Solve the following pair of equations: (a neq | vedclass

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(D) Let $\frac{1}{a} = x$ and $\frac{1}{b} = y$. The equations become:$5x + \frac{3}{2}y = 1$ and $\frac{1}{2}x - 3y = 1$Multiplying the first equatio

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$(\frac{7}{2}, -\frac{7}{2})$

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(D) Let $\frac{1}{a} = x$ and $\frac{1}{b} = y$. The equations become:$5x + \frac{3}{2}y = 1$ and $\frac{1}{2}x - 3y = 1$Multiplying the first equation by $2$,we get $10x + 3y = 2$ ... $(1)$Multiplying the second equation by $2$,we get $x - 6y = 2$ ... $(2)$To eliminate $y$,multiply equation $(1)$ by $2$ and add to equation $(2)$:$20x + 6y = 4$$x - 6y = 2$Adding these gives $21x = 6$,so $x = \frac{6}{21} = \frac{2}{7}$.Substitute $x = \frac{2}{7}$ into equation $(1)$:$10(\frac{2}{7}) + 3y = 2$$\frac{20}{7} + 3y = 2$$3y = 2 - \frac{20}{7} = \frac{14 - 20}{7} = -\frac{6}{7}$$y = -\frac{6}{7} 	imes \frac{1}{3} = -\frac{2}{7}$.Since $x = \frac{1}{a} = \frac{2}{7}$,we have $a = \frac{7}{2}$.Since $y = \frac{1}{b} = -\frac{2}{7}$,we have $b = -\frac{7}{2}$.Thus,the solution is $(a, b) = (\frac{7}{2}, -\frac{7}{2})$.

Solve the following pair of equations: $(a \neq 0, b \neq 0)$
$\frac{5}{a} + \frac{3}{2b} = 1$
$\frac{1}{2a} - \frac{3}{b} = 1$

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Solve the following pair of equations: $(a \neq 0, b \neq 0)$$\frac{5}{a} + \frac{3}{2b} = 1$$\frac{1}{2a} - \frac{3}{b} = 1$