Solve 3*(-52/9+4/6)*23geq4+c*(5/3+3/8)

$$3 \cdot ( - 5 \frac { 2 } { 9 } + \frac { 4 } { 6 } ) \cdot 23 \geq 4 + c \cdot ( \frac { 5 } { 3 } + \frac { 3 } { 8 } )$$

Solve for c

$$c\leq -\frac{7640}{49}$$

Solution Steps

Multiply both sides of the equation by 72, the least common multiple of 9,6,3,8. Since 72 is positive, the inequality direction remains the same.

$$216\left(-\frac{5\times 9+2}{9}+\frac{4}{6}\right)\times 23\geq 288+72c\left(\frac{5}{3}+\frac{3}{8}\right)$$

Multiply 5 and 9 to get 45.

$$216\left(-\frac{45+2}{9}+\frac{4}{6}\right)\times 23\geq 288+72c\left(\frac{5}{3}+\frac{3}{8}\right)$$

Add 45 and 2 to get 47.

$$216\left(-\frac{47}{9}+\frac{4}{6}\right)\times 23\geq 288+72c\left(\frac{5}{3}+\frac{3}{8}\right)$$

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

$$216\left(-\frac{47}{9}+\frac{2}{3}\right)\times 23\geq 288+72c\left(\frac{5}{3}+\frac{3}{8}\right)$$

Least common multiple of 9 and 3 is 9. Convert -\frac{47}{9} and \frac{2}{3} to fractions with denominator 9.

$$216\left(-\frac{47}{9}+\frac{6}{9}\right)\times 23\geq 288+72c\left(\frac{5}{3}+\frac{3}{8}\right)$$

Since -\frac{47}{9} and \frac{6}{9} have the same denominator, add them by adding their numerators.

$$216\times \frac{-47+6}{9}\times 23\geq 288+72c\left(\frac{5}{3}+\frac{3}{8}\right)$$

Add -47 and 6 to get -41.

$$216\left(-\frac{41}{9}\right)\times 23\geq 288+72c\left(\frac{5}{3}+\frac{3}{8}\right)$$

Express 216\left(-\frac{41}{9}\right) as a single fraction.

$$\frac{216\left(-41\right)}{9}\times 23\geq 288+72c\left(\frac{5}{3}+\frac{3}{8}\right)$$

Multiply 216 and -41 to get -8856.

$$\frac{-8856}{9}\times 23\geq 288+72c\left(\frac{5}{3}+\frac{3}{8}\right)$$

Divide -8856 by 9 to get -984.

$$-984\times 23\geq 288+72c\left(\frac{5}{3}+\frac{3}{8}\right)$$

Multiply -984 and 23 to get -22632.

$$-22632\geq 288+72c\left(\frac{5}{3}+\frac{3}{8}\right)$$

Least common multiple of 3 and 8 is 24. Convert \frac{5}{3} and \frac{3}{8} to fractions with denominator 24.

$$-22632\geq 288+72c\left(\frac{40}{24}+\frac{9}{24}\right)$$

Since \frac{40}{24} and \frac{9}{24} have the same denominator, add them by adding their numerators.

$$-22632\geq 288+72c\times \frac{40+9}{24}$$

Add 40 and 9 to get 49.

$$-22632\geq 288+72c\times \frac{49}{24}$$

Express 72\times \frac{49}{24} as a single fraction.

$$-22632\geq 288+\frac{72\times 49}{24}c$$

Multiply 72 and 49 to get 3528.

$$-22632\geq 288+\frac{3528}{24}c$$

Divide 3528 by 24 to get 147.

$$-22632\geq 288+147c$$

Swap sides so that all variable terms are on the left hand side. This changes the sign direction.

$$288+147c\leq -22632$$

Subtract 288 from both sides.

$$147c\leq -22632-288$$

Subtract 288 from -22632 to get -22920.

$$147c\leq -22920$$

Divide both sides by 147. Since 147 is positive, the inequality direction remains the same.

$$c\leq \frac{-22920}{147}$$

Reduce the fraction \frac{-22920}{147} to lowest terms by extracting and canceling out 3.

$$c\leq -\frac{7640}{49}$$