Question

Solve the following simultaneous equations.
(1) $\frac{2}{x}+\frac{2}{3y}=\frac{1}{6};\frac{3}{x}+\frac{2}{y}=0$
(2) $\frac{7}{2x+1}+\frac{13}{y+2}=27;\frac{13}{2x+1}+\frac{7}{y+2}=33$
(3) $\frac{148}{x}+\frac{231}{y}=\frac{527}{xy};\frac{231}{x}+\frac{148}{y}=\frac{610}{xy}$
(4) $\frac{7x-2y}{xy}=5;\frac{8x+7y}{xy}=15$
(5) $\frac{1}{2\left(3x+4y\right)}+\frac{1}{5\left(2x-3y\right)}=\frac{1}{4};\frac{5}{\left(3x+4y\right)}-\frac{2}{\left(2x-3y\right)}=-\frac{3}{2}$

Answer 1

(1) $\frac{2}{x}+\frac{2}{3y}=\frac{1}{6};\frac{3}{x}+\frac{2}{y}=0$
Let $\frac{1}{x}=u\mathrm{and}\frac{1}{y}=v$
$$\begin{gathered} 2u+\frac{2}{3}v=\frac{1}{6} \\ 12u+4v=1.....\left(\mathrm{I}\right) \\ 3u+2v=0.....\left(\mathrm{II}\right) \end{gathered}$$
Multiply (II) with 2
$6u+4v=0.....\left(\mathrm{III}\right)$
$\left(\mathrm{I}\right)-\left(\mathrm{III}\right)$
$$\begin{gathered} 6u=1 \\ \Rightarrow u=\frac{1}{6} \end{gathered}$$
Putting the value of u in II.
$$\begin{gathered} 3\times \frac{1}{6}+2v=0 \\ \Rightarrow \frac{1}{2}+2v=0 \\ \Rightarrow v=\frac{-1}{4} \end{gathered}$$
$$\begin{gathered} \frac{1}{x}=u \\ \Rightarrow x=6 \\ \frac{1}{y}=v \\ \Rightarrow y=-4 \\ \left(x,y\right)=\left(6,-4\right) \end{gathered}$$

(2) $\frac{7}{2x+1}+\frac{13}{y+2}=27;\frac{13}{2x+1}+\frac{7}{y+2}=33$
Let $\frac{1}{2x+1}=u\mathrm{and}\frac{1}{y+2}=v$
$$\begin{gathered} 7u+13v=27.....\left(\mathrm{I}\right) \\ 13u+7v=33.....\left(\mathrm{II}\right) \end{gathered}$$
(I) + (II)
$$\begin{gathered} 20u+20v=60 \\ u+v=3.....\left(\mathrm{III}\right) \end{gathered}$$
(II) − (I)
$$\begin{gathered} 6u-6v=6 \\ u-v=1.....\left(\mathrm{IV}\right) \end{gathered}$$
(III) + (IV)
$$\begin{gathered} 2u=4 \\ \Rightarrow u=2 \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}u\mathrm{in}\left(\mathrm{IV}\right) \\ 2-v=1 \\ \Rightarrow v=1 \end{gathered}$$
$$\begin{gathered} \frac{1}{2x+1}=u=2 \\ \Rightarrow 2x+1=\frac{1}{2} \\ \Rightarrow x=\frac{-1}{4} \\ \mathrm{and}\frac{1}{y+2}=v=1 \\ \Rightarrow y+2=1 \\ \Rightarrow y=-1 \\ \left(x,y\right)=\left(\frac{-1}{4},-1\right) \end{gathered}$$

