Solve Matrix

$$\left. \begin{array} { l } { \text { If } \frac { \sqrt { 32 } + \sqrt { 48 } } { \sqrt { 8 } + \sqrt { 12 } } = x , \text { then the vare } } \\ { \left. \begin{array} { l l } { \text { (a) } \sqrt { 2 } } & { \text { (b) } 2 } \end{array} \right. } \end{array} \right.$$

Solve for x

$$x=2$$

Steps for Solving Linear Equation

Factor 32=4^{2}\times 2. Rewrite the square root of the product \sqrt{4^{2}\times 2} as the product of square roots \sqrt{4^{2}}\sqrt{2}. Take the square root of 4^{2}.

$$\frac{4\sqrt{2}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}=x$$

Factor 48=4^{2}\times 3. Rewrite the square root of the product \sqrt{4^{2}\times 3} as the product of square roots \sqrt{4^{2}}\sqrt{3}. Take the square root of 4^{2}.

$$\frac{4\sqrt{2}+4\sqrt{3}}{\sqrt{8}+\sqrt{12}}=x$$

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}.

$$\frac{4\sqrt{2}+4\sqrt{3}}{2\sqrt{2}+\sqrt{12}}=x$$

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

$$\frac{4\sqrt{2}+4\sqrt{3}}{2\sqrt{2}+2\sqrt{3}}=x$$

Rationalize the denominator of \frac{4\sqrt{2}+4\sqrt{3}}{2\sqrt{2}+2\sqrt{3}} by multiplying numerator and denominator by 2\sqrt{2}-2\sqrt{3}.

$$\frac{\left(4\sqrt{2}+4\sqrt{3}\right)\left(2\sqrt{2}-2\sqrt{3}\right)}{\left(2\sqrt{2}+2\sqrt{3}\right)\left(2\sqrt{2}-2\sqrt{3}\right)}=x$$

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

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

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

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

Calculate 2 to the power of 2 and get 4.

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

The square of \sqrt{2} is 2.

$$\frac{\left(4\sqrt{2}+4\sqrt{3}\right)\left(2\sqrt{2}-2\sqrt{3}\right)}{4\times 2-\left(2\sqrt{3}\right)^{2}}=x$$

Multiply 4 and 2 to get 8.

$$\frac{\left(4\sqrt{2}+4\sqrt{3}\right)\left(2\sqrt{2}-2\sqrt{3}\right)}{8-\left(2\sqrt{3}\right)^{2}}=x$$

Expand \left(2\sqrt{3}\right)^{2}.

$$\frac{\left(4\sqrt{2}+4\sqrt{3}\right)\left(2\sqrt{2}-2\sqrt{3}\right)}{8-2^{2}\left(\sqrt{3}\right)^{2}}=x$$

Calculate 2 to the power of 2 and get 4.

$$\frac{\left(4\sqrt{2}+4\sqrt{3}\right)\left(2\sqrt{2}-2\sqrt{3}\right)}{8-4\left(\sqrt{3}\right)^{2}}=x$$

The square of \sqrt{3} is 3.

$$\frac{\left(4\sqrt{2}+4\sqrt{3}\right)\left(2\sqrt{2}-2\sqrt{3}\right)}{8-4\times 3}=x$$

Multiply 4 and 3 to get 12.

$$\frac{\left(4\sqrt{2}+4\sqrt{3}\right)\left(2\sqrt{2}-2\sqrt{3}\right)}{8-12}=x$$

Subtract 12 from 8 to get -4.

$$\frac{\left(4\sqrt{2}+4\sqrt{3}\right)\left(2\sqrt{2}-2\sqrt{3}\right)}{-4}=x$$

Apply the distributive property by multiplying each term of 4\sqrt{2}+4\sqrt{3} by each term of 2\sqrt{2}-2\sqrt{3}.

$$\frac{8\left(\sqrt{2}\right)^{2}-8\sqrt{3}\sqrt{2}+8\sqrt{3}\sqrt{2}-8\left(\sqrt{3}\right)^{2}}{-4}=x$$

The square of \sqrt{2} is 2.

$$\frac{8\times 2-8\sqrt{3}\sqrt{2}+8\sqrt{3}\sqrt{2}-8\left(\sqrt{3}\right)^{2}}{-4}=x$$

Multiply 8 and 2 to get 16.

$$\frac{16-8\sqrt{3}\sqrt{2}+8\sqrt{3}\sqrt{2}-8\left(\sqrt{3}\right)^{2}}{-4}=x$$

To multiply \sqrt{3} and \sqrt{2}, multiply the numbers under the square root.

$$\frac{16-8\sqrt{6}+8\sqrt{3}\sqrt{2}-8\left(\sqrt{3}\right)^{2}}{-4}=x$$

To multiply \sqrt{3} and \sqrt{2}, multiply the numbers under the square root.

$$\frac{16-8\sqrt{6}+8\sqrt{6}-8\left(\sqrt{3}\right)^{2}}{-4}=x$$

Combine -8\sqrt{6} and 8\sqrt{6} to get 0.

$$\frac{16-8\left(\sqrt{3}\right)^{2}}{-4}=x$$

The square of \sqrt{3} is 3.

$$\frac{16-8\times 3}{-4}=x$$

Multiply -8 and 3 to get -24.

$$\frac{16-24}{-4}=x$$

Subtract 24 from 16 to get -8.

$$\frac{-8}{-4}=x$$

Divide -8 by -4 to get 2.

$$2=x$$

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

$$x=2$$