Solve sqrt{18}-3u=6-xsqrt{2}

$$\sqrt { 18 } - 3 u = 6 - x \sqrt { 2 }$$

Solve for u

$$u=\frac{\sqrt{2}\left(x+3\right)-6}{3}$$

Steps for Solving Linear Equation

Factor 18=3^{2}\times 2. Rewrite the square root of the product \sqrt{3^{2}\times 2} as the product of square roots \sqrt{3^{2}}\sqrt{2}. Take the square root of 3^{2}.

$$3\sqrt{2}-3u=6-x\sqrt{2}$$

Subtract 3\sqrt{2} from both sides.

$$-3u=6-x\sqrt{2}-3\sqrt{2}$$

Reorder the terms.

$$-3u=-\sqrt{2}x-3\sqrt{2}+6$$

The equation is in standard form.

$$-3u=-\sqrt{2}x+6-3\sqrt{2}$$

Divide both sides by -3.

$$\frac{-3u}{-3}=\frac{-\sqrt{2}x+6-3\sqrt{2}}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$u=\frac{-\sqrt{2}x+6-3\sqrt{2}}{-3}$$

Divide -\sqrt{2}x-3\sqrt{2}+6 by -3.

$$u=\frac{\sqrt{2}x}{3}+\sqrt{2}-2$$

Solve for x

$$x=\frac{3\sqrt{2}\left(u+2-\sqrt{2}\right)}{2}$$

Steps for Solving Linear Equation

Factor 18=3^{2}\times 2. Rewrite the square root of the product \sqrt{3^{2}\times 2} as the product of square roots \sqrt{3^{2}}\sqrt{2}. Take the square root of 3^{2}.

$$3\sqrt{2}-3u=6-x\sqrt{2}$$

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

$$6-x\sqrt{2}=3\sqrt{2}-3u$$

Subtract 6 from both sides.

$$-x\sqrt{2}=3\sqrt{2}-3u-6$$

The equation is in standard form.

$$\left(-\sqrt{2}\right)x=-3u+3\sqrt{2}-6$$

Divide both sides by -\sqrt{2}.

$$\frac{\left(-\sqrt{2}\right)x}{-\sqrt{2}}=\frac{-3u+3\sqrt{2}-6}{-\sqrt{2}}$$

Dividing by -\sqrt{2} undoes the multiplication by -\sqrt{2}.

$$x=\frac{-3u+3\sqrt{2}-6}{-\sqrt{2}}$$

Divide 3\sqrt{2}-3u-6 by -\sqrt{2}.

$$x=-\frac{3\sqrt{2}\left(-u+\sqrt{2}-2\right)}{2}$$