Solve the following pairs of equations:frac{2 | vedclass

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(A) Given pair of equations is:$\frac{2xy}{x+y} = \frac{3}{2} \implies \frac{x+y}{2xy} = \frac{2}{3} \implies \frac{1}{y} + \frac{1}{x} = \frac{4}{3}

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$\frac{1}{2}, \frac{-3}{2}$

Vedclass · Answer

(A) Given pair of equations is:$\frac{2xy}{x+y} = \frac{3}{2} \implies \frac{x+y}{2xy} = \frac{2}{3} \implies \frac{1}{y} + \frac{1}{x} = \frac{4}{3} \quad ...(i)$$\frac{xy}{2x-y} = \frac{-3}{10} \implies \frac{2x-y}{xy} = \frac{-10}{3} \implies \frac{2}{y} - \frac{1}{x} = \frac{-10}{3} \quad ...(ii)$Let $\frac{1}{x} = u$ and $\frac{1}{y} = v$. Then the equations become:$v + u = \frac{4}{3} \quad ...(iii)$$2v - u = \frac{-10}{3} \quad ...(iv)$Adding $(iii)$ and $(iv)$:$(v + u) + (2v - u) = \frac{4}{3} - \frac{10}{3}$$3v = \frac{-6}{3} = -2 \implies v = \frac{-2}{3}$Substituting $v = \frac{-2}{3}$ in $(iii)$:$u + (\frac{-2}{3}) = \frac{4}{3} \implies u = \frac{4}{3} + \frac{2}{3} = \frac{6}{3} = 2$Since $u = \frac{1}{x} = 2$,we get $x = \frac{1}{2}$.Since $v = \frac{1}{y} = \frac{-2}{3}$,we get $y = \frac{-3}{2}$.Thus,the solution is $x = \frac{1}{2}, y = \frac{-3}{2}$.

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:
$\frac{2xy}{x+y} = \frac{3}{2}$
$\frac{xy}{2x-y} = \frac{-3}{10}, \quad x+y \neq 0, 2x-y \neq 0$

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Solve the following pairs of equations:$\frac{2xy}{x+y} = \frac{3}{2}$$\frac{xy}{2x-y} = \frac{-3}{10}, \quad x+y \neq 0, 2x-y \neq 0$