Solve frac{sqrt{7-1}}{sqrt{7+1}}-frac{sqrt{7+4}}{sqrt{7-1}}=a+bsqrt{7}

$$\frac { \sqrt { 7 - 1 } } { \sqrt { 7 + 1 } } - \frac { \sqrt { 7 + 4 } } { \sqrt { 7 - 1 } } = a + b \sqrt { 7 }$$

Solve for b

$$b=-\frac{\sqrt{7}\left(6a+\sqrt{66}-3\sqrt{3}\right)}{42}$$

Steps for Solving Linear Equation

Subtract 1 from 7 to get 6.

$$\frac{\sqrt{6}}{\sqrt{7+1}}-\frac{\sqrt{7+4}}{\sqrt{7-1}}=a+b\sqrt{7}$$

Add 7 and 1 to get 8.

$$\frac{\sqrt{6}}{\sqrt{8}}-\frac{\sqrt{7+4}}{\sqrt{7-1}}=a+b\sqrt{7}$$

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

$$\frac{\sqrt{6}}{2\sqrt{2}}-\frac{\sqrt{7+4}}{\sqrt{7-1}}=a+b\sqrt{7}$$

Rationalize the denominator of \frac{\sqrt{6}}{2\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

$$\frac{\sqrt{6}\sqrt{2}}{2\left(\sqrt{2}\right)^{2}}-\frac{\sqrt{7+4}}{\sqrt{7-1}}=a+b\sqrt{7}$$

The square of \sqrt{2} is 2.

$$\frac{\sqrt{6}\sqrt{2}}{2\times 2}-\frac{\sqrt{7+4}}{\sqrt{7-1}}=a+b\sqrt{7}$$

Factor 6=2\times 3. Rewrite the square root of the product \sqrt{2\times 3} as the product of square roots \sqrt{2}\sqrt{3}.

$$\frac{\sqrt{2}\sqrt{3}\sqrt{2}}{2\times 2}-\frac{\sqrt{7+4}}{\sqrt{7-1}}=a+b\sqrt{7}$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$\frac{2\sqrt{3}}{2\times 2}-\frac{\sqrt{7+4}}{\sqrt{7-1}}=a+b\sqrt{7}$$

Multiply 2 and 2 to get 4.

$$\frac{2\sqrt{3}}{4}-\frac{\sqrt{7+4}}{\sqrt{7-1}}=a+b\sqrt{7}$$

Divide 2\sqrt{3} by 4 to get \frac{1}{2}\sqrt{3}.

$$\frac{1}{2}\sqrt{3}-\frac{\sqrt{7+4}}{\sqrt{7-1}}=a+b\sqrt{7}$$

Add 7 and 4 to get 11.

$$\frac{1}{2}\sqrt{3}-\frac{\sqrt{11}}{\sqrt{7-1}}=a+b\sqrt{7}$$

Subtract 1 from 7 to get 6.

$$\frac{1}{2}\sqrt{3}-\frac{\sqrt{11}}{\sqrt{6}}=a+b\sqrt{7}$$

Rationalize the denominator of \frac{\sqrt{11}}{\sqrt{6}} by multiplying numerator and denominator by \sqrt{6}.

$$\frac{1}{2}\sqrt{3}-\frac{\sqrt{11}\sqrt{6}}{\left(\sqrt{6}\right)^{2}}=a+b\sqrt{7}$$

The square of \sqrt{6} is 6.

$$\frac{1}{2}\sqrt{3}-\frac{\sqrt{11}\sqrt{6}}{6}=a+b\sqrt{7}$$

To multiply \sqrt{11} and \sqrt{6}, multiply the numbers under the square root.

$$\frac{1}{2}\sqrt{3}-\frac{\sqrt{66}}{6}=a+b\sqrt{7}$$

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

$$a+b\sqrt{7}=\frac{1}{2}\sqrt{3}-\frac{\sqrt{66}}{6}$$

Subtract a from both sides.

$$b\sqrt{7}=\frac{1}{2}\sqrt{3}-\frac{\sqrt{66}}{6}-a$$

Multiply both sides of the equation by 6, the least common multiple of 2,6.

$$6b\sqrt{7}=3\sqrt{3}-\sqrt{66}-6a$$

The equation is in standard form.

$$6\sqrt{7}b=-6a+3\sqrt{3}-\sqrt{66}$$

Divide both sides by 6\sqrt{7}.

$$\frac{6\sqrt{7}b}{6\sqrt{7}}=\frac{-6a+3\sqrt{3}-\sqrt{66}}{6\sqrt{7}}$$

Dividing by 6\sqrt{7} undoes the multiplication by 6\sqrt{7}.

$$b=\frac{-6a+3\sqrt{3}-\sqrt{66}}{6\sqrt{7}}$$

Divide 3\sqrt{3}-\sqrt{66}-6a by 6\sqrt{7}.

$$b=\frac{\sqrt{7}\left(-6a+3\sqrt{3}-\sqrt{66}\right)}{42}$$

Solve for a

$$a=-\frac{3\sqrt{7}b}{3}+\frac{\sqrt{3}}{2}-\frac{\sqrt{66}}{6}$$