Question

Solve the following simultaneous equations.

(i) $\frac{x}{3}+\frac{y}{4}=4;\frac{x}{2}-\frac{y}{4}=1$

(ii) $\frac{x}{3}+5y=13;2x+\frac{y}{2}=19$

(iii) $\frac{2}{x}+\frac{3}{y}=13;\frac{5}{x}-\frac{4}{y}=-2$

Answer 1

(i) $\frac{x}{3}+\frac{y}{4}=4$ .....(I)
$\frac{x}{2}-\frac{y}{4}=1$ .....(II)
Multiplying (I) with LCM of 3 and 4 which is 12 we get
$4x+3y=48$ .....(III)
Multiplying (II) with LCM of 2 and 4 which is 4 we get
$2x-y=4$ .....(IV)
Multiplying (IV) with 2
$4x-2y=8$ .....(V)
Subtracting (V) from (III)
$$\begin{gathered} 5y=40 \\ \Rightarrow y=8 \end{gathered}$$
Putting this value of y in (IV) we get
$$\begin{gathered} 2x-8=4 \\ \Rightarrow 2x=12 \\ \Rightarrow x=6 \end{gathered}$$

(ii)
$$\begin{gathered} \frac{x}{3}+5y=13.....\left(\mathrm{I}\right) \\ 2x+\frac{y}{2}=19.....\left(\mathrm{II}\right) \end{gathered}$$
$$\begin{gathered} \Rightarrow x+15y=39.....\left(\mathrm{III}\right) \\ 4x+y=38.....\left(\mathrm{IV}\right) \end{gathered}$$
Multiplying (III) with 4 we get
$4x+60y=156$ .....(V)
Subtracting (IV) from (V)
$$\begin{gathered} 59y=118 \\ \Rightarrow y=2 \end{gathered}$$
Putting the value of y in (IV) we get
$$\begin{gathered} 4x+2=38 \\ \Rightarrow 4x=36 \\ \Rightarrow x=9 \end{gathered}$$

(iii) $\frac{2}{x}+\frac{3}{y}=13;\frac{5}{x}-\frac{4}{y}=-2$
Let $\frac{1}{x}=u\mathrm{and}\frac{1}{y}=v$
So, the equations obtained are
$$\begin{gathered} 2u+3v=13.....\left(\mathrm{I}\right)\left(\times 5\right) \\ 5u-4v=-2.....\left(\mathrm{II}\right)\left(\times 2\right) \\ 10u+15v=65.....\left(\mathrm{III}\right) \\ 10u-8v=-4.....\left(\mathrm{IV}\right) \\ \mathrm{Subtracting}\left(\mathrm{IV}\right)\mathrm{from}\left(\mathrm{III}\right) \\ 23v=69 \\ \Rightarrow v=3 \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}v\mathrm{in}\left(\mathrm{I}\right) \\ 2u+3\times 3=13 \\ \Rightarrow 2u=4 \\ \Rightarrow u=2 \\ \frac{1}{x}=u\Rightarrow x=\frac{1}{u}=\frac{1}{2} \\ \frac{1}{y}=v\Rightarrow y=\frac{1}{v}=\frac{1}{3} \end{gathered}$$