3$, 2 + 3$\sqrt 3$), so the answer is confirmed.
