Solve 21/4x+2leq9/12+13/2

$$\frac { 21 } { 4 } x + 2 \leq \frac { 9 } { 12 } + \frac { 13 } { 2 }$$

Solve for x

$$x\leq 1$$

Solution Steps

Reduce the fraction \frac{9}{12} to lowest terms by extracting and canceling out 3.

$$\frac{21}{4}x+2\leq \frac{3}{4}+\frac{13}{2}$$

Least common multiple of 4 and 2 is 4. Convert \frac{3}{4} and \frac{13}{2} to fractions with denominator 4.

$$\frac{21}{4}x+2\leq \frac{3}{4}+\frac{26}{4}$$

Since \frac{3}{4} and \frac{26}{4} have the same denominator, add them by adding their numerators.

$$\frac{21}{4}x+2\leq \frac{3+26}{4}$$

Add 3 and 26 to get 29.

$$\frac{21}{4}x+2\leq \frac{29}{4}$$

Subtract 2 from both sides.

$$\frac{21}{4}x\leq \frac{29}{4}-2$$

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

$$\frac{21}{4}x\leq \frac{29}{4}-\frac{8}{4}$$

Since \frac{29}{4} and \frac{8}{4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{21}{4}x\leq \frac{29-8}{4}$$

Subtract 8 from 29 to get 21.

$$\frac{21}{4}x\leq \frac{21}{4}$$

Multiply both sides by \frac{4}{21}, the reciprocal of \frac{21}{4}. Since \frac{21}{4} is positive, the inequality direction remains the same.

$$x\leq \frac{21}{4}\times \frac{4}{21}$$

Cancel out \frac{21}{4} and its reciprocal \frac{4}{21}.

$$x\leq 1$$