Solve frac{frac{sqrt{2}}{2}x}{5-frac{sqrt{2}}{2}x}=y/3sqrt{2/2x}

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

Solve for y

$$y=-\frac{3x^{2}}{\sqrt{2}x-10}$$

$$x\neq 5\sqrt{2}\text{ and }x\neq 0$$

Steps for Solving Linear Equation

Express \frac{\sqrt{2}}{2}x as a single fraction.

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

Express \frac{\sqrt{2}}{2}x as a single fraction.

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

Express \frac{3\sqrt{2}}{2}x as a single fraction.

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

Divide y by \frac{3\sqrt{2}x}{2} by multiplying y by the reciprocal of \frac{3\sqrt{2}x}{2}.

$$\frac{\frac{\sqrt{2}x}{2}}{5-\frac{\sqrt{2}x}{2}}=\frac{y\times 2}{3\sqrt{2}x}$$

Factor the expressions that are not already factored in \frac{y\times 2}{3\sqrt{2}x}.

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

Cancel out \sqrt{2} in both numerator and denominator.

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

Express \frac{\frac{\sqrt{2}x}{2}}{5-\frac{\sqrt{2}x}{2}} as a single fraction.

$$\frac{\sqrt{2}x}{2\left(5-\frac{\sqrt{2}x}{2}\right)}=\frac{\sqrt{2}y}{3x}$$

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

$$\frac{\sqrt{2}y}{3x}=\frac{\sqrt{2}x}{2\left(5-\frac{\sqrt{2}x}{2}\right)}$$

Use the distributive property to multiply 2 by 5-\frac{\sqrt{2}x}{2}.

$$\frac{\sqrt{2}y}{3x}=\frac{\sqrt{2}x}{10+2\left(-\frac{\sqrt{2}x}{2}\right)}$$

Express 2\left(-\frac{\sqrt{2}x}{2}\right) as a single fraction.

$$\frac{\sqrt{2}y}{3x}=\frac{\sqrt{2}x}{10+\frac{-2\sqrt{2}x}{2}}$$

Cancel out 2 and 2.

$$\frac{\sqrt{2}y}{3x}=\frac{\sqrt{2}x}{10-\sqrt{2}x}$$

Rationalize the denominator of \frac{\sqrt{2}x}{10-\sqrt{2}x} by multiplying numerator and denominator by 10+\sqrt{2}x.

$$\frac{\sqrt{2}y}{3x}=\frac{\sqrt{2}x\left(10+\sqrt{2}x\right)}{\left(10-\sqrt{2}x\right)\left(10+\sqrt{2}x\right)}$$

Consider \left(10-\sqrt{2}x\right)\left(10+\sqrt{2}x\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\sqrt{2}y}{3x}=\frac{\sqrt{2}x\left(10+\sqrt{2}x\right)}{10^{2}-\left(\sqrt{2}x\right)^{2}}$$

Calculate 10 to the power of 2 and get 100.

$$\frac{\sqrt{2}y}{3x}=\frac{\sqrt{2}x\left(10+\sqrt{2}x\right)}{100-\left(\sqrt{2}x\right)^{2}}$$

Expand \left(\sqrt{2}x\right)^{2}.

$$\frac{\sqrt{2}y}{3x}=\frac{\sqrt{2}x\left(10+\sqrt{2}x\right)}{100-\left(\sqrt{2}\right)^{2}x^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{\sqrt{2}y}{3x}=\frac{\sqrt{2}x\left(10+\sqrt{2}x\right)}{100-2x^{2}}$$

Factor the expressions that are not already factored in \frac{\sqrt{2}x\left(10+\sqrt{2}x\right)}{100-2x^{2}}.

$$\frac{\sqrt{2}y}{3x}=\frac{\sqrt{2}x\left(\sqrt{2}x+10\right)}{2\left(-x^{2}+50\right)}$$

Cancel out \sqrt{2} in both numerator and denominator.

$$\frac{\sqrt{2}y}{3x}=\frac{x\left(\sqrt{2}x+10\right)}{\sqrt{2}\left(-x^{2}+50\right)}$$

Multiply both sides of the equation by 3x.

