Solve y=5pi^3/17-sqrt{2x}

$$y = \frac { 5 \pi ^ { 3 } } { 17 } - \sqrt { 2 x }$$

Solve for x

$$x=\frac{\left(5\pi ^{3}-17y\right)^{2}}{578}$$

$$y\leq \frac{5\pi ^{3}}{17}$$

Steps for Solving Radical Equations

Swap sides so that all variable terms are on the left hand side.

$$\frac{5\pi ^{3}}{17}-\sqrt{2x}=y$$

Subtract \frac{5\pi ^{3}}{17} from both sides.

$$-\sqrt{2x}=y-\frac{5\pi ^{3}}{17}$$

Multiply both sides of the equation by 17.

$$-17\sqrt{2x}=17y-5\pi ^{3}$$

Divide both sides by -17.

$$\frac{-17\sqrt{2x}}{-17}=\frac{17y-5\pi ^{3}}{-17}$$

Dividing by -17 undoes the multiplication by -17.

$$\sqrt{2x}=\frac{17y-5\pi ^{3}}{-17}$$

Divide 17y-5\pi ^{3} by -17.

$$\sqrt{2x}=-y+\frac{5\pi ^{3}}{17}$$

Square both sides of the equation.

$$2x=\frac{\left(5\pi ^{3}-17y\right)^{2}}{289}$$

Divide both sides by 2.

$$\frac{2x}{2}=\frac{\left(5\pi ^{3}-17y\right)^{2}}{2\times 289}$$

Dividing by 2 undoes the multiplication by 2.

$$x=\frac{\left(5\pi ^{3}-17y\right)^{2}}{2\times 289}$$

Divide \frac{\left(-17y+5\pi ^{3}\right)^{2}}{289} by 2.

$$x=\frac{\left(5\pi ^{3}-17y\right)^{2}}{578}$$

Solve for y

$$y=-\sqrt{2x}+\frac{5\pi ^{3}}{17}$$

$$x\geq 0$$