Question

Solve the following simultaneous equations.

$$\begin{gathered} \left(1\right)\frac{2}{x}-\frac{3}{y}=15;\frac{8}{x}+\frac{5}{y}=77 \\ \left(2\right)\frac{10}{x+y}+\frac{2}{x-y}=4;\frac{15}{x+y}-\frac{5}{x-y}=-2 \\ \left(3\right)\frac{27}{x-2}+\frac{31}{y+3}=85;\frac{31}{x-2}+\frac{27}{y+3}=89 \\ \left(4\right)\frac{1}{3x+y}+\frac{2}{3x-y}=\frac{3}{4};\frac{1}{2\left(3x+y\right)}-\frac{1}{2\left(3x-y\right)}=-\frac{1}{8} \end{gathered}$$

Answer 1

$\left(1\right)\frac{2}{x}-\frac{3}{y}=15;\frac{8}{x}+\frac{5}{y}=77$
$\mathrm{Let}\frac{1}{x}=u\mathrm{and}\frac{1}{y}=v$
So, the equation becomes
$$\begin{gathered} 2u-3v=15.....\left(\mathrm{I}\right) \\ 8u+5v=77.....\left(\mathrm{II}\right) \end{gathered}$$
Multiply (I) with 4 we get
$8u-12v=60.....\left(\mathrm{III}\right)$
(II) − (III)
$$\begin{gathered} 8u-8u+5v-\left(-12v\right)=77-60 \\ \Rightarrow 17v=17 \\ \Rightarrow v=1 \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}v\mathrm{in}\left(\mathrm{I}\right) \\ 2\mathrm{u}-3\left(1\right)=15 \\ \Rightarrow 2\mathrm{u}=15+3=18 \\ \Rightarrow \mathrm{u}=9 \end{gathered}$$
Thus,
$$\begin{gathered} \frac{1}{x}=u=9 \\ \Rightarrow x=\frac{1}{9} \\ \frac{1}{y}=v=1 \\ \Rightarrow y=1 \\ \left(x,y\right)=\left(\frac{1}{9},1\right) \end{gathered}$$

$\left(2\right)\frac{10}{x+y}+\frac{2}{x-y}=4;\frac{15}{x+y}-\frac{5}{x-y}=-2$
Let $\frac{1}{x+y}=u\mathrm{and}\frac{1}{x-y}=v$
So, the equation becomes
$$\begin{gathered} 10u+2v=4.....\left(\mathrm{I}\right) \\ 15u-5v=-2.....\left(\mathrm{II}\right) \end{gathered}$$
Multiplying (I) with 5 and (II) with 2 we get
$$\begin{gathered} 50u+10v=20.....\left(\mathrm{III}\right) \\ 30u-10v=-4.....\left(\mathrm{IV}\right) \end{gathered}$$
Adding (III) and (IV) we get
$u=\frac{16}{80}=\frac{1}{5}$
Putting this value in (I)
$$\begin{gathered} 10\times \frac{1}{5}+2v=4 \\ \Rightarrow 2+2v=4 \\ \Rightarrow v=1 \end{gathered}$$

$$\begin{gathered} \frac{1}{x+y}=\frac{1}{5}\mathrm{and}\frac{1}{x-y}=1 \\ \Rightarrow x+y=5\mathrm{and}x-y=1 \\ \mathrm{Solving}\mathrm{these}\mathrm{equations}\mathrm{we}\mathrm{get} \\ x=3\mathrm{and}y=2 \end{gathered}$$

$\left(3\right)\frac{27}{x-2}+\frac{31}{y+3}=85;\frac{31}{x-2}+\frac{27}{y+3}=89$
$\mathrm{Let}\frac{1}{x-2}=u\mathrm{and}\frac{1}{y+3}=v$
$$\begin{gathered} 27u+31v=85.....\left(\mathrm{I}\right) \\ 31u+27v=89.....\left(\mathrm{II}\right) \\ \mathrm{Adding}\left(\mathrm{I}\right)\mathrm{and}\left(\mathrm{II}\right) \\ 58u+58v=174 \\ u+v=3.....\left(\mathrm{III}\right) \\ \mathrm{Subtracting}\left(\mathrm{II}\right)\mathrm{from}\left(\mathrm{I}\right) \\ 4u-4v=4 \\ \Rightarrow u-v=1.....\left(\mathrm{IV}\right) \end{gathered}$$
Adding (III) and (IV) we get
$$\begin{gathered} 2u=4 \\ \Rightarrow u=2 \end{gathered}$$
Putting the value of u in III
$$\begin{gathered} 2+v=3 \\ \Rightarrow v=1 \end{gathered}$$
$$\begin{gathered} \frac{1}{x-2}=u=2 \\ \Rightarrow x-2=\frac{1}{2} \\ \Rightarrow x=\frac{5}{2} \end{gathered}$$
$$\begin{gathered} \frac{1}{y+3}=1 \\ \Rightarrow y+3=1 \\ \Rightarrow y=-2 \end{gathered}$$
$\left(x,y\right)=\left(\frac{5}{2},-2\right)$

$\left(4\right)\frac{1}{3x+y}+\frac{2}{3x-y}=\frac{3}{4};\frac{1}{2\left(3x+y\right)}-\frac{1}{2\left(3x-y\right)}=-\frac{1}{8}$
Let $\frac{1}{3x+y}=u\mathrm{and}\frac{1}{3x-y}=v$
$u+2v=\frac{3}{4}\mathrm{and}\frac{1}{2}u-\frac{1}{2}v=\frac{-1}{8}$
So, the equations become
$$\begin{gathered} 4u+4v=3.....\left(\mathrm{I}\right) \\ 4u-4v=1.....\left(\mathrm{II}\right) \end{gathered}$$
Adding (I) and (II)
$$\begin{gathered} 8u=4 \\ \Rightarrow u=\frac{1}{2} \end{gathered}$$
Putting the value of u in (I)
$$\begin{gathered} \frac{1}{2}+2v=\frac{3}{4} \\ \Rightarrow v=\frac{1}{4} \end{gathered}$$
$$\begin{gathered} \frac{1}{3x+y}=u\mathrm{and}\frac{1}{3x-y}=v \\ \Rightarrow \frac{1}{3x+y}=\frac{1}{2} \\ 3x+y=2.....\left(\mathrm{III}\right) \\ \mathrm{Also},\frac{1}{3x-y}=\frac{1}{4} \\ \Rightarrow 3x-y=4.....\left(\mathrm{IV}\right) \end{gathered}$$
(III) + (IV) we get
$$\begin{gathered} 6x=6 \\ \Rightarrow x=1 \\ y=-1 \end{gathered}$$