Solve 3/8(7+2x)+3geq18

$$\frac { 3 } { 8 } ( 7 + 2 x ) + 3 \geq 18$$

Solve for x

$$x\geq \frac{33}{2}$$

Solution Steps

Use the distributive property to multiply \frac{3}{8} by 7+2x.

$$\frac{3}{8}\times 7+\frac{3}{8}\times 2x+3\geq 18$$

Express \frac{3}{8}\times 7 as a single fraction.

$$\frac{3\times 7}{8}+\frac{3}{8}\times 2x+3\geq 18$$

Multiply 3 and 7 to get 21.

$$\frac{21}{8}+\frac{3}{8}\times 2x+3\geq 18$$

Express \frac{3}{8}\times 2 as a single fraction.

$$\frac{21}{8}+\frac{3\times 2}{8}x+3\geq 18$$

Multiply 3 and 2 to get 6.

$$\frac{21}{8}+\frac{6}{8}x+3\geq 18$$

Reduce the fraction \frac{6}{8} to lowest terms by extracting and canceling out 2.

$$\frac{21}{8}+\frac{3}{4}x+3\geq 18$$

Convert 3 to fraction \frac{24}{8}.

$$\frac{21}{8}+\frac{3}{4}x+\frac{24}{8}\geq 18$$

Since \frac{21}{8} and \frac{24}{8} have the same denominator, add them by adding their numerators.

$$\frac{21+24}{8}+\frac{3}{4}x\geq 18$$

Add 21 and 24 to get 45.

$$\frac{45}{8}+\frac{3}{4}x\geq 18$$

Subtract \frac{45}{8} from both sides.

$$\frac{3}{4}x\geq 18-\frac{45}{8}$$

Convert 18 to fraction \frac{144}{8}.

$$\frac{3}{4}x\geq \frac{144}{8}-\frac{45}{8}$$

Since \frac{144}{8} and \frac{45}{8} have the same denominator, subtract them by subtracting their numerators.

$$\frac{3}{4}x\geq \frac{144-45}{8}$$

Subtract 45 from 144 to get 99.

$$\frac{3}{4}x\geq \frac{99}{8}$$

Multiply both sides by \frac{4}{3}, the reciprocal of \frac{3}{4}. Since \frac{3}{4} is positive, the inequality direction remains the same.

$$x\geq \frac{99}{8}\times \frac{4}{3}$$

Multiply \frac{99}{8} times \frac{4}{3} by multiplying numerator times numerator and denominator times denominator.

$$x\geq \frac{99\times 4}{8\times 3}$$

Do the multiplications in the fraction \frac{99\times 4}{8\times 3}.

$$x\geq \frac{396}{24}$$

Reduce the fraction \frac{396}{24} to lowest terms by extracting and canceling out 12.

$$x\geq \frac{33}{2}$$