Answer
The solution is $(1, 3)$.
Work Step by Step
To solve this system of equations, we set the two equations equal to one another because they are both equal to the same number, $y$:
$3x^2 - x + 1 = 3x^2 + x - 1$
Subtract $3x^2$ from both sides to gather this term on the left side of the equation. These terms cancel each other out:
$3x^2 - x + 1-3x^2 = 3x^2 + x - 1-3x^2$
$- x + 1 = x - 1$
Subtract $x$ from both sides to move variables to the left side of the equation:
$- x + 1-x = x - 1-x$
$- x - x + 1 = -1$
$-2x + 1 = -1$
Subtract $1$ from each side of the equation to isolate the constants to the right side of the equation:
$-2x + 1-1 = -1-1$
$-2x = -2$
Divide both sides by $-2$ to solve for $x$:
$x = 1$
Now that we have the value for $x$, we can plug this value into either of the original equations to find the value for $y$. Let's use the first equation:
$y=3x^2-x+1\\
y = 3(1)^2 - (1) + 1\\
y=3(1)-1+1$
$y = 3 - 1 + 1$
$y = 3$
Thus, the solution is $(1, 3)$.
