Solve 4+3/4(x+2)<3/8x+1

$$4+ \frac{ 3 }{ 4 } (x+2) < \frac{ 3 }{ 8 } x+1$$

Solve for x

$$x<-12$$

Solution Steps

Use the distributive property to multiply \frac{3}{4} by x+2.

$$4+\frac{3}{4}x+\frac{3}{4}\times 2<\frac{3}{8}x+1$$

Express \frac{3}{4}\times 2 as a single fraction.

$$4+\frac{3}{4}x+\frac{3\times 2}{4}<\frac{3}{8}x+1$$

Multiply 3 and 2 to get 6.

$$4+\frac{3}{4}x+\frac{6}{4}<\frac{3}{8}x+1$$

Reduce the fraction \frac{6}{4} to lowest terms by extracting and canceling out 2.

$$4+\frac{3}{4}x+\frac{3}{2}<\frac{3}{8}x+1$$

Convert 4 to fraction \frac{8}{2}.

$$\frac{8}{2}+\frac{3}{4}x+\frac{3}{2}<\frac{3}{8}x+1$$

Since \frac{8}{2} and \frac{3}{2} have the same denominator, add them by adding their numerators.

$$\frac{8+3}{2}+\frac{3}{4}x<\frac{3}{8}x+1$$

Add 8 and 3 to get 11.

$$\frac{11}{2}+\frac{3}{4}x<\frac{3}{8}x+1$$

Subtract \frac{3}{8}x from both sides.

$$\frac{11}{2}+\frac{3}{4}x-\frac{3}{8}x<1$$

Combine \frac{3}{4}x and -\frac{3}{8}x to get \frac{3}{8}x.

$$\frac{11}{2}+\frac{3}{8}x<1$$

Subtract \frac{11}{2} from both sides.

$$\frac{3}{8}x<1-\frac{11}{2}$$

Convert 1 to fraction \frac{2}{2}.

$$\frac{3}{8}x<\frac{2}{2}-\frac{11}{2}$$

Since \frac{2}{2} and \frac{11}{2} have the same denominator, subtract them by subtracting their numerators.

$$\frac{3}{8}x<\frac{2-11}{2}$$

Subtract 11 from 2 to get -9.

$$\frac{3}{8}x<-\frac{9}{2}$$

Multiply both sides by \frac{8}{3}, the reciprocal of \frac{3}{8}. Since \frac{3}{8} is positive, the inequality direction remains the same.

$$x<-\frac{9}{2}\times \frac{8}{3}$$

Multiply -\frac{9}{2} times \frac{8}{3} by multiplying numerator times numerator and denominator times denominator.

$$x<\frac{-9\times 8}{2\times 3}$$

Do the multiplications in the fraction \frac{-9\times 8}{2\times 3}.

$$x<\frac{-72}{6}$$

Divide -72 by 6 to get -12.

$$x<-12$$