Solve 7/3(x+1)geq4/5(6x-5)

$$\frac{ 7 }{ 3 } (x+1) \geq \frac{ 4 }{ 5 } (6x-5)$$

Solve for x

$$x\leq \frac{95}{37}$$

Solution Steps

Use the distributive property to multiply \frac{7}{3} by x+1.

$$\frac{7}{3}x+\frac{7}{3}\geq \frac{4}{5}\left(6x-5\right)$$

Use the distributive property to multiply \frac{4}{5} by 6x-5.

$$\frac{7}{3}x+\frac{7}{3}\geq \frac{4}{5}\times 6x+\frac{4}{5}\left(-5\right)$$

Express \frac{4}{5}\times 6 as a single fraction.

$$\frac{7}{3}x+\frac{7}{3}\geq \frac{4\times 6}{5}x+\frac{4}{5}\left(-5\right)$$

Multiply 4 and 6 to get 24.

$$\frac{7}{3}x+\frac{7}{3}\geq \frac{24}{5}x+\frac{4}{5}\left(-5\right)$$

Express \frac{4}{5}\left(-5\right) as a single fraction.

$$\frac{7}{3}x+\frac{7}{3}\geq \frac{24}{5}x+\frac{4\left(-5\right)}{5}$$

Multiply 4 and -5 to get -20.

$$\frac{7}{3}x+\frac{7}{3}\geq \frac{24}{5}x+\frac{-20}{5}$$

Divide -20 by 5 to get -4.

$$\frac{7}{3}x+\frac{7}{3}\geq \frac{24}{5}x-4$$

Subtract \frac{24}{5}x from both sides.

$$\frac{7}{3}x+\frac{7}{3}-\frac{24}{5}x\geq -4$$

Combine \frac{7}{3}x and -\frac{24}{5}x to get -\frac{37}{15}x.

$$-\frac{37}{15}x+\frac{7}{3}\geq -4$$

Subtract \frac{7}{3} from both sides.

$$-\frac{37}{15}x\geq -4-\frac{7}{3}$$

Convert -4 to fraction -\frac{12}{3}.

$$-\frac{37}{15}x\geq -\frac{12}{3}-\frac{7}{3}$$

Since -\frac{12}{3} and \frac{7}{3} have the same denominator, subtract them by subtracting their numerators.

$$-\frac{37}{15}x\geq \frac{-12-7}{3}$$

Subtract 7 from -12 to get -19.

$$-\frac{37}{15}x\geq -\frac{19}{3}$$

Multiply both sides by -\frac{15}{37}, the reciprocal of -\frac{37}{15}. Since -\frac{37}{15} is negative, the inequality direction is changed.

$$x\leq -\frac{19}{3}\left(-\frac{15}{37}\right)$$

Multiply -\frac{19}{3} times -\frac{15}{37} by multiplying numerator times numerator and denominator times denominator.

$$x\leq \frac{-19\left(-15\right)}{3\times 37}$$

Do the multiplications in the fraction \frac{-19\left(-15\right)}{3\times 37}.

$$x\leq \frac{285}{111}$$

Reduce the fraction \frac{285}{111} to lowest terms by extracting and canceling out 3.

$$x\leq \frac{95}{37}$$