Solve frac{3-sqrt{5}}{3+2sqrt{5}}=xsqrt{5}-y

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

Solve for x

$$x=\frac{\sqrt{5}y}{5}-\frac{19\sqrt{5}}{55}+\frac{9}{11}$$

Steps for Solving Linear Equation

Rationalize the denominator of \frac{3-\sqrt{5}}{3+2\sqrt{5}} by multiplying numerator and denominator by 3-2\sqrt{5}.

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

Consider \left(3+2\sqrt{5}\right)\left(3-2\sqrt{5}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Calculate 3 to the power of 2 and get 9.

$$\frac{\left(3-\sqrt{5}\right)\left(3-2\sqrt{5}\right)}{9-\left(2\sqrt{5}\right)^{2}}=x\sqrt{5}-y$$

Expand \left(2\sqrt{5}\right)^{2}.

$$\frac{\left(3-\sqrt{5}\right)\left(3-2\sqrt{5}\right)}{9-2^{2}\left(\sqrt{5}\right)^{2}}=x\sqrt{5}-y$$

Calculate 2 to the power of 2 and get 4.

$$\frac{\left(3-\sqrt{5}\right)\left(3-2\sqrt{5}\right)}{9-4\left(\sqrt{5}\right)^{2}}=x\sqrt{5}-y$$

The square of \sqrt{5} is 5.

$$\frac{\left(3-\sqrt{5}\right)\left(3-2\sqrt{5}\right)}{9-4\times 5}=x\sqrt{5}-y$$

Multiply 4 and 5 to get 20.

$$\frac{\left(3-\sqrt{5}\right)\left(3-2\sqrt{5}\right)}{9-20}=x\sqrt{5}-y$$

Subtract 20 from 9 to get -11.

$$\frac{\left(3-\sqrt{5}\right)\left(3-2\sqrt{5}\right)}{-11}=x\sqrt{5}-y$$

Use the distributive property to multiply 3-\sqrt{5} by 3-2\sqrt{5} and combine like terms.

$$\frac{9-9\sqrt{5}+2\left(\sqrt{5}\right)^{2}}{-11}=x\sqrt{5}-y$$

The square of \sqrt{5} is 5.

$$\frac{9-9\sqrt{5}+2\times 5}{-11}=x\sqrt{5}-y$$

Multiply 2 and 5 to get 10.

$$\frac{9-9\sqrt{5}+10}{-11}=x\sqrt{5}-y$$

Add 9 and 10 to get 19.

$$\frac{19-9\sqrt{5}}{-11}=x\sqrt{5}-y$$

Multiply both numerator and denominator by -1.

$$\frac{-19+9\sqrt{5}}{11}=x\sqrt{5}-y$$

Divide each term of -19+9\sqrt{5} by 11 to get -\frac{19}{11}+\frac{9}{11}\sqrt{5}.

$$-\frac{19}{11}+\frac{9}{11}\sqrt{5}=x\sqrt{5}-y$$

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

$$x\sqrt{5}-y=-\frac{19}{11}+\frac{9}{11}\sqrt{5}$$

Add y to both sides.

$$x\sqrt{5}=-\frac{19}{11}+\frac{9}{11}\sqrt{5}+y$$

The equation is in standard form.

$$\sqrt{5}x=y+\frac{9\sqrt{5}}{11}-\frac{19}{11}$$

Divide both sides by \sqrt{5}.

$$\frac{\sqrt{5}x}{\sqrt{5}}=\frac{y+\frac{9\sqrt{5}}{11}-\frac{19}{11}}{\sqrt{5}}$$

Dividing by \sqrt{5} undoes the multiplication by \sqrt{5}.

$$x=\frac{y+\frac{9\sqrt{5}}{11}-\frac{19}{11}}{\sqrt{5}}$$

Divide \frac{9\sqrt{5}}{11}+y-\frac{19}{11} by \sqrt{5}.

$$x=\frac{\sqrt{5}y}{5}-\frac{19\sqrt{5}}{55}+\frac{9}{11}$$

Solve for y

$$y=\frac{\sqrt{5}\left(11x-9\right)+19}{11}$$

Steps for Solving Linear Equation

Rationalize the denominator of \frac{3-\sqrt{5}}{3+2\sqrt{5}} by multiplying numerator and denominator by 3-2\sqrt{5}.

