Solve X=-(-(-2/3))+(2/3);quadY=1+(2/5-4/3)-(frac{2}{5}-frac{4}{3}-1)

$$X = - ( - ( - \frac { 2 } { 3 } ) ) + ( \frac { 2 } { 3 } ) ; \quad Y = 1 + ( \frac { 2 } { 5 } - \frac { 4 } { 3 } ) - ( \frac { 2 } { 5 } - \frac { 4 } { 3 } - 1 )$$

Solve for X, Y

$$X=0$$

$$Y=2$$

Solution Steps

Consider the first equation. The opposite of -\frac{2}{3} is \frac{2}{3}.

$$X=-\frac{2}{3}+\frac{2}{3}$$

Add -\frac{2}{3} and \frac{2}{3} to get 0.

$$X=0$$

Consider the second equation. Add 1 and \frac{2}{5} to get \frac{7}{5}.

$$Y=\frac{7}{5}-\frac{4}{3}-\left(\frac{2}{5}-\frac{4}{3}-1\right)$$

Subtract \frac{4}{3} from \frac{7}{5} to get \frac{1}{15}.

$$Y=\frac{1}{15}-\left(\frac{2}{5}-\frac{4}{3}-1\right)$$

Subtract \frac{4}{3} from \frac{2}{5} to get -\frac{14}{15}.

$$Y=\frac{1}{15}-\left(-\frac{14}{15}-1\right)$$

Subtract 1 from -\frac{14}{15} to get -\frac{29}{15}.

$$Y=\frac{1}{15}-\left(-\frac{29}{15}\right)$$

The opposite of -\frac{29}{15} is \frac{29}{15}.

$$Y=\frac{1}{15}+\frac{29}{15}$$

Add \frac{1}{15} and \frac{29}{15} to get 2.

$$Y=2$$

The system is now solved.

$$X=0 Y=2$$