2/x + 3/y = 4
3/x -2/y = 19
so I get x = 0.2 and y = -0.5
is x = 0 and y = 0 also a solution?
Hello, CarmineCortez!
. . $\displaystyle \begin{array}{ccc}\dfrac{2}{x} + \dfrac{3}{y} &=& 4 \\ \\ [-3mm] \dfrac{3}{x} -\dfrac{2}{y} &=& 19 \end{array}$
So I get: .$\displaystyle x = \frac{1}{5},\;y = -\frac{1}{2}$ . . . . Correct!
Is $\displaystyle x = 0,\:y = 0$ also a solution? . . . . no
In the original equations, you see that neither $\displaystyle x$ nor $\displaystyle y$ can be zero.