Solve 6-5/3x>4x+10

$$6 - \frac { 5 } { 3 } x > 4 x + 10$$

Solve for x

$$x<-\frac{12}{17}$$

Solution Steps

Subtract 4x from both sides.

$$6-\frac{5}{3}x-4x>10$$

Combine -\frac{5}{3}x and -4x to get -\frac{17}{3}x.

$$6-\frac{17}{3}x>10$$

Subtract 6 from both sides.

$$-\frac{17}{3}x>10-6$$

Subtract 6 from 10 to get 4.

$$-\frac{17}{3}x>4$$

Multiply both sides by -\frac{3}{17}, the reciprocal of -\frac{17}{3}. Since -\frac{17}{3} is negative, the inequality direction is changed.

$$x<4\left(-\frac{3}{17}\right)$$

Express 4\left(-\frac{3}{17}\right) as a single fraction.

$$x<\frac{4\left(-3\right)}{17}$$

Multiply 4 and -3 to get -12.

$$x<\frac{-12}{17}$$

Fraction \frac{-12}{17} can be rewritten as -\frac{12}{17} by extracting the negative sign.

$$x<-\frac{12}{17}$$