$$\sqrt{2}y=\frac{3}{2}x\left(-x^{2}+50\right)^{-1}\times 2^{\frac{1}{2}}x\left(\sqrt{2}x+10\right)$$

Reorder the terms.

$$\sqrt{2}y=\frac{3}{2}\sqrt{2}\times \frac{1}{-x^{2}+50}xx\left(\sqrt{2}x+10\right)$$

Multiply both sides of the equation by 2\left(x^{2}-50\right), the least common multiple of 2,-x^{2}+50.

$$\sqrt{2}y\times 2\left(x^{2}-50\right)=\frac{3}{2}\sqrt{2}\left(-2\right)\times 1xx\left(\sqrt{2}x+10\right)$$

Multiply x and x to get x^{2}.

$$\sqrt{2}y\times 2\left(x^{2}-50\right)=\frac{3}{2}\sqrt{2}\left(-2\right)\times 1x^{2}\left(\sqrt{2}x+10\right)$$

Use the distributive property to multiply \sqrt{2}y\times 2 by x^{2}-50.

$$2\sqrt{2}yx^{2}-50\sqrt{2}y\times 2=\frac{3}{2}\sqrt{2}\left(-2\right)\times 1x^{2}\left(\sqrt{2}x+10\right)$$

Multiply -50 and 2 to get -100.

$$2\sqrt{2}yx^{2}-100\sqrt{2}y=\frac{3}{2}\sqrt{2}\left(-2\right)\times 1x^{2}\left(\sqrt{2}x+10\right)$$

Multiply \frac{3}{2} and -2 to get -3.

$$2\sqrt{2}yx^{2}-100\sqrt{2}y=-3\sqrt{2}x^{2}\left(\sqrt{2}x+10\right)$$

Use the distributive property to multiply -3\sqrt{2}x^{2} by \sqrt{2}x+10.

$$2\sqrt{2}yx^{2}-100\sqrt{2}y=-3\left(\sqrt{2}\right)^{2}x^{3}-30x^{2}\sqrt{2}$$

The square of \sqrt{2} is 2.

$$2\sqrt{2}yx^{2}-100\sqrt{2}y=-3\times 2x^{3}-30x^{2}\sqrt{2}$$

Multiply -3 and 2 to get -6.

$$2\sqrt{2}yx^{2}-100\sqrt{2}y=-6x^{3}-30x^{2}\sqrt{2}$$

Combine all terms containing y.

$$\left(2\sqrt{2}x^{2}-100\sqrt{2}\right)y=-6x^{3}-30x^{2}\sqrt{2}$$

The equation is in standard form.

$$\left(2\sqrt{2}x^{2}-100\sqrt{2}\right)y=-6x^{3}-30\sqrt{2}x^{2}$$

Divide both sides by 2\sqrt{2}x^{2}-100\sqrt{2}.

$$\frac{\left(2\sqrt{2}x^{2}-100\sqrt{2}\right)y}{2\sqrt{2}x^{2}-100\sqrt{2}}=-\frac{6\left(x+5\sqrt{2}\right)x^{2}}{2\sqrt{2}x^{2}-100\sqrt{2}}$$

Dividing by 2\sqrt{2}x^{2}-100\sqrt{2} undoes the multiplication by 2\sqrt{2}x^{2}-100\sqrt{2}.

$$y=-\frac{6\left(x+5\sqrt{2}\right)x^{2}}{2\sqrt{2}x^{2}-100\sqrt{2}}$$

Divide -6\left(x+5\sqrt{2}\right)x^{2} by 2\sqrt{2}x^{2}-100\sqrt{2}.

$$y=-\frac{3\sqrt{2}\left(x+5\sqrt{2}\right)x^{2}}{2\left(x^{2}-50\right)}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{2}\left(\sqrt{y\left(y+60\right)}-y\right)}{6}$$

$$x=-\frac{\sqrt{2}\left(\sqrt{y\left(y+60\right)}+y\right)}{6}\text{, }y\neq 0$$

Solve for x

$$x=\frac{\sqrt{2}\left(\sqrt{y\left(y+60\right)}-y\right)}{6}$$

$$x=-\frac{\sqrt{2}\left(\sqrt{y\left(y+60\right)}+y\right)}{6}\text{, }y>0\text{ or }y\leq -60$$