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\left(3-\sqrt{2}\right)\left(\frac{k\sqrt{2}}{\left(\sqrt{2}\right)^{2}}-3\right)=k
Rationalize the denominator of \frac{k}{\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.
\left(3-\sqrt{2}\right)\left(\frac{k\sqrt{2}}{2}-3\right)=k
The square of \sqrt{2} is 2.
3\times \frac{k\sqrt{2}}{2}-9-\sqrt{2}\times \frac{k\sqrt{2}}{2}+3\sqrt{2}=k
Apply the distributive property by multiplying each term of 3-\sqrt{2} by each term of \frac{k\sqrt{2}}{2}-3.
\frac{3k\sqrt{2}}{2}-9-\sqrt{2}\times \frac{k\sqrt{2}}{2}+3\sqrt{2}=k
Express 3\times \frac{k\sqrt{2}}{2} as a single fraction.
\frac{3k\sqrt{2}}{2}-9-\frac{\sqrt{2}k\sqrt{2}}{2}+3\sqrt{2}=k
Express \sqrt{2}\times \frac{k\sqrt{2}}{2} as a single fraction.
\frac{3k\sqrt{2}-\sqrt{2}k\sqrt{2}}{2}-9+3\sqrt{2}=k
Since \frac{3k\sqrt{2}}{2} and \frac{\sqrt{2}k\sqrt{2}}{2} have the same denominator, subtract them by subtracting their numerators.
\frac{3k\sqrt{2}-2k}{2}-9+3\sqrt{2}=k
Do the multiplications in 3k\sqrt{2}-\sqrt{2}k\sqrt{2}.
\frac{3k\sqrt{2}-2k}{2}-9+3\sqrt{2}-k=0
Subtract k from both sides.
\frac{3k\sqrt{2}-2k}{2}+3\sqrt{2}-k=9
Add 9 to both sides. Anything plus zero gives itself.
\frac{3k\sqrt{2}-2k}{2}-k=9-3\sqrt{2}
Subtract 3\sqrt{2} from both sides.
3k\sqrt{2}-2k-2k=18-6\sqrt{2}
Multiply both sides of the equation by 2.
3\sqrt{2}k-2k-2k=-6\sqrt{2}+18
Reorder the terms.
3\sqrt{2}k-4k=-6\sqrt{2}+18
Combine -2k and -2k to get -4k.
\left(3\sqrt{2}-4\right)k=-6\sqrt{2}+18
Combine all terms containing k.
\left(3\sqrt{2}-4\right)k=18-6\sqrt{2}
The equation is in standard form.
\frac{\left(3\sqrt{2}-4\right)k}{3\sqrt{2}-4}=\frac{18-6\sqrt{2}}{3\sqrt{2}-4}
Divide both sides by 3\sqrt{2}-4.
k=\frac{18-6\sqrt{2}}{3\sqrt{2}-4}
Dividing by 3\sqrt{2}-4 undoes the multiplication by 3\sqrt{2}-4.
k=15\sqrt{2}+18
Divide -6\sqrt{2}+18 by 3\sqrt{2}-4.