Solve frac{3-sqrt{5}}{3+2sqrt{5}}=-19/11+asqrt{5}

$$\frac { 3 - \sqrt { 5 } } { 3 + 2 \sqrt { 5 } } = - \frac { 19 } { 11 } + a \sqrt { 5 }$$

Solve for a

$$a=\frac{9}{11}\approx 0.818181818$$

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}.

$$\frac{\left(3-\sqrt{5}\right)\left(3-2\sqrt{5}\right)}{\left(3+2\sqrt{5}\right)\left(3-2\sqrt{5}\right)}=-\frac{19}{11}+a\sqrt{5}$$

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}.

$$\frac{\left(3-\sqrt{5}\right)\left(3-2\sqrt{5}\right)}{3^{2}-\left(2\sqrt{5}\right)^{2}}=-\frac{19}{11}+a\sqrt{5}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{\left(3-\sqrt{5}\right)\left(3-2\sqrt{5}\right)}{9-\left(2\sqrt{5}\right)^{2}}=-\frac{19}{11}+a\sqrt{5}$$

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

$$\frac{\left(3-\sqrt{5}\right)\left(3-2\sqrt{5}\right)}{9-2^{2}\left(\sqrt{5}\right)^{2}}=-\frac{19}{11}+a\sqrt{5}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{\left(3-\sqrt{5}\right)\left(3-2\sqrt{5}\right)}{9-4\left(\sqrt{5}\right)^{2}}=-\frac{19}{11}+a\sqrt{5}$$

The square of \sqrt{5} is 5.

$$\frac{\left(3-\sqrt{5}\right)\left(3-2\sqrt{5}\right)}{9-4\times 5}=-\frac{19}{11}+a\sqrt{5}$$

Multiply 4 and 5 to get 20.

$$\frac{\left(3-\sqrt{5}\right)\left(3-2\sqrt{5}\right)}{9-20}=-\frac{19}{11}+a\sqrt{5}$$

Subtract 20 from 9 to get -11.

$$\frac{\left(3-\sqrt{5}\right)\left(3-2\sqrt{5}\right)}{-11}=-\frac{19}{11}+a\sqrt{5}$$

Apply the distributive property by multiplying each term of 3-\sqrt{5} by each term of 3-2\sqrt{5}.

$$\frac{9-6\sqrt{5}-3\sqrt{5}+2\left(\sqrt{5}\right)^{2}}{-11}=-\frac{19}{11}+a\sqrt{5}$$

Combine -6\sqrt{5} and -3\sqrt{5} to get -9\sqrt{5}.

$$\frac{9-9\sqrt{5}+2\left(\sqrt{5}\right)^{2}}{-11}=-\frac{19}{11}+a\sqrt{5}$$

The square of \sqrt{5} is 5.

$$\frac{9-9\sqrt{5}+2\times 5}{-11}=-\frac{19}{11}+a\sqrt{5}$$

Multiply 2 and 5 to get 10.

$$\frac{9-9\sqrt{5}+10}{-11}=-\frac{19}{11}+a\sqrt{5}$$

Add 9 and 10 to get 19.

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

Multiply both numerator and denominator by -1.

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

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

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

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

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

Add \frac{19}{11} to both sides.

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

Add -\frac{19}{11} and \frac{19}{11} to get 0.

$$a\sqrt{5}=\frac{9}{11}\sqrt{5}$$

The equation is in standard form.

$$\sqrt{5}a=\frac{9\sqrt{5}}{11}$$

Divide both sides by \sqrt{5}.

$$\frac{\sqrt{5}a}{\sqrt{5}}=\frac{9\sqrt{5}}{11\sqrt{5}}$$

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

$$a=\frac{9\sqrt{5}}{11\sqrt{5}}$$

Divide \frac{9\sqrt{5}}{11} by \sqrt{5}.

$$a=\frac{9}{11}$$