Solve (1/6+3/9x):8/9=13/2-frac{14}{3}

$$( \frac { 1 } { 6 } + \frac { 3 } { 9 } x ) : \frac { 8 } { 9 } = \frac { 13 } { 2 } - \frac { 14 } { 3 }$$

Solve for x

$$x = \frac{79}{18} = 4\frac{7}{18} \approx 4.388888889$$

Steps for Solving Linear Equation

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

$$\frac{\frac{1}{6}+\frac{1}{3}x}{\frac{8}{9}}=\frac{13}{2}-\frac{14}{3}$$

Least common multiple of 2 and 3 is 6. Convert \frac{13}{2} and \frac{14}{3} to fractions with denominator 6.

$$\frac{\frac{1}{6}+\frac{1}{3}x}{\frac{8}{9}}=\frac{39}{6}-\frac{28}{6}$$

Since \frac{39}{6} and \frac{28}{6} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{1}{6}+\frac{1}{3}x}{\frac{8}{9}}=\frac{39-28}{6}$$

Subtract 28 from 39 to get 11.

$$\frac{\frac{1}{6}+\frac{1}{3}x}{\frac{8}{9}}=\frac{11}{6}$$

Divide each term of \frac{1}{6}+\frac{1}{3}x by \frac{8}{9} to get \frac{\frac{1}{6}}{\frac{8}{9}}+\frac{\frac{1}{3}x}{\frac{8}{9}}.

$$\frac{\frac{1}{6}}{\frac{8}{9}}+\frac{\frac{1}{3}x}{\frac{8}{9}}=\frac{11}{6}$$

Divide \frac{1}{6} by \frac{8}{9} by multiplying \frac{1}{6} by the reciprocal of \frac{8}{9}.

$$\frac{1}{6}\times \frac{9}{8}+\frac{\frac{1}{3}x}{\frac{8}{9}}=\frac{11}{6}$$

Multiply \frac{1}{6} times \frac{9}{8} by multiplying numerator times numerator and denominator times denominator.

$$\frac{1\times 9}{6\times 8}+\frac{\frac{1}{3}x}{\frac{8}{9}}=\frac{11}{6}$$

Do the multiplications in the fraction \frac{1\times 9}{6\times 8}.

$$\frac{9}{48}+\frac{\frac{1}{3}x}{\frac{8}{9}}=\frac{11}{6}$$

Reduce the fraction \frac{9}{48} to lowest terms by extracting and canceling out 3.

$$\frac{3}{16}+\frac{\frac{1}{3}x}{\frac{8}{9}}=\frac{11}{6}$$

Divide \frac{1}{3}x by \frac{8}{9} to get \frac{3}{8}x.

$$\frac{3}{16}+\frac{3}{8}x=\frac{11}{6}$$

Subtract \frac{3}{16} from both sides.

$$\frac{3}{8}x=\frac{11}{6}-\frac{3}{16}$$

Least common multiple of 6 and 16 is 48. Convert \frac{11}{6} and \frac{3}{16} to fractions with denominator 48.

$$\frac{3}{8}x=\frac{88}{48}-\frac{9}{48}$$

Since \frac{88}{48} and \frac{9}{48} have the same denominator, subtract them by subtracting their numerators.

$$\frac{3}{8}x=\frac{88-9}{48}$$

Subtract 9 from 88 to get 79.

$$\frac{3}{8}x=\frac{79}{48}$$

Multiply both sides by \frac{8}{3}, the reciprocal of \frac{3}{8}.

$$x=\frac{79}{48}\times \frac{8}{3}$$

Multiply \frac{79}{48} times \frac{8}{3} by multiplying numerator times numerator and denominator times denominator.

$$x=\frac{79\times 8}{48\times 3}$$

Do the multiplications in the fraction \frac{79\times 8}{48\times 3}.

$$x=\frac{632}{144}$$

Reduce the fraction \frac{632}{144} to lowest terms by extracting and canceling out 8.

$$x=\frac{79}{18}$$