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 $$\frac{8 x+7 y}{x y}=15 ,b>(vi) 6 x+3 y=6 x y$$ 2 x+4 y=5 x y$4(vii) $\frac{10}{x+y}+\frac{2}{x-y}=4 $$\frac{15}{x+y}-\frac{5}{x-y}=-2 (viii) \frac{1}{3 x+y}+\frac{1}{3 x-y}=\frac{3}{4}$$ \frac{1}{2(3 x+y)}-\frac{1}{2(3 x-y)}=\frac{-1}{8}$.

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To do:

We have to solve the given pairs of equations by reducing them to a pair of linear equations.

Solution:

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

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

Put $\frac{1}{x}=u$ and $\frac{1}{y}=v$ in equations (i) and (ii), we get,

$\frac{1}{2} u+\frac{1}{3} v=2$

$\Rightarrow \frac{3 u+2 v}{6}=2$

$\Rightarrow 3 u+2 v=12$.....(iii)

From equation (ii),

$\frac{1}{3} u+\frac{1}{2} v=\frac{13}{6}$

$\Rightarrow \frac{2 u+3 v}{6}=\frac{13}{6}$

$2u+3v=13$......(iv)
Solving (iii) and (iv), we get,

By cross multiplication method,

$\frac{u}{2(13)-3(12)} =\frac{v}{12(2)-13(3)}=\frac{-1}{3(3)-2(2)}$

$\frac{u}{-10}=\frac{v}{-15}=\frac{-1}{5}$

$\frac{u}{-10}=\frac{-1}{5}$ and $\frac{v}{-15}=\frac{-1}{5}$

$u=2, v=3$

Therefore,

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

$y=\frac{1}{v}=\frac{1}{3}$

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

$\frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}=-1$.........(ii) Let $\frac{1}{\sqrt{x}}=u$ and  $\frac{1}{\sqrt{y}}=v$

Put the above values in and (ii),

$2 u+3 v=2$......(iii)

$4u-9v=-1$........(iv)

$\frac{u}{-3+18}=\frac{v}{8+2}=\frac{-1}{-18-12}$

$\frac{u}{15}=\frac{v}{10}=\frac{1}{30}$

$u=\frac{1}{2}$ and $v=\frac{1}{3}$

$u=\frac{1}{\sqrt{x}}=\frac{1}{2}$

This implies,

$x=2^2$

$x=4$

$v=\frac{1}{\sqrt{y}}=\frac{1}{3}$

This implies,

$y=3^2$

$y=9$

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

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

Let $\frac{1}{x}=u$

Put the above values in equations (i) and (ii),

$4u+3y=14$...........(iii)

$3u-4y=23$..........(iv)

$\frac{u}{69+56}=\frac{y}{42-92}=\frac{-1}{-16-9}$

$\frac{u}{125}=\frac{y}{-50}=\frac{-1}{-25}$

$u=\frac{125}{25}$

$u=5$

$y=\frac{-50}{25}$

$y=-2$

This implies,

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

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

$y=-2$

(iv) Let $\frac{1}{x-1}=u$ and $\frac{1}{y-2}=v$

This implies, the given system of equations can be written as,

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

$5u+v=2$

$5u+v-2=0$---(i)

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

$6u-3v=1$

$6u=1+3v$

$u=\frac{1+3v}{6}$---(ii)

Substituting $u=\frac{1+3v}{6}$ in equation (i), we get,

$5(\frac{1+3v}{6})+v-2=0$

$\frac{5(1+3v)}{6}=2-v$

$5+15v=6(2-v)$

$5+15v=12-6v$

$15v+6v=12-5$

$21v=7$

$v=\frac{7}{21}$

$v=\frac{1}{3}$

Using $v=\frac{1}{3}$ in equation (i), we get,

$5u+(\frac{1}{3})-2=0$

$5u+\frac{1-2\times3}{3}=0$

$5u+\frac{-5}{3}=0$

$5u=\frac{5}{3}$

$u=\frac{5}{5\times3}$

$u=\frac{1}{3}$

Using the values of $u$ and $v$, we get,

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

$\Rightarrow x-1=3$

$\Rightarrow x=3+1$

$\Rightarrow x=4$

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

$\Rightarrow y-2=3$

$\Rightarrow y=3+2$

$\Rightarrow y=5$

Therefore, the solution of the given system of equations is $x=4$ and $y=5$.

