Solve for x
x=\frac{y^{2}+1}{3}
y\geq 0
Solve for x (complex solution)
x=\frac{y^{2}+1}{3}
arg(y)<\pi \text{ or }y=0
Solve for y
y=\sqrt{3x-1}
x\geq \frac{1}{3}
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\sqrt{3x-1}=y
Swap sides so that all variable terms are on the left hand side.
3x-1=y^{2}
Square both sides of the equation.
3x-1-\left(-1\right)=y^{2}-\left(-1\right)
Add 1 to both sides of the equation.
3x=y^{2}-\left(-1\right)
Subtracting -1 from itself leaves 0.
3x=y^{2}+1
Subtract -1 from y^{2}.
\frac{3x}{3}=\frac{y^{2}+1}{3}
Divide both sides by 3.
x=\frac{y^{2}+1}{3}
Dividing by 3 undoes the multiplication by 3.
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Limits
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