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

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

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

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

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

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

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

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

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

Solution:

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

Substituting $\quad \frac{\mathbf{1}}{\mathbf{x}}=\mathbf{u a n d} \frac{\mathbf{1}}{\mathbf{v}}=\mathbf{v}$

We get $\frac{1}{2} \mathbf{u}+\frac{1}{3} \mathbf{v}=\mathbf{2}, \frac{1}{3} \mathbf{u}+\frac{1}{8} \mathbf{v}=\frac{13}{6}$

Multiplying by 6 on both sides, we get

$\Rightarrow 3 u+2 v=12$…(ii)

$2 \mathrm{u}+3 \mathrm{v}=13$..(ii)

Multiplying (i) by 3 and (ii) by 2, then subtracting later from first, we get

$3(3 u+2 v)-2(2 u+3 v)=3 \times 12-2 \times 13$

$\Rightarrow 9 \mathrm{u}-4 \mathrm{u}=36-26 \quad \Rightarrow \mathrm{u}=2$

Then substituting $u=2$ in (i), we get

$6+2 v=12$

$\Rightarrow v=3$

Now, $u=2$ and $v=3$

$\Rightarrow \frac{\mathbf{1}}{\mathbf{x}}=2$ and $\frac{\mathbf{1}}{\mathbf{y}}=3 \quad \Rightarrow x=\frac{\mathbf{1}}{\mathbf{2}}$ and $y=\frac{\mathbf{1}}{\mathbf{3}}$

(ii) $\frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}=2$ ..(i)

$\frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}=-1$…(ii)

Take $\frac{\mathbf{1}}{\sqrt{\mathbf{x}}}=\mathrm{a}, \frac{\mathbf{1}}{\sqrt{\mathbf{y}}}=\mathrm{b}$, we get

$2 a+3 b=2$ …(iii)

$4 a-9 b=-1$ …(iv)

Multiplying (iii) by 3 , we get

$6 a+9 b=6$ …(v)

Adding (iv) and (v), we get

$10 a=5$

$\Rightarrow a=\frac{1}{2}$ Substituting $a=\frac{1}{2}$ in (iii), we get

$3 b=1$

$\Rightarrow \mathrm{b}=\frac{\mathbf{1}}{\mathbf{3}}$

Now, $\frac{1}{\sqrt{x}}=a=\frac{1}{2}$ and $\frac{1}{\sqrt{y}}=b=\frac{1}{3}$

$\Rightarrow \sqrt{\mathbf{x}}=2, \sqrt{\mathbf{y}}=3$

Squaring, we get

$x=4, y=9$

(iii) $\frac{4}{x}+3 y=14$ ..(i)

$\frac{3}{x}-4 y=23$ ..(ii)

Take, $\frac{\mathbf{1}}{\mathbf{x}}=\mathrm{a}$

$4 a+3 y=14$ …(iii)

$3 a-4 y=23$ …(iv)

Multiplying (iii) by 4 and (iv) by 3

$16 a+12 y=56$

$9 a-12 y=69$

$25 a=125$

$\Rightarrow a=5$

Substituting a in (iii), we get

$20+3 y=14$

$\Rightarrow 3 y=-6$

$\Rightarrow y=-2$

As, $\frac{1}{x}=a=5$

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

Hence, $x=\frac{1}{5}$ and $y=-2$

(iv) $\frac{5}{(x-1)}+\frac{1}{(y-2)}=2$ …(i)

\frac{6}{(x-1)}-\frac{3}{(y-2)}=1 …(ii)

Take, $\frac{\mathbf{1}}{(\mathbf{x}-\mathbf{1})}=\mathrm{a}$ and $\frac{\mathbf{1}}{(\mathbf{y}-\boldsymbol{z})}=\mathrm{b}$

$5 a+b=2$ ..(iii)

$6 a-3 b=1$ …(iv)

Multiplying (iii) by 3 , we get

$15 a+3 b=6$

$6 a-3 b=1$

$21 a=7$

$a=1 / 3$

So, by solving, $b=1 / 3$

As, $a=\frac{1}{3}=\frac{1}{x-1} \Rightarrow x-1=3 \Rightarrow x=4$

and $\mathrm{b}=\frac{\mathbf{1}}{\mathbf{3}}=\frac{\mathbf{1}}{\mathbf{v}-\mathbf{2}} \Rightarrow \mathrm{y}-2=3$

