Question

Solve the following pairs of equations by reducing them to a pair of linear equations:

(i) $\frac{1}{2x}+\frac{1}{3y}=2;\frac{1}{3x}+\frac{1}{2y}=\frac{13}{6}$

(ii) $\frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}=2;\frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}=-1$

(iii) $\frac{4}{x}+3y=14;\frac{3}{x}-4y=23$

(iv) $\frac{5}{x-1}+\frac{1}{y-2}=2;\frac{6}{x-1}-\frac{3}{y-2}=1$

(v) $\frac{7x-2y}{xy}=5;\frac{8x+7y}{xy}=15$

(vi) $6x+3y=6xy;2x+4y=5xy$

(vii) $\frac{10}{x+y}+\frac{2}{x-y}=4;\frac{15}{x+y}-\frac{5}{x-y}=-2$

(viii) $\frac{1}{3x+y}+\frac{1}{3x-y}=\frac{3}{4};\frac{1}{2(3x+y)}-\frac{1}{2(3x-y)}=\frac{-1}{8}$

Answer 1

(i)

$\frac{1}{2x}+\frac{1}{3y}=2$

$\frac{1}{3x}+\frac{1}{2y}=\frac{13}{6}$

Lets assume $\frac{1}{x}=p$ and $\frac{1}{y}=q$, both equations will become

$\frac{p}{2}+\frac{q}{3}=2$

$\Rightarrow 3p+2q=12$ ... (A)

$\frac{p}{3}+\frac{q}{2}=\frac{13}{6}$

$\Rightarrow 2p+3q=13$ ...(B)

Multiply (A) by $3$ and (B) by $2$, we get

$9p+6q=36$

$4p+6q=26$

On Subtracting these both, we get

$9p+6q-(4p+6q)=36-26$

$\Rightarrow 5p=10$

$\Rightarrow p=2$

$\Rightarrow x=\frac{1}{p}=\frac{1}{2}$

$\therefore 4\times \frac{1}{2}+6q=26$

$q=\frac{26-2}{6}=4$

$\Rightarrow y=\frac{1}{q}=\frac{1}{4}$

(ii)

$\frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}=2$

$\frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}=-1$

Lets assume $\frac{1}{\sqrt{x}}=p$ and $\frac{1}{\sqrt{y}}=q$,both equations will become

$2p+3q=2$ ...(A)

$4p-9q=-1$ ...(B)

Multiply (A) by $3$, we get

$6p+9q=6$

$4p-9q=-1$

Adding these both, we get

$6p+9q+4p-9q=6+(-1)$

$10p=5$

$\Rightarrow p=\frac{1}{2}$

$\Rightarrow x=\frac{1}{p^2}=4$ [$\because \frac{1}{\sqrt{x}}=p,(\frac{1}{\sqrt{x}}{)}^2=p^2,\frac{1}{x}=p^2,x=\frac{1}{p^2}$]

$\therefore 4\times \frac{1}{2}-9q=-1$

$\Rightarrow -9q=-3$

$\Rightarrow q=\frac{1}{3}$

$\Rightarrow y=\frac{1}{q^2}=9$

(iii)

$\frac{4}{x}+3y=14$

$\frac{3}{x}-4y=23$

Lets assume $\frac{1}{x}=p$, both equations become

$4p+3y=14$ ...(A)

$3p-4y=23$ ...(B)

Multiply (A) by $4$ and (B) by $3$, we get

$16p+12y=56$

$9p-12y=69$

Adding these both, we get

$25p=125$

$\Rightarrow p=5$

$\Rightarrow x=\frac{1}{p}=\frac{1}{5}$

$\Rightarrow 3\times 5-4y=23$

$\Rightarrow -4y=23-15=8$

$\Rightarrow y=-2$

(iv)

$\frac{5}{x-1}+\frac{1}{y-2}=2$

$\frac{6}{x-1}-\frac{3}{y-2}=1$

Lets assume $\frac{1}{x-1}=p$ and $\frac{1}{y-2}=q$, both equations become

$5p+q=2$

$6p-3q=1$

Multiply (A) by $3$, we get

$15p+3q=6$

$6p-3q=1$

Adding both of these, we get

$21p=7$

$\Rightarrow p=\frac{1}{3}$

$\Rightarrow \frac{1}{x-1}=\frac{1}{3}$

$\Rightarrow x-1=3$

$\Rightarrow x=4$

$\Rightarrow 6\times \frac{1}{3}-3q=1$

$\Rightarrow -3q=-1$

$\therefore q=\frac{1}{3}$

$\Rightarrow \frac{1}{y-2}=\frac{1}{3}$

$\Rightarrow y-2=3$

$\Rightarrow y=5$

(v)

$\frac{7x-2y}{xy}=5$

$\Rightarrow \frac{7}{y}-\frac{2}{x}=5$

$\frac{8x+7y}{xy}=15$

$\Rightarrow \frac{8}{y}+\frac{7}{x}=15$

Let $t=\frac{1}{x}$ and $r=\frac{1}{y}$ equation will be

$7r-2t=5....\left(A\right);8r+7t=\mathrm{15......}\left(B\right)$

on multiplying (A) by $7$ and (B) by $2$, then adding these, we get

$\Rightarrow (49r-14t)+(16r+14)=35+30$

$\Rightarrow 65r=65$

On solving we get
$t=1r=1$
$\therefore x=1andy=1$
(vi) $6x+3y=6xy;2x+4y=5xy$
Dividing both side by $xy$ then we have
$\frac{6}{y}+\frac{3}{x}=6;\frac{2}{y}+\frac{4}{x}=5$
Let $t=\frac{1}{x}andr=\frac{1}{y}$ equation will be
$6r+3t=\mathrm{6....}\left(A\right);2r+4t=\mathrm{5......}\left(B\right)$

On multiplying (B) by $3$ and then subtracting (B) from (A), we get

$\Rightarrow 6r+12t-(6r+3t)=15-6$

$\Rightarrow t=1$

on solving we get
$t=1,r=\frac{1}{2}$
$\therefore x=1andy=2$
(vii) $\frac{10}{x+y}+\frac{2}{x-y}=4;\frac{15}{x+y}-\frac{5}{x-y}=-2$
Let $t=\frac{1}{x+y}andr=\frac{1}{x-y}$ then equation will be
$10t+2r=\mathrm{4....}\left(A\right);15t-5r=-\mathrm{2....}\left(B\right)$

On multiplying (A) by $5$ and (B) by $2$, and then adding, we get

$\Rightarrow 50t+10r+30t-10r=20-4$

$\Rightarrow 80t=16$

on solving we have
$t=\frac{1}{5}andr=1$
$\therefore x+y=5andx-y=1$
Hence, $x=4,y=1$
(viii) $\frac{1}{3x+y}+\frac{1}{3x-y}=\frac{3}{4};\frac{1}{2(3x+y)}-\frac{1}{2(3x-y)}=\frac{-1}{8}$
Let $t=\frac{1}{3x+y}andr=\frac{1}{3x-y}$ then equation will be
$t+r=\frac{3}{4};\frac{t}{2}-\frac{r}{2}=\frac{-1}{8}$

$4t+4r=\mathrm{3....}\left(A\right)4t-4r=-\mathrm{1....}\left(B\right)$

on adding (A) and (B), we get

$4t+4r+4t-4r=3-1$

$8t=2$

On solving we have
$t=\frac{1}{4}andr=\frac{1}{2}$
$\therefore 3x+y=4and3x-y=2$
Hence, $x=1,y=1$