Solve sqrt{3}*(3sqrt{3}x-4)-6=2(3-2sqrt{3}x)

$$\sqrt { 3 } \times ( 3 \sqrt { 3 } x - 4 ) - 6 = 2 ( 3 - 2 \sqrt { 3 } x )$$

Solve for x

$$x = \frac{20 - 4 \sqrt{3}}{11} \approx 1.188345161$$

Steps for Solving Linear Equation

Use the distributive property to multiply \sqrt{3} by 3\sqrt{3}x-4.

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

The square of \sqrt{3} is 3.

$$3x\times 3-4\sqrt{3}-6=2\left(3-2\sqrt{3}x\right)$$

Multiply 3 and 3 to get 9.

$$9x-4\sqrt{3}-6=2\left(3-2\sqrt{3}x\right)$$

Subtract 2\left(3-2\sqrt{3}x\right) from both sides.

$$9x-4\sqrt{3}-6-2\left(3-2\sqrt{3}x\right)=0$$

Use the distributive property to multiply -2 by 3-2\sqrt{3}x.

$$9x-4\sqrt{3}-6-6+4\sqrt{3}x=0$$

Subtract 6 from -6 to get -12.

$$9x-4\sqrt{3}-12+4\sqrt{3}x=0$$

Add 4\sqrt{3} to both sides. Anything plus zero gives itself.

$$9x-12+4\sqrt{3}x=4\sqrt{3}$$

Add 12 to both sides.

$$9x+4\sqrt{3}x=4\sqrt{3}+12$$

Combine all terms containing x.

$$\left(9+4\sqrt{3}\right)x=4\sqrt{3}+12$$

The equation is in standard form.

$$\left(4\sqrt{3}+9\right)x=4\sqrt{3}+12$$

Divide both sides by 9+4\sqrt{3}.

$$\frac{\left(4\sqrt{3}+9\right)x}{4\sqrt{3}+9}=\frac{4\sqrt{3}+12}{4\sqrt{3}+9}$$

Dividing by 9+4\sqrt{3} undoes the multiplication by 9+4\sqrt{3}.

$$x=\frac{4\sqrt{3}+12}{4\sqrt{3}+9}$$

Divide 4\sqrt{3}+12 by 9+4\sqrt{3}.

$$x=\frac{20-4\sqrt{3}}{11}$$