Solve 3/4(x+1)+x<6

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

Solve for x

$$x<3$$

Solution Steps

Use the distributive property to multiply \frac{3}{4} by x+1.

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

Combine \frac{3}{4}x and x to get \frac{7}{4}x.

$$\frac{7}{4}x+\frac{3}{4}<6$$

Subtract \frac{3}{4} from both sides.

$$\frac{7}{4}x<6-\frac{3}{4}$$

Convert 6 to fraction \frac{24}{4}.

$$\frac{7}{4}x<\frac{24}{4}-\frac{3}{4}$$

Since \frac{24}{4} and \frac{3}{4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{7}{4}x<\frac{24-3}{4}$$

Subtract 3 from 24 to get 21.

$$\frac{7}{4}x<\frac{21}{4}$$

Multiply both sides by \frac{4}{7}, the reciprocal of \frac{7}{4}. Since \frac{7}{4} is positive, the inequality direction remains the same.

$$x<\frac{21}{4}\times \frac{4}{7}$$

Multiply \frac{21}{4} times \frac{4}{7} by multiplying numerator times numerator and denominator times denominator.

$$x<\frac{21\times 4}{4\times 7}$$

Cancel out 4 in both numerator and denominator.

$$x<\frac{21}{7}$$

Divide 21 by 7 to get 3.

$$x<3$$