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(B) Given equations are:$1) \frac{x}{2} + \frac{3y}{5} = -1$$2) \frac{x}{2} + \frac{y}{3} = \frac{1}{3}$Subtracting equation $(2)$ from equation $(1)$

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$(4, -5)$

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(B) Given equations are:$1) \frac{x}{2} + \frac{3y}{5} = -1$$2) \frac{x}{2} + \frac{y}{3} = \frac{1}{3}$Subtracting equation $(2)$ from equation $(1)$ to eliminate $x$:$(\frac{x}{2} + \frac{3y}{5}) - (\frac{x}{2} + \frac{y}{3}) = -1 - \frac{1}{3}$$\frac{3y}{5} - \frac{y}{3} = -\frac{4}{3}$$\frac{9y - 5y}{15} = -\frac{4}{3}$$\frac{4y}{15} = -\frac{4}{3}$$y = -\frac{4}{3} 	imes \frac{15}{4} = -5$Substitute $y = -5$ into equation $(2)$:$\frac{x}{2} + \frac{-5}{3} = \frac{1}{3}$$\frac{x}{2} = \frac{1}{3} + \frac{5}{3} = \frac{6}{3} = 2$$x = 4$Thus,the solution is $(x, y) = (4, -5)$.

Solve the following pairs of equations by the method of elimination:
$\frac{x}{2} + \frac{3y}{5} + 1 = 0$
$\frac{x}{2} + \frac{y}{3} - \frac{1}{3} = 0$

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Solve the following pairs of equations by the method of elimination:$\frac{x}{2} + \frac{3y}{5} + 1 = 0$$\frac{x}{2} + \frac{y}{3} - \frac{1}{3} = 0$