Solve |s_{20}|-2/3(3/2-5/2+7/2)=-frac{2}{3}

$$| s _ { 20 } | - \frac { 2 } { 3 } ( \frac { 3 } { 2 } - \frac { 5 } { 2 } + \frac { 7 } { 2 } ) = - \frac { 2 } { 3 }$$

Solve for s_20

$$s_{20}=-1$$

$$s_{20}=1$$

Steps Using Definition of Absolute Value

Since \frac{3}{2} and \frac{5}{2} have the same denominator, subtract them by subtracting their numerators.

$$|s_{20}|-\frac{2}{3}\left(\frac{3-5}{2}+\frac{7}{2}\right)=-\frac{2}{3}$$

Subtract 5 from 3 to get -2.

$$|s_{20}|-\frac{2}{3}\left(\frac{-2}{2}+\frac{7}{2}\right)=-\frac{2}{3}$$

Divide -2 by 2 to get -1.

$$|s_{20}|-\frac{2}{3}\left(-1+\frac{7}{2}\right)=-\frac{2}{3}$$

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

$$|s_{20}|-\frac{2}{3}\left(-\frac{2}{2}+\frac{7}{2}\right)=-\frac{2}{3}$$

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

$$|s_{20}|-\frac{2}{3}\times \frac{-2+7}{2}=-\frac{2}{3}$$

Add -2 and 7 to get 5.

$$|s_{20}|-\frac{2}{3}\times \frac{5}{2}=-\frac{2}{3}$$

Multiply \frac{2}{3} times \frac{5}{2} by multiplying numerator times numerator and denominator times denominator.

$$|s_{20}|-\frac{2\times 5}{3\times 2}=-\frac{2}{3}$$

Cancel out 2 in both numerator and denominator.

$$|s_{20}|-\frac{5}{3}=-\frac{2}{3}$$

Add \frac{5}{3} to both sides.

$$|s_{20}|=-\frac{2}{3}+\frac{5}{3}$$

Since -\frac{2}{3} and \frac{5}{3} have the same denominator, add them by adding their numerators.

$$|s_{20}|=\frac{-2+5}{3}$$

Add -2 and 5 to get 3.

$$|s_{20}|=\frac{3}{3}$$

Divide 3 by 3 to get 1.

$$|s_{20}|=1$$

Use the definition of absolute value.

$$s_{20}=1 s_{20}=-1$$