Solve 3*(2/3)^2+2/3m-8=0

$$3 \times ( \frac { 2 } { 3 } ) ^ { 2 } + \frac { 2 } { 3 } m - 8 = 0$$

Solve for m

$$m=10$$

Steps for Solving Linear Equation

Calculate \frac{2}{3} to the power of 2 and get \frac{4}{9}.

$$3\times \frac{4}{9}+\frac{2}{3}m-8=0$$

Express 3\times \frac{4}{9} as a single fraction.

$$\frac{3\times 4}{9}+\frac{2}{3}m-8=0$$

Multiply 3 and 4 to get 12.

$$\frac{12}{9}+\frac{2}{3}m-8=0$$

Reduce the fraction \frac{12}{9} to lowest terms by extracting and canceling out 3.

$$\frac{4}{3}+\frac{2}{3}m-8=0$$

Convert 8 to fraction \frac{24}{3}.

$$\frac{4}{3}+\frac{2}{3}m-\frac{24}{3}=0$$

Since \frac{4}{3} and \frac{24}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{4-24}{3}+\frac{2}{3}m=0$$

Subtract 24 from 4 to get -20.

$$-\frac{20}{3}+\frac{2}{3}m=0$$

Add \frac{20}{3} to both sides. Anything plus zero gives itself.

$$\frac{2}{3}m=\frac{20}{3}$$

Multiply both sides by \frac{3}{2}, the reciprocal of \frac{2}{3}.

$$m=\frac{20}{3}\times \frac{3}{2}$$

Multiply \frac{20}{3} times \frac{3}{2} by multiplying numerator times numerator and denominator times denominator.

$$m=\frac{20\times 3}{3\times 2}$$

Cancel out 3 in both numerator and denominator.

$$m=\frac{20}{2}$$

Divide 20 by 2 to get 10.

$$m=10$$