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

Consider \left(3+2\sqrt{5}\right)\left(3-2\sqrt{5}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Calculate 3 to the power of 2 and get 9.

$$\frac{\left(3-\sqrt{5}\right)\left(3-2\sqrt{5}\right)}{9-\left(2\sqrt{5}\right)^{2}}=x\sqrt{5}-y$$

Expand \left(2\sqrt{5}\right)^{2}.

$$\frac{\left(3-\sqrt{5}\right)\left(3-2\sqrt{5}\right)}{9-2^{2}\left(\sqrt{5}\right)^{2}}=x\sqrt{5}-y$$

Calculate 2 to the power of 2 and get 4.

$$\frac{\left(3-\sqrt{5}\right)\left(3-2\sqrt{5}\right)}{9-4\left(\sqrt{5}\right)^{2}}=x\sqrt{5}-y$$

The square of \sqrt{5} is 5.

$$\frac{\left(3-\sqrt{5}\right)\left(3-2\sqrt{5}\right)}{9-4\times 5}=x\sqrt{5}-y$$

Multiply 4 and 5 to get 20.

$$\frac{\left(3-\sqrt{5}\right)\left(3-2\sqrt{5}\right)}{9-20}=x\sqrt{5}-y$$

Subtract 20 from 9 to get -11.

$$\frac{\left(3-\sqrt{5}\right)\left(3-2\sqrt{5}\right)}{-11}=x\sqrt{5}-y$$

Use the distributive property to multiply 3-\sqrt{5} by 3-2\sqrt{5} and combine like terms.

$$\frac{9-9\sqrt{5}+2\left(\sqrt{5}\right)^{2}}{-11}=x\sqrt{5}-y$$

The square of \sqrt{5} is 5.

$$\frac{9-9\sqrt{5}+2\times 5}{-11}=x\sqrt{5}-y$$

Multiply 2 and 5 to get 10.

$$\frac{9-9\sqrt{5}+10}{-11}=x\sqrt{5}-y$$

Add 9 and 10 to get 19.

$$\frac{19-9\sqrt{5}}{-11}=x\sqrt{5}-y$$

Multiply both numerator and denominator by -1.

$$\frac{-19+9\sqrt{5}}{11}=x\sqrt{5}-y$$

Divide each term of -19+9\sqrt{5} by 11 to get -\frac{19}{11}+\frac{9}{11}\sqrt{5}.

$$-\frac{19}{11}+\frac{9}{11}\sqrt{5}=x\sqrt{5}-y$$

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

$$x\sqrt{5}-y=-\frac{19}{11}+\frac{9}{11}\sqrt{5}$$

Subtract x\sqrt{5} from both sides.

$$-y=-\frac{19}{11}+\frac{9}{11}\sqrt{5}-x\sqrt{5}$$

Reorder the terms.

$$-y=-\sqrt{5}x+\frac{9}{11}\sqrt{5}-\frac{19}{11}$$

The equation is in standard form.

$$-y=-\sqrt{5}x+\frac{9\sqrt{5}}{11}-\frac{19}{11}$$

Divide both sides by -1.

$$\frac{-y}{-1}=\frac{-\sqrt{5}x+\frac{9\sqrt{5}}{11}-\frac{19}{11}}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$y=\frac{-\sqrt{5}x+\frac{9\sqrt{5}}{11}-\frac{19}{11}}{-1}$$

Divide \frac{9\sqrt{5}}{11}-\sqrt{5}x-\frac{19}{11} by -1.

$$y=\sqrt{5}x-\frac{9\sqrt{5}}{11}+\frac{19}{11}$$