Solve 10*(29/60-x)=18*(25/60-x)

$$10 \times ( \frac{ 29 }{ 60 } -x)=18 \times ( \frac{ 25 }{ 60 } -x)$$

Solve for x

$$x=\frac{1}{3}\approx 0.333333333$$

Steps for Solving Linear Equation

Use the distributive property to multiply 10 by \frac{29}{60}-x.

$$10\times \frac{29}{60}-10x=18\left(\frac{25}{60}-x\right)$$

Express 10\times \frac{29}{60} as a single fraction.

$$\frac{10\times 29}{60}-10x=18\left(\frac{25}{60}-x\right)$$

Multiply 10 and 29 to get 290.

$$\frac{290}{60}-10x=18\left(\frac{25}{60}-x\right)$$

Reduce the fraction \frac{290}{60} to lowest terms by extracting and canceling out 10.

$$\frac{29}{6}-10x=18\left(\frac{25}{60}-x\right)$$

Reduce the fraction \frac{25}{60} to lowest terms by extracting and canceling out 5.

$$\frac{29}{6}-10x=18\left(\frac{5}{12}-x\right)$$

Use the distributive property to multiply 18 by \frac{5}{12}-x.

$$\frac{29}{6}-10x=18\times \frac{5}{12}-18x$$

Express 18\times \frac{5}{12} as a single fraction.

$$\frac{29}{6}-10x=\frac{18\times 5}{12}-18x$$

Multiply 18 and 5 to get 90.

$$\frac{29}{6}-10x=\frac{90}{12}-18x$$

Reduce the fraction \frac{90}{12} to lowest terms by extracting and canceling out 6.

$$\frac{29}{6}-10x=\frac{15}{2}-18x$$

Add 18x to both sides.

$$\frac{29}{6}-10x+18x=\frac{15}{2}$$

Combine -10x and 18x to get 8x.

$$\frac{29}{6}+8x=\frac{15}{2}$$

Subtract \frac{29}{6} from both sides.

$$8x=\frac{15}{2}-\frac{29}{6}$$

Least common multiple of 2 and 6 is 6. Convert \frac{15}{2} and \frac{29}{6} to fractions with denominator 6.

$$8x=\frac{45}{6}-\frac{29}{6}$$

Since \frac{45}{6} and \frac{29}{6} have the same denominator, subtract them by subtracting their numerators.

$$8x=\frac{45-29}{6}$$

Subtract 29 from 45 to get 16.

$$8x=\frac{16}{6}$$

Reduce the fraction \frac{16}{6} to lowest terms by extracting and canceling out 2.

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

Divide both sides by 8.

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

Express \frac{\frac{8}{3}}{8} as a single fraction.

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

Cancel out 8 in both numerator and denominator.

$$x=\frac{1}{3}$$