Solve 1-1+x/3<x

$$1 - \frac { 1 + x } { 3 } < x$$

Solve for x

$$x>\frac{1}{2}$$

Solution Steps

Divide each term of 1+x by 3 to get \frac{1}{3}+\frac{1}{3}x.

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

To find the opposite of \frac{1}{3}+\frac{1}{3}x, find the opposite of each term.

$$1-\frac{1}{3}-\frac{1}{3}x<x$$

Convert 1 to fraction \frac{3}{3}.

$$\frac{3}{3}-\frac{1}{3}-\frac{1}{3}x<x$$

Since \frac{3}{3} and \frac{1}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{3-1}{3}-\frac{1}{3}x<x$$

Subtract 1 from 3 to get 2.

$$\frac{2}{3}-\frac{1}{3}x<x$$

Subtract x from both sides.

$$\frac{2}{3}-\frac{1}{3}x-x<0$$

Combine -\frac{1}{3}x and -x to get -\frac{4}{3}x.

$$\frac{2}{3}-\frac{4}{3}x<0$$

Subtract \frac{2}{3} from both sides. Anything subtracted from zero gives its negation.

$$-\frac{4}{3}x<-\frac{2}{3}$$

Multiply both sides by -\frac{3}{4}, the reciprocal of -\frac{4}{3}. Since -\frac{4}{3} is negative, the inequality direction is changed.

$$x>-\frac{2}{3}\left(-\frac{3}{4}\right)$$

Multiply -\frac{2}{3} times -\frac{3}{4} by multiplying numerator times numerator and denominator times denominator.

$$x>\frac{-2\left(-3\right)}{3\times 4}$$

Do the multiplications in the fraction \frac{-2\left(-3\right)}{3\times 4}.

$$x>\frac{6}{12}$$

Reduce the fraction \frac{6}{12} to lowest terms by extracting and canceling out 6.

$$x>\frac{1}{2}$$