Solve 100+x/(280-40)-x/280geq0.6

$$\frac { 100 + x } { ( 280 - 40 ) } - \frac { x } { 280 } \geq 0.6$$

Solve for x

$$x\geq 308$$

Solution Steps

Multiply both sides of the equation by 280. Since 280 is positive, the inequality direction remains the same.

$$\frac{7}{6}\left(100+x\right)-x\geq 168$$

Use the distributive property to multiply \frac{7}{6} by 100+x.

$$\frac{7}{6}\times 100+\frac{7}{6}x-x\geq 168$$

Express \frac{7}{6}\times 100 as a single fraction.

$$\frac{7\times 100}{6}+\frac{7}{6}x-x\geq 168$$

Multiply 7 and 100 to get 700.

$$\frac{700}{6}+\frac{7}{6}x-x\geq 168$$

Reduce the fraction \frac{700}{6} to lowest terms by extracting and canceling out 2.

$$\frac{350}{3}+\frac{7}{6}x-x\geq 168$$

Subtract \frac{350}{3} from both sides.

$$\frac{7}{6}x-x\geq 168-\frac{350}{3}$$

Convert 168 to fraction \frac{504}{3}.

$$\frac{7}{6}x-x\geq \frac{504}{3}-\frac{350}{3}$$

Since \frac{504}{3} and \frac{350}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{7}{6}x-x\geq \frac{504-350}{3}$$

Subtract 350 from 504 to get 154.

$$\frac{7}{6}x-x\geq \frac{154}{3}$$

Combine \frac{7}{6}x and -x to get \frac{1}{6}x.

$$\frac{1}{6}x\geq \frac{154}{3}$$

Multiply both sides by 6, the reciprocal of \frac{1}{6}. Since \frac{1}{6} is positive, the inequality direction remains the same.

$$x\geq \frac{154}{3}\times 6$$

Express \frac{154}{3}\times 6 as a single fraction.

$$x\geq \frac{154\times 6}{3}$$

Multiply 154 and 6 to get 924.

$$x\geq \frac{924}{3}$$

Divide 924 by 3 to get 308.

$$x\geq 308$$