Solve 30%(x+1)<51%+x

$$30 \% ( x + 1 ) < 51 \% + x$$

Solve for x

$$x>-\frac{3}{10}$$

Solution Steps

Reduce the fraction \frac{30}{100} to lowest terms by extracting and canceling out 10.

$$\frac{3}{10}\left(x+1\right)<\frac{51}{100}+x$$

Use the distributive property to multiply \frac{3}{10} by x+1.

$$\frac{3}{10}x+\frac{3}{10}<\frac{51}{100}+x$$

Subtract x from both sides.

$$\frac{3}{10}x+\frac{3}{10}-x<\frac{51}{100}$$

Combine \frac{3}{10}x and -x to get -\frac{7}{10}x.

$$-\frac{7}{10}x+\frac{3}{10}<\frac{51}{100}$$

Subtract \frac{3}{10} from both sides.

$$-\frac{7}{10}x<\frac{51}{100}-\frac{3}{10}$$

Least common multiple of 100 and 10 is 100. Convert \frac{51}{100} and \frac{3}{10} to fractions with denominator 100.

$$-\frac{7}{10}x<\frac{51}{100}-\frac{30}{100}$$

Since \frac{51}{100} and \frac{30}{100} have the same denominator, subtract them by subtracting their numerators.

$$-\frac{7}{10}x<\frac{51-30}{100}$$

Subtract 30 from 51 to get 21.

$$-\frac{7}{10}x<\frac{21}{100}$$

Multiply both sides by -\frac{10}{7}, the reciprocal of -\frac{7}{10}. Since -\frac{7}{10} is negative, the inequality direction is changed.

$$x>\frac{21}{100}\left(-\frac{10}{7}\right)$$

Multiply \frac{21}{100} times -\frac{10}{7} by multiplying numerator times numerator and denominator times denominator.

$$x>\frac{21\left(-10\right)}{100\times 7}$$

Do the multiplications in the fraction \frac{21\left(-10\right)}{100\times 7}.

$$x>\frac{-210}{700}$$

Reduce the fraction \frac{-210}{700} to lowest terms by extracting and canceling out 70.

$$x>-\frac{3}{10}$$