Solve 4/4+x<7/3*-3/2

$$\frac { 4 } { 4 } + x < \frac { 7 } { 3 } \times \frac { - 3 } { 2 }$$

Solve for x

$$x<-\frac{9}{2}$$

Solution Steps

Divide 4 by 4 to get 1.

$$1+x<\frac{7}{3}\times \frac{-3}{2}$$

Fraction \frac{-3}{2} can be rewritten as -\frac{3}{2} by extracting the negative sign.

$$1+x<\frac{7}{3}\left(-\frac{3}{2}\right)$$

Multiply \frac{7}{3} times -\frac{3}{2} by multiplying numerator times numerator and denominator times denominator.

$$1+x<\frac{7\left(-3\right)}{3\times 2}$$

Do the multiplications in the fraction \frac{7\left(-3\right)}{3\times 2}.

$$1+x<\frac{-21}{6}$$

Reduce the fraction \frac{-21}{6} to lowest terms by extracting and canceling out 3.

$$1+x<-\frac{7}{2}$$

Subtract 1 from both sides.

$$x<-\frac{7}{2}-1$$

Convert 1 to fraction \frac{2}{2}.

$$x<-\frac{7}{2}-\frac{2}{2}$$

Since -\frac{7}{2} and \frac{2}{2} have the same denominator, subtract them by subtracting their numerators.

$$x<\frac{-7-2}{2}$$

Subtract 2 from -7 to get -9.

$$x<-\frac{9}{2}$$