Solve {l}{text{(1)}(+2/5)-(-3/5)=1}{text{(2)}(-2/5)-(-3/5)=frac{1}{5}}

$$\left. \begin{array} { l } { \text { (1) } ( + \frac { 2 } { 5 } ) - ( - \frac { 3 } { 5 } ) = 1 } \\ { \text { (2) } ( - \frac { 2 } { 5 } ) - ( - \frac { 3 } { 5 } ) = \frac { 1 } { 5 } } \end{array} \right.$$

Verify

Solution Steps

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

$$\frac{2}{5}-\left(-\frac{3}{5}\right)=1\text{ and }2\left(-\frac{2}{5}\right)-\left(-\frac{3}{5}\right)=\frac{1}{5}$$

The opposite of -\frac{3}{5} is \frac{3}{5}.

$$\frac{2}{5}+\frac{3}{5}=1\text{ and }2\left(-\frac{2}{5}\right)-\left(-\frac{3}{5}\right)=\frac{1}{5}$$

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

$$\frac{2+3}{5}=1\text{ and }2\left(-\frac{2}{5}\right)-\left(-\frac{3}{5}\right)=\frac{1}{5}$$

Add 2 and 3 to get 5.

$$\frac{5}{5}=1\text{ and }2\left(-\frac{2}{5}\right)-\left(-\frac{3}{5}\right)=\frac{1}{5}$$

Divide 5 by 5 to get 1.

$$1=1\text{ and }2\left(-\frac{2}{5}\right)-\left(-\frac{3}{5}\right)=\frac{1}{5}$$

Compare 1 and 1.

$$\text{true}\text{ and }2\left(-\frac{2}{5}\right)-\left(-\frac{3}{5}\right)=\frac{1}{5}$$

Express 2\left(-\frac{2}{5}\right) as a single fraction.

$$\text{true}\text{ and }\frac{2\left(-2\right)}{5}-\left(-\frac{3}{5}\right)=\frac{1}{5}$$

Multiply 2 and -2 to get -4.

$$\text{true}\text{ and }\frac{-4}{5}-\left(-\frac{3}{5}\right)=\frac{1}{5}$$

Fraction \frac{-4}{5} can be rewritten as -\frac{4}{5} by extracting the negative sign.

$$\text{true}\text{ and }-\frac{4}{5}-\left(-\frac{3}{5}\right)=\frac{1}{5}$$

The opposite of -\frac{3}{5} is \frac{3}{5}.

$$\text{true}\text{ and }-\frac{4}{5}+\frac{3}{5}=\frac{1}{5}$$

Since -\frac{4}{5} and \frac{3}{5} have the same denominator, add them by adding their numerators.

$$\text{true}\text{ and }\frac{-4+3}{5}=\frac{1}{5}$$

Add -4 and 3 to get -1.

$$\text{true}\text{ and }-\frac{1}{5}=\frac{1}{5}$$

Compare -\frac{1}{5} and \frac{1}{5}.

$$\text{true}\text{ and }\text{false}$$

The conjunction of \text{true} and \text{false} is \text{false}.

$$\text{false}$$