Solve 5x+1/3x<2/3+4+25-31

$$5x+ \frac{ 1 }{ 3 } x < \frac{ 2 }{ 3 } +4+25-31$$

Solve for x

$$x<-\frac{1}{4}$$

Solution Steps

Combine 5x and \frac{1}{3}x to get \frac{16}{3}x.

$$\frac{16}{3}x<\frac{2}{3}+4+25-31$$

Convert 4 to fraction \frac{12}{3}.

$$\frac{16}{3}x<\frac{2}{3}+\frac{12}{3}+25-31$$

Since \frac{2}{3} and \frac{12}{3} have the same denominator, add them by adding their numerators.

$$\frac{16}{3}x<\frac{2+12}{3}+25-31$$

Add 2 and 12 to get 14.

$$\frac{16}{3}x<\frac{14}{3}+25-31$$

Convert 25 to fraction \frac{75}{3}.

$$\frac{16}{3}x<\frac{14}{3}+\frac{75}{3}-31$$

Since \frac{14}{3} and \frac{75}{3} have the same denominator, add them by adding their numerators.

$$\frac{16}{3}x<\frac{14+75}{3}-31$$

Add 14 and 75 to get 89.

$$\frac{16}{3}x<\frac{89}{3}-31$$

Convert 31 to fraction \frac{93}{3}.

$$\frac{16}{3}x<\frac{89}{3}-\frac{93}{3}$$

Since \frac{89}{3} and \frac{93}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{16}{3}x<\frac{89-93}{3}$$

Subtract 93 from 89 to get -4.

$$\frac{16}{3}x<-\frac{4}{3}$$

Multiply both sides by \frac{3}{16}, the reciprocal of \frac{16}{3}. Since \frac{16}{3} is positive, the inequality direction remains the same.

$$x<-\frac{4}{3}\times \frac{3}{16}$$

Multiply -\frac{4}{3} times \frac{3}{16} by multiplying numerator times numerator and denominator times denominator.

$$x<\frac{-4\times 3}{3\times 16}$$

Cancel out 3 in both numerator and denominator.

$$x<\frac{-4}{16}$$

Reduce the fraction \frac{-4}{16} to lowest terms by extracting and canceling out 4.

$$x<-\frac{1}{4}$$