Solve (1/8+1/12)*3+(1/12+1/24)X=1

$$( \frac { 1 } { 8 } + \frac { 1 } { 12 } ) \times 3 + ( \frac { 1 } { 12 } + \frac { 1 } { 24 } ) X = 1$$

Solve for X

$$X=3$$

Steps for Solving Linear Equation

Least common multiple of 8 and 12 is 24. Convert \frac{1}{8} and \frac{1}{12} to fractions with denominator 24.

$$\left(\frac{3}{24}+\frac{2}{24}\right)\times 3+\left(\frac{1}{12}+\frac{1}{24}\right)X=1$$

Since \frac{3}{24} and \frac{2}{24} have the same denominator, add them by adding their numerators.

$$\frac{3+2}{24}\times 3+\left(\frac{1}{12}+\frac{1}{24}\right)X=1$$

Add 3 and 2 to get 5.

$$\frac{5}{24}\times 3+\left(\frac{1}{12}+\frac{1}{24}\right)X=1$$

Express \frac{5}{24}\times 3 as a single fraction.

$$\frac{5\times 3}{24}+\left(\frac{1}{12}+\frac{1}{24}\right)X=1$$

Multiply 5 and 3 to get 15.

$$\frac{15}{24}+\left(\frac{1}{12}+\frac{1}{24}\right)X=1$$

Reduce the fraction \frac{15}{24} to lowest terms by extracting and canceling out 3.

$$\frac{5}{8}+\left(\frac{1}{12}+\frac{1}{24}\right)X=1$$

Least common multiple of 12 and 24 is 24. Convert \frac{1}{12} and \frac{1}{24} to fractions with denominator 24.

$$\frac{5}{8}+\left(\frac{2}{24}+\frac{1}{24}\right)X=1$$

Since \frac{2}{24} and \frac{1}{24} have the same denominator, add them by adding their numerators.

$$\frac{5}{8}+\frac{2+1}{24}X=1$$

Add 2 and 1 to get 3.

$$\frac{5}{8}+\frac{3}{24}X=1$$

Reduce the fraction \frac{3}{24} to lowest terms by extracting and canceling out 3.

$$\frac{5}{8}+\frac{1}{8}X=1$$

Subtract \frac{5}{8} from both sides.

$$\frac{1}{8}X=1-\frac{5}{8}$$

Convert 1 to fraction \frac{8}{8}.

$$\frac{1}{8}X=\frac{8}{8}-\frac{5}{8}$$

Since \frac{8}{8} and \frac{5}{8} have the same denominator, subtract them by subtracting their numerators.

$$\frac{1}{8}X=\frac{8-5}{8}$$

Subtract 5 from 8 to get 3.

$$\frac{1}{8}X=\frac{3}{8}$$

Multiply both sides by 8, the reciprocal of \frac{1}{8}.

$$X=\frac{3}{8}\times 8$$

Cancel out 8 and 8.

$$X=3$$