Solve 4/3x+3/2>4x+9

$$\frac { 4 } { 3 } x + \frac { 3 } { 2 } > 4 x + 9$$

Solve for x

$$x<-\frac{45}{16}$$

Solution Steps

Subtract 4x from both sides.

$$\frac{4}{3}x+\frac{3}{2}-4x>9$$

Combine \frac{4}{3}x and -4x to get -\frac{8}{3}x.

$$-\frac{8}{3}x+\frac{3}{2}>9$$

Subtract \frac{3}{2} from both sides.

$$-\frac{8}{3}x>9-\frac{3}{2}$$

Convert 9 to fraction \frac{18}{2}.

$$-\frac{8}{3}x>\frac{18}{2}-\frac{3}{2}$$

Since \frac{18}{2} and \frac{3}{2} have the same denominator, subtract them by subtracting their numerators.

$$-\frac{8}{3}x>\frac{18-3}{2}$$

Subtract 3 from 18 to get 15.

$$-\frac{8}{3}x>\frac{15}{2}$$

Multiply both sides by -\frac{3}{8}, the reciprocal of -\frac{8}{3}. Since -\frac{8}{3} is negative, the inequality direction is changed.

$$x<\frac{15}{2}\left(-\frac{3}{8}\right)$$

Multiply \frac{15}{2} times -\frac{3}{8} by multiplying numerator times numerator and denominator times denominator.

$$x<\frac{15\left(-3\right)}{2\times 8}$$

Do the multiplications in the fraction \frac{15\left(-3\right)}{2\times 8}.

$$x<\frac{-45}{16}$$

Fraction \frac{-45}{16} can be rewritten as -\frac{45}{16} by extracting the negative sign.

$$x<-\frac{45}{16}$$