Solve 9%(x-100)+100>95%(x-50)+50

$$9 \% ( x - 100 ) + 100 > 95 \% ( x - 50 ) + 50$$

Solve for x

$$x<\frac{4425}{43}$$

Solution Steps

Use the distributive property to multiply \frac{9}{100} by x-100.

$$\frac{9}{100}x+\frac{9}{100}\left(-100\right)+100>\frac{95}{100}\left(x-50\right)+50$$

Express \frac{9}{100}\left(-100\right) as a single fraction.

$$\frac{9}{100}x+\frac{9\left(-100\right)}{100}+100>\frac{95}{100}\left(x-50\right)+50$$

Multiply 9 and -100 to get -900.

$$\frac{9}{100}x+\frac{-900}{100}+100>\frac{95}{100}\left(x-50\right)+50$$

Divide -900 by 100 to get -9.

$$\frac{9}{100}x-9+100>\frac{95}{100}\left(x-50\right)+50$$

Add -9 and 100 to get 91.

$$\frac{9}{100}x+91>\frac{95}{100}\left(x-50\right)+50$$

Reduce the fraction \frac{95}{100} to lowest terms by extracting and canceling out 5.

$$\frac{9}{100}x+91>\frac{19}{20}\left(x-50\right)+50$$

Use the distributive property to multiply \frac{19}{20} by x-50.

$$\frac{9}{100}x+91>\frac{19}{20}x+\frac{19}{20}\left(-50\right)+50$$

Express \frac{19}{20}\left(-50\right) as a single fraction.

$$\frac{9}{100}x+91>\frac{19}{20}x+\frac{19\left(-50\right)}{20}+50$$

Multiply 19 and -50 to get -950.

$$\frac{9}{100}x+91>\frac{19}{20}x+\frac{-950}{20}+50$$

Reduce the fraction \frac{-950}{20} to lowest terms by extracting and canceling out 10.

$$\frac{9}{100}x+91>\frac{19}{20}x-\frac{95}{2}+50$$

Convert 50 to fraction \frac{100}{2}.

$$\frac{9}{100}x+91>\frac{19}{20}x-\frac{95}{2}+\frac{100}{2}$$

Since -\frac{95}{2} and \frac{100}{2} have the same denominator, add them by adding their numerators.

$$\frac{9}{100}x+91>\frac{19}{20}x+\frac{-95+100}{2}$$

Add -95 and 100 to get 5.

$$\frac{9}{100}x+91>\frac{19}{20}x+\frac{5}{2}$$

Subtract \frac{19}{20}x from both sides.

$$\frac{9}{100}x+91-\frac{19}{20}x>\frac{5}{2}$$

Combine \frac{9}{100}x and -\frac{19}{20}x to get -\frac{43}{50}x.

$$-\frac{43}{50}x+91>\frac{5}{2}$$

Subtract 91 from both sides.

$$-\frac{43}{50}x>\frac{5}{2}-91$$

Convert 91 to fraction \frac{182}{2}.

$$-\frac{43}{50}x>\frac{5}{2}-\frac{182}{2}$$

Since \frac{5}{2} and \frac{182}{2} have the same denominator, subtract them by subtracting their numerators.

$$-\frac{43}{50}x>\frac{5-182}{2}$$

Subtract 182 from 5 to get -177.

$$-\frac{43}{50}x>-\frac{177}{2}$$

Multiply both sides by -\frac{50}{43}, the reciprocal of -\frac{43}{50}. Since -\frac{43}{50} is negative, the inequality direction is changed.

$$x<-\frac{177}{2}\left(-\frac{50}{43}\right)$$

Multiply -\frac{177}{2} times -\frac{50}{43} by multiplying numerator times numerator and denominator times denominator.

$$x<\frac{-177\left(-50\right)}{2\times 43}$$

Do the multiplications in the fraction \frac{-177\left(-50\right)}{2\times 43}.

$$x<\frac{8850}{86}$$

Reduce the fraction \frac{8850}{86} to lowest terms by extracting and canceling out 2.

$$x<\frac{4425}{43}$$