(3) $\frac{148}{x}+\frac{231}{y}=\frac{527}{xy};\frac{231}{x}+\frac{148}{y}=\frac{610}{xy}$
Multiply by xy
$$\begin{gathered} 148y+231x=527.....\left(\mathrm{I}\right) \\ 231y+148x=610.....\left(\mathrm{II}\right) \\ \mathrm{Adding}\left(\mathrm{I}\right)\mathrm{and}\left(\mathrm{II}\right) \\ 379y+379x=1137 \\ \Rightarrow x+y=3.....\left(\mathrm{III}\right) \\ \left(\mathrm{II}\right)-\left(\mathrm{I}\right) \\ 83y-83x=83 \\ \Rightarrow y-x=1.....\left(\mathrm{IV}\right) \\ \left(\mathrm{III}\right)+\left(\mathrm{IV}\right) \\ 2y=4 \\ \Rightarrow y=2 \end{gathered}$$
Putting the value of y in (IV)
$$\begin{gathered} 2-x=1 \\ \Rightarrow x=1 \\ \left(x,y\right)=\left(1,2\right) \end{gathered}$$

(4) $\frac{7x-2y}{xy}=5;\frac{8x+7y}{xy}=15$
$\Rightarrow \frac{7}{y}-\frac{2}{x}=5\mathrm{and}\frac{8}{y}+\frac{7}{x}=15$
Let $\frac{1}{x}=u,\frac{1}{y}=v$
$$\begin{gathered} 7v-2u=5.....\left(\mathrm{I}\right) \\ 8v+7u=15.....\left(\mathrm{II}\right) \end{gathered}$$
Multiply (I) with 7 and (II) with 2
$$\begin{gathered} 49v-14u=35.....\left(\mathrm{III}\right) \\ 16v+14u=30.....\left(\mathrm{IV}\right) \end{gathered}$$
Adding (III) and (IV)
$$\begin{gathered} 65v=65 \\ \Rightarrow v=1 \\ \mathrm{And}\frac{1}{y}=v=1 \\ \Rightarrow y=1 \end{gathered}$$
Putting the value of v in (I)
$$\begin{gathered} 7\left(1\right)-2u=5 \\ \Rightarrow u=1 \\ \frac{1}{x}=u=1 \\ \Rightarrow x=1 \\ \left(x,y\right)=\left(1,1\right) \end{gathered}$$

(5) $\frac{1}{2\left(3x+4y\right)}+\frac{1}{5\left(2x-3y\right)}=\frac{1}{4};\frac{5}{\left(3x+4y\right)}-\frac{2}{\left(2x-3y\right)}=-\frac{3}{2}$
$$\begin{gathered} \frac{1}{3x+4y}=u,\frac{1}{2x-3y}=v \\ \frac{1}{2}u+\frac{1}{5}v=\frac{1}{4} \\ \Rightarrow 10u+4v=5.....\left(\mathrm{I}\right) \\ 5u-2v=\frac{-3}{2} \\ \Rightarrow 10u-4v=-3.....\left(\mathrm{II}\right) \end{gathered}$$
(I) + (II)
$$\begin{gathered} 20u=2 \\ \Rightarrow u=\frac{1}{10} \end{gathered}$$
Putting the value of u in (II)
$$\begin{gathered} 10\times \frac{1}{10}-4v=-3 \\ \Rightarrow 1+3=4v \\ \Rightarrow v=1 \end{gathered}$$
$$\begin{gathered} \frac{1}{3x+4y}=u=\frac{1}{10} \\ \Rightarrow 3x+4y=10.....\left(\mathrm{III}\right) \\ \frac{1}{2x-3y}=v=1 \\ \Rightarrow 2x-3y=1.....\left(\mathrm{IV}\right) \end{gathered}$$
Multiply (III) with 2 and (IV) with 3
$$\begin{gathered} 6x+8y=20.....\left(\mathrm{V}\right) \\ 6x-9y=3.....\left(\mathrm{VI}\right) \end{gathered}$$
(V) − (VI)
$$\begin{gathered} 17y=17 \\ \Rightarrow y=1 \end{gathered}$$
Putting the value of y in (VI)
$$\begin{gathered} 6x-9=3 \\ \Rightarrow 6x=12 \\ \Rightarrow x=2 \\ \left(x,y\right)=\left(2,1\right) \end{gathered}$$