Solve sqrt{2/3-5x}-sqrt{3x+1/2}=0.x>0

$$\sqrt { \frac { 2 } { 3 } - 5 x } - \sqrt { 3 x + \frac { 1 } { 2 } } = 0 . x > 0$$

Solve for x

$$x=\frac{1}{48}\approx 0.020833333$$

Solution Steps

Subtract -\sqrt{3x+\frac{1}{2}} from both sides of the equation.

$$\sqrt{\frac{2}{3}-5x}=\sqrt{3x+\frac{1}{2}}$$

Square both sides of the equation.

$$\left(\sqrt{\frac{2}{3}-5x}\right)^{2}=\left(\sqrt{3x+\frac{1}{2}}\right)^{2}$$

Calculate \sqrt{\frac{2}{3}-5x} to the power of 2 and get \frac{2}{3}-5x.

$$\frac{2}{3}-5x=\left(\sqrt{3x+\frac{1}{2}}\right)^{2}$$

Calculate \sqrt{3x+\frac{1}{2}} to the power of 2 and get 3x+\frac{1}{2}.

$$\frac{2}{3}-5x=3x+\frac{1}{2}$$

Subtract 3x from both sides.

$$\frac{2}{3}-5x-3x=\frac{1}{2}$$

Combine -5x and -3x to get -8x.

$$\frac{2}{3}-8x=\frac{1}{2}$$

Subtract \frac{2}{3} from both sides.

$$-8x=\frac{1}{2}-\frac{2}{3}$$

Least common multiple of 2 and 3 is 6. Convert \frac{1}{2} and \frac{2}{3} to fractions with denominator 6.

$$-8x=\frac{3}{6}-\frac{4}{6}$$

Since \frac{3}{6} and \frac{4}{6} have the same denominator, subtract them by subtracting their numerators.

$$-8x=\frac{3-4}{6}$$

Subtract 4 from 3 to get -1.

$$-8x=-\frac{1}{6}$$

Divide both sides by -8.

$$x=\frac{-\frac{1}{6}}{-8}$$

Express \frac{-\frac{1}{6}}{-8} as a single fraction.

$$x=\frac{-1}{6\left(-8\right)}$$

Multiply 6 and -8 to get -48.

$$x=\frac{-1}{-48}$$

Fraction \frac{-1}{-48} can be simplified to \frac{1}{48} by removing the negative sign from both the numerator and the denominator.

$$x=\frac{1}{48}$$

Substitute \frac{1}{48} for x in the equation \sqrt{\frac{2}{3}-5x}-\sqrt{3x+\frac{1}{2}}=0.

$$\sqrt{\frac{2}{3}-5\times \frac{1}{48}}-\sqrt{3\times \frac{1}{48}+\frac{1}{2}}=0$$

Simplify. The value x=\frac{1}{48} satisfies the equation.

$$0=0$$

Equation \sqrt{\frac{2}{3}-5x}=\sqrt{3x+\frac{1}{2}} has a unique solution.

$$x=\frac{1}{48}$$