Solve 8+xsqrt{12}=8x/sqrt{3}

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

Solve for x

$$x=4\sqrt{3}\approx 6.92820323$$

Steps for Solving Linear Equation

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}.

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

Rationalize the denominator of \frac{8x}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

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

The square of \sqrt{3} is 3.

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

Subtract \frac{8x\sqrt{3}}{3} from both sides.

$$8+x\times 2\sqrt{3}-\frac{8x\sqrt{3}}{3}=0$$

Subtract 8 from both sides. Anything subtracted from zero gives its negation.

$$x\times 2\sqrt{3}-\frac{8x\sqrt{3}}{3}=-8$$

Multiply both sides of the equation by 3.

$$3x\times 2\sqrt{3}-8x\sqrt{3}=-24$$

Multiply 3 and 2 to get 6.

$$6x\sqrt{3}-8x\sqrt{3}=-24$$

Combine 6x\sqrt{3} and -8x\sqrt{3} to get -2x\sqrt{3}.

$$-2x\sqrt{3}=-24$$

The equation is in standard form.

$$\left(-2\sqrt{3}\right)x=-24$$

Divide both sides by -2\sqrt{3}.

$$\frac{\left(-2\sqrt{3}\right)x}{-2\sqrt{3}}=-\frac{24}{-2\sqrt{3}}$$

Dividing by -2\sqrt{3} undoes the multiplication by -2\sqrt{3}.

$$x=-\frac{24}{-2\sqrt{3}}$$

Divide -24 by -2\sqrt{3}.

$$x=4\sqrt{3}$$