Solve 1/5(x-2)-[1+2x-(x+1/2)]leq1

$$\frac { 1 } { 5 } ( x - 2 ) - [ 1 + 2 x - ( x + \frac { 1 } { 2 } ) ] \leq 1$$

Solve for x

$$x\geq -\frac{19}{8}$$

Solution Steps

Use the distributive property to multiply \frac{1}{5} by x-2.

$$\frac{1}{5}x+\frac{1}{5}\left(-2\right)-\left(1+2x-\left(x+\frac{1}{2}\right)\right)\leq 1$$

Multiply \frac{1}{5} and -2 to get \frac{-2}{5}.

$$\frac{1}{5}x+\frac{-2}{5}-\left(1+2x-\left(x+\frac{1}{2}\right)\right)\leq 1$$

Fraction \frac{-2}{5} can be rewritten as -\frac{2}{5} by extracting the negative sign.

$$\frac{1}{5}x-\frac{2}{5}-\left(1+2x-\left(x+\frac{1}{2}\right)\right)\leq 1$$

To find the opposite of x+\frac{1}{2}, find the opposite of each term.

$$\frac{1}{5}x-\frac{2}{5}-\left(1+2x-x-\frac{1}{2}\right)\leq 1$$

Combine 2x and -x to get x.

$$\frac{1}{5}x-\frac{2}{5}-\left(1+x-\frac{1}{2}\right)\leq 1$$

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

$$\frac{1}{5}x-\frac{2}{5}-\left(\frac{2}{2}+x-\frac{1}{2}\right)\leq 1$$

Since \frac{2}{2} and \frac{1}{2} have the same denominator, subtract them by subtracting their numerators.

$$\frac{1}{5}x-\frac{2}{5}-\left(\frac{2-1}{2}+x\right)\leq 1$$

Subtract 1 from 2 to get 1.

$$\frac{1}{5}x-\frac{2}{5}-\left(\frac{1}{2}+x\right)\leq 1$$

To find the opposite of \frac{1}{2}+x, find the opposite of each term.

$$\frac{1}{5}x-\frac{2}{5}-\frac{1}{2}-x\leq 1$$

Least common multiple of 5 and 2 is 10. Convert -\frac{2}{5} and \frac{1}{2} to fractions with denominator 10.

$$\frac{1}{5}x-\frac{4}{10}-\frac{5}{10}-x\leq 1$$

Since -\frac{4}{10} and \frac{5}{10} have the same denominator, subtract them by subtracting their numerators.

$$\frac{1}{5}x+\frac{-4-5}{10}-x\leq 1$$

Subtract 5 from -4 to get -9.

$$\frac{1}{5}x-\frac{9}{10}-x\leq 1$$

Combine \frac{1}{5}x and -x to get -\frac{4}{5}x.

$$-\frac{4}{5}x-\frac{9}{10}\leq 1$$

Add \frac{9}{10} to both sides.

$$-\frac{4}{5}x\leq 1+\frac{9}{10}$$

Convert 1 to fraction \frac{10}{10}.

$$-\frac{4}{5}x\leq \frac{10}{10}+\frac{9}{10}$$

Since \frac{10}{10} and \frac{9}{10} have the same denominator, add them by adding their numerators.

$$-\frac{4}{5}x\leq \frac{10+9}{10}$$

Add 10 and 9 to get 19.

$$-\frac{4}{5}x\leq \frac{19}{10}$$

Multiply both sides by -\frac{5}{4}, the reciprocal of -\frac{4}{5}. Since -\frac{4}{5} is negative, the inequality direction is changed.

$$x\geq \frac{19}{10}\left(-\frac{5}{4}\right)$$

Multiply \frac{19}{10} times -\frac{5}{4} by multiplying numerator times numerator and denominator times denominator.

$$x\geq \frac{19\left(-5\right)}{10\times 4}$$

Do the multiplications in the fraction \frac{19\left(-5\right)}{10\times 4}.

$$x\geq \frac{-95}{40}$$

Reduce the fraction \frac{-95}{40} to lowest terms by extracting and canceling out 5.

$$x\geq -\frac{19}{8}$$