Solve sqrt{3x}-8+sqrt{3x}-14=9

$$\sqrt { 3 x } - 8 + \sqrt { 3 x } - 14 = 9$$

Solve for x

$$x = \frac{961}{12} = 80\frac{1}{12} \approx 80.083333333$$

Solution Steps

Subtract -8+\sqrt{3x}-14 from both sides of the equation.

$$\sqrt{3x}=9-\left(-8+\sqrt{3x}-14\right)$$

Subtract 14 from -8 to get -22.

$$\sqrt{3x}=9-\left(-22+\sqrt{3x}\right)$$

To find the opposite of -22+\sqrt{3x}, find the opposite of each term.

$$\sqrt{3x}=9-\left(-22\right)-\sqrt{3x}$$

The opposite of -22 is 22.

$$\sqrt{3x}=9+22-\sqrt{3x}$$

Add 9 and 22 to get 31.

$$\sqrt{3x}=31-\sqrt{3x}$$

Square both sides of the equation.

$$\left(\sqrt{3x}\right)^{2}=\left(31-\sqrt{3x}\right)^{2}$$

Calculate \sqrt{3x} to the power of 2 and get 3x.

$$3x=\left(31-\sqrt{3x}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(31-\sqrt{3x}\right)^{2}.

$$3x=961-62\sqrt{3x}+\left(\sqrt{3x}\right)^{2}$$

Calculate \sqrt{3x} to the power of 2 and get 3x.

$$3x=961-62\sqrt{3x}+3x$$

Add 62\sqrt{3x} to both sides.

$$3x+62\sqrt{3x}=961+3x$$

Subtract 3x from both sides.

$$3x+62\sqrt{3x}-3x=961$$

Combine 3x and -3x to get 0.

$$62\sqrt{3x}=961$$

Divide both sides by 62.

$$\sqrt{3x}=\frac{961}{62}$$

Reduce the fraction \frac{961}{62} to lowest terms by extracting and canceling out 31.

$$\sqrt{3x}=\frac{31}{2}$$

Square both sides of the equation.

$$3x=\frac{961}{4}$$

Divide both sides by 3.

$$\frac{3x}{3}=\frac{\frac{961}{4}}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x=\frac{\frac{961}{4}}{3}$$

Divide \frac{961}{4} by 3.

$$x=\frac{961}{12}$$

Substitute \frac{961}{12} for x in the equation \sqrt{3x}-8+\sqrt{3x}-14=9.

$$\sqrt{3\times \frac{961}{12}}-8+\sqrt{3\times \frac{961}{12}}-14=9$$

Simplify. The value x=\frac{961}{12} satisfies the equation.

$$9=9$$

Equation \sqrt{3x}=-\sqrt{3x}+31 has a unique solution.

$$x=\frac{961}{12}$$