$\Rightarrow y=5$

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

By solving, we get

$\frac{7}{y}-\frac{2}{x}=5$ (i)

$\frac{8}{y}+\frac{7}{x}=15$ …(ii)

Taking $\frac{\mathbf{1}}{\mathbf{y}}=\mathrm{u}, \frac{\mathbf{1}}{\mathbf{x}}=\mathrm{v}$

$7 u-2 v=5$ …(iii)

$8 u+7 v=15$ …(iv)

Multiplying (iii) by 7 and (iv) by 2 , we get

$49 u-14 v=35$

$16 u+14 v=30$

$65 u=65$

$u=1$

By solving, we get $\mathrm{v}=1$

As, $u=1=\frac{\mathbf{1}}{\mathbf{y}} \Rightarrow y=1$

and $\mathrm{v}=1=\frac{\mathbf{1}}{\mathbf{x}} \Rightarrow \mathrm{x}=1$

(vi) $6 x+3 y=6 x y$

$2 x+4 y=5 x y$

By solving, we get

$\frac{6}{v}+\frac{3}{z}=6$ …(i)

$\frac{2}{y}+\frac{4}{x}=5$ ..(ii)

Take, $\frac{1}{\mathbf{y}}=\mathrm{u}, \frac{\mathbf{1}}{\mathbf{x}}=\mathrm{v}$, we get

$6 u+3 v=6$ ..(iii)

$2 u+4 v=5$ ..(iv)

Multiply (iv) by 3 , we get

$6 u+12 v=15$ …(v)

Subtract (iii) from (v), we get

$9 v=9 v=1$

By solving we get $\mathrm{u}=1 / 2$

$\mathrm{As}, \frac{\mathbf{1}}{\mathbf{x}}=\mathrm{v}=1 \Rightarrow \mathrm{x}=1$

and $\frac{\mathbf{1}}{\mathbf{y}}=\mathrm{u}=\frac{\mathbf{1}}{\mathbf{2}} \Rightarrow \mathrm{y}=2$

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

Take $\frac{\mathbf{1}}{\mathbf{x}+\mathbf{y}}=\mathrm{u}$ and $\frac{\mathbf{1}}{\mathbf{x}-\mathbf{y}}=\mathrm{v}$

$10 u+2 v=4$ …(i)

$15 u-5 v=-2$ …(ii)

Multiply (i) by 5 and (ii) by 2 , we get

$50 u+10 v=20$

$30 \mathrm{u}-10 \mathrm{v}=-4$

$80 \mathrm{u}=16$

$\mathrm{u}=1 / 5$

By solving, we get $\mathrm{v}=1$

$\mathrm{As}, \frac{1}{\mathrm{x}+\mathrm{y}}=\mathrm{u}=\frac{1}{5}$

$\Rightarrow x+y=5$ …(iii)

and $\frac{\mathbf{1}}{\mathbf{x}-\mathbf{y}}=\mathrm{v}=1$

$\Rightarrow x-y=1$ ..(iv)

Adding (iii) and (iv), we get

$2 x=6$

$\Rightarrow x=3$ and $y=2$

Hence, $x=3, y=2$

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

$\frac{1}{2(3 x+y)}-\frac{1}{2(3 x-y)}=-\frac{1}{8}$ …(ii)

Let, $\frac{\mathbf{1}}{\mathbf{3 x}+\mathbf{y}}=\mathrm{u}, \quad \frac{\mathbf{1}}{\mathbf{3 x}-\mathbf{y}}=\mathrm{v}$

$u+v=\frac{3}{4}$ ..(iii)

$\frac{\mathbf{u}}{\mathbf{2}}-\frac{\mathbf{v}}{\mathbf{2}}=-\frac{\mathbf{1}}{\mathbf{8}}$ …(iv)

From (iv), we get

$u-v=-\frac{1}{4}$ …(v)

Solving (iii) and (v), we get

$2 \mathrm{u}=\frac{1}{2}$

$u=\frac{1}{4}, \quad v=\frac{1}{2}$

So, $\frac{1}{3 x+y}=\frac{1}{4} \Rightarrow 3 x+y=4$

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

Solving, we get

$x=1, y=1$
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