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

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(A) Given equations are:$\frac{2}{a} + \frac{3}{b} = -1$  --- $(1)$$\frac{4}{a} - \frac{9}{b} = -7$  --- $(2)$Let $\frac{1}{a} = x$ and $\frac{1}{b} =

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$(-1, 3)$

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(A) Given equations are:$\frac{2}{a} + \frac{3}{b} = -1$  --- $(1)$$\frac{4}{a} - \frac{9}{b} = -7$  --- $(2)$Let $\frac{1}{a} = x$ and $\frac{1}{b} = y$. Then the equations become:$2x + 3y = -1$  --- $(3)$$4x - 9y = -7$  --- $(4)$Multiply equation $(3)$ by $3$ to eliminate $y$:$6x + 9y = -3$  --- $(5)$Adding equation $(4)$ and $(5)$:$(4x - 9y) + (6x + 9y) = -7 + (-3)$$10x = -10$$x = -1$Since $x = \frac{1}{a}$,we have $\frac{1}{a} = -1$,so $a = -1$.Substitute $x = -1$ into equation $(3)$:$2(-1) + 3y = -1$$-2 + 3y = -1$$3y = 1$$y = \frac{1}{3}$Since $y = \frac{1}{b}$,we have $\frac{1}{b} = \frac{1}{3}$,so $b = 3$.Thus,the solution is $(a, b) = (-1, 3)$.

Solve the following pair of equations: $(a \neq 0, b \neq 0)$
$\frac{2}{a} = -1 - \frac{3}{b}$
$\frac{9}{b} = \frac{4}{a} + 7$

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Solve the following pair of equations: $(a \neq 0, b \neq 0)$$\frac{2}{a} = -1 - \frac{3}{b}$$\frac{9}{b} = \frac{4}{a} + 7$