Solve the following pairs of equations:frac{1 | vedclass

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(B) Given pair of linear equations is:$\frac{1}{2x} - \frac{1}{y} = -1$ ........ $(i)$$\frac{1}{x} + \frac{1}{2y} = 8$ ........ $(ii)$Let $u = \frac{1

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$\frac{1}{6}, \frac{1}{4}$

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(B) Given pair of linear equations is:$\frac{1}{2x} - \frac{1}{y} = -1$ ........ $(i)$$\frac{1}{x} + \frac{1}{2y} = 8$ ........ $(ii)$Let $u = \frac{1}{x}$ and $v = \frac{1}{y}$. Substituting these in the equations:$\frac{u}{2} - v = -1 \Rightarrow u - 2v = -2$ ........ $(iii)$$u + \frac{v}{2} = 8 \Rightarrow 2u + v = 16$ ........ $(iv)$Multiply equation $(iv)$ by $2$ to eliminate $v$:$4u + 2v = 32$ ........ $(v)$Adding $(iii)$ and $(v)$:$(u - 2v) + (4u + 2v) = -2 + 32$$5u = 30 \Rightarrow u = 6$Substitute $u = 6$ in equation $(iv)$:$2(6) + v = 16 \Rightarrow 12 + v = 16 \Rightarrow v = 4$Since $u = \frac{1}{x} = 6$,we get $x = \frac{1}{6}$.Since $v = \frac{1}{y} = 4$,we get $y = \frac{1}{4}$.Thus,the solution is $x = \frac{1}{6}$ and $y = \frac{1}{4}$.

Part $I$	Part $II$
$1. x+2y=3, 2x+4y=5$	$a. \text{Solution set is a singleton set.}$
$2. 2x+3y=6, x+\frac{3}{2}y=3$	$b. \text{Solution set is an empty set.}$
$3. 3x-y=0, x-y+6=0$	$c. \text{Solution set is an infinite set.}$
	$d. \text{Solution set contains two members.}$

Solve the following pairs of equations:
$\frac{1}{2x} - \frac{1}{y} = -1$
$\frac{1}{x} + \frac{1}{2y} = 8, \quad x, y \neq 0$

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Solve the following pairs of equations:$\frac{1}{2x} - \frac{1}{y} = -1$$\frac{1}{x} + \frac{1}{2y} = 8, \quad x, y \neq 0$