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

$\frac{8 x+7 y}{x y}=15$........(ii)

Let $\frac{1}{y}=u$ and $\frac{1}{x}=v$

Substitute the above values in (i) and (ii)

$7 u-2 v=5$........(iii)

$8 u+7 v=15$..........(iv)

$\frac{u}{-30-35}=\frac{y}{40-105}=\frac{-1}{49+16}$

$\frac{u}{-65}=\frac{y}{-65}=\frac{-1}{65}$

$u=\frac{65}{65}$

$u=1$

$v=\frac{65}{65}$

$v=1$

This implies,

$x=\frac{1}{v}=1$

$y=\frac{1}{u}=1$

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

Dividing by $xy$ on both sides,

$\frac{6x}{xy}+\frac{3y}{xy}=\frac{6xy}{xy}$

$\frac{6}{y}+\frac{3}{x}=6$......(i)

$2 x+4 y=5 x y$

Dividing by $xy$ on both sides,

$\frac{2x}{xy}+\frac{4y}{xy}=\frac{5xy}{xy}$

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

Let $\frac{1}{y}=u$ and $\frac{1}{x}=v$

Substitute the above values in (i) and (ii),

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

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

By cross multiplication method, we get,

$\frac{u}{15-24}=\frac{v}{12-30}=\frac{-1}{24-6}$

$\frac{u}{-9}=\frac{v}{-18}=\frac{-1}{18}$

$u=\frac{9}{18}$

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

$v=\frac{18}{18}$

$v=1$

This implies,

$x=\frac{1}{v}=1$

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

(vii) Let $\frac{1}{x+y}=u$ and $\frac{1}{x-y}=v$

This implies, the given system of equations can be written as,

$\frac{10}{x+y} +\frac{2}{x-y}=4$

$10u+2v=4$

$10u+2v-4=0$---(i)

$\frac{15}{x+y}-\frac{5}{x-y}=-2$

$15u-5v=-2$

$15u=5v-2$

$u=\frac{5v-2}{15}$---(ii)

Substituting $u=\frac{5v-2}{15}$ in equation (i), we get,

$10(\frac{5v-2}{15})+2v-4=0$

$\frac{2(5v-2)}{3}=4-2v$

$10v-4=3(4-2v)$

$10v-4=12-6v$

$10v+6v=12+4$

$16v=16$

$v=\frac{16}{16}$

$v=1$

Using $v=1$ in equation (i), we get,

$10u+2(1)-4=0$

$10u+2-4=0$

$10u-2=0$

$10u=2$

$u=\frac{2}{10}$

$u=\frac{1}{5}$

Using the values of $u$ and $v$, we get,

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

$\Rightarrow x+y=5$....(iii)

$\frac{1}{x-y}=1$

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

Adding  equations (iii) and (iv), we get,

$x+y+x-y=5+1$

$\Rightarrow 2x=6$

$\Rightarrow x=\frac{6}{2}$

$\Rightarrow x=3$

Substituting the value of $x$ in (iii), we get,

$3+y=5$

$\Rightarrow y=5-3$

$\Rightarrow y=2$

Therefore, the solution of the given system of equations is $x=3$ and $y=2$.

(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 $\frac{1}{3x+y}=X$ and $\frac{1}{3x-y}=Y$

Substitute the above values in (i) and (ii),

$X+Y=\frac{3}{4}$

$4X+4Y=3$.......(iii)

$\frac{X}{2}-\frac{Y}{2}=\frac{-1}{8}$

$4X-4Y=-1$........(iv)

Adding (iii) and (iv), we get,

$8X=2$

$X=\frac{1}{4}$

This implies,

$4\frac{1}{4}-4Y=-1$

$1-4Y=-1$

$4Y=2$

$Y=\frac{1}{2}$

$X=\frac{1}{3x+y}=\frac{1}{4}$

$3x+y=4$.......(v)

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

$3x-y=2$.......(vi)

Adding (v) and (vi), we get,

$6x=6$

$x=1$

This implies,

$3(1)-y=2$

$y=3-2=1$.

Updated on 10-Oct-2022 13:19:57