Solve the following pairs of linear equations | vedclass

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(B) Given equations are:$(1)$ $\frac{2}{x} + \frac{3}{y} = \frac{9}{xy}$$(2)$ $\frac{4}{x} + \frac{9}{y} = \frac{21}{xy}$Multiply both equations by $x

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$(1, 3)$

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(B) Given equations are:$(1)$ $\frac{2}{x} + \frac{3}{y} = \frac{9}{xy}$$(2)$ $\frac{4}{x} + \frac{9}{y} = \frac{21}{xy}$Multiply both equations by $xy$ (since $x 
eq 0, y 
eq 0$):$(1)$ $2y + 3x = 9 \implies 3x + 2y = 9$$(2)$ $4y + 9x = 21 \implies 9x + 4y = 21$Multiply the first equation by $2$ to eliminate $y$:$2(3x + 2y) = 2(9) \implies 6x + 4y = 18$Subtract this from the second equation:$(9x + 4y) - (6x + 4y) = 21 - 18$$3x = 3 \implies x = 1$Substitute $x = 1$ into $3x + 2y = 9$:$3(1) + 2y = 9$$3 + 2y = 9 \implies 2y = 6 \implies y = 3$Thus,the solution is $(x, y) = (1, 3)$.

Solve the following pairs of linear equations: $\frac{2}{x} + \frac{3}{y} = \frac{9}{xy}, \quad \frac{4}{x} + \frac{9}{y} = \frac{21}{xy} \quad (x \neq 0, y \neq 0)$

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Solve the following pairs of linear equations: $\frac{2}{x} + \frac{3}{y} = \frac{9}{xy}, \quad \frac{4}{x} + \frac{9}{y} = \frac{21}{xy} \quad (x \neq 0, y \neq 0)$

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