Solve the following pairs of equations:
$ x+y=3.3 $
$ \frac{0.6}{3 x-2 y}=-1,3 x-2 y ≠ 0 $

To do: 

We have to solve the given pairs of equations.

Solution: 

(i) $x+y=3.3$

$\Rightarrow y=3.3-x$......(i)

$\frac{0.6}{3 x-2 y}=-1$

$0.6=-1(3x-2y)$

$0.6=-3x+2y$

$3x=2y-0.6$

$3x=2(3.3-x)-0.6$          [From (i)]

$3x=6.6-2x-0.6$

$3x+2x=6$

$x=\frac{6}{5}$

$x=1.2$

This implies,

$y=3.3-1.2$

$y=2.1$

Hence, the solution of the given pair of equations is $x=1.2$ and $y=2.1$.

(ii) $\frac{x}{3}+\frac{y}{4}=4$

$\Rightarrow \frac{4x+3y}{12}=4$

$4x+3y=12(4)$

$3y=48-4x$......(i)

$\frac{5x}{6}-\frac{y}{8}=4$

$\Rightarrow \frac{4(5x)-3(y)}{24}=4$

$20x-3y=24(4)$

$20x=96+3y$

$20x=96+48-4x$          [From (i)]

$20x+4x=144$

$24x=144$

$x=\frac{144}{24}$

$x=6$

This implies,

$y=\frac{48-4(6)}{3}$

$y=\frac{24}{3}$

$y=8$

Hence, the solution of the given pair of equations is $x=6$ and $y=8$.

(iii) $4 x+\frac{6}{y}=15$

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

This implies,

$4x+6u=15$......(i)

$6 x-\frac{8}{y}=14$

$6x-8u=14$.......(ii)

Multiplying (i) by 8 and (ii) by 6, we get,

$8(4x+6u)=8(15)$

$32x+48u=120$.......(iii)

$6(6x-8u)=6(14)$

$36x-48u=84$.........(iv)

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

$32x+36x+48u-48u=120+84$

$68x=204$

$x=\frac{204}{68}$

$x=3$

This implies,

$4(3)+6u=15$

$6u=15-12$

$u=\frac{3}{6}$

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

This implies,

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

$y=2$

Hence, the solution of the given pair of equations is $x=3$ and $y=2$. 

(iv) $\frac{1}{2 x}-\frac{1}{y}=-1$........(i)

$\frac{1}{x}+\frac{1}{2 y}=8$........(ii)

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

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

$\Rightarrow \frac{u-2 v}{2}=-1$

$\Rightarrow u-2 v=-2$.....(iii)

From equation (ii),

$u+\frac{1}{2} v=8$

$\Rightarrow \frac{2 u+v}{2}=8$

$2u+v=16$......(iv)

Multiplying (iv) by 2 and adding the result with (iii), we get,

$u-2v+2(2u+v)=-2+2(16)$

$u+4u-2v+2v=-2+32$

$5u=30$

$u=6$

This implies,

$6-2v=-2$

$2v=6+2$

$v=\frac{8}{2}$

$v=4$

Therefore,

$x=\frac{1}{u}=\frac{1}{6}$

$y=\frac{1}{v}=\frac{1}{4}$

(v) \( 43 x+67 y=-24 \)......(i)

\( 67 x+43 y=24 \).........(ii)

Multiplying (i) by 43 and (ii) by 67 and subtracting the results, we get,

$43(43x+67y)=43(-24)$

$43^2x+43(67)y=24(-43)$.........(iii)

$67(67x+43y)=67(24)$

$67^2x+43(67)y=24(67)$.......(iv)

Subtracting (iii) from (iv), we get,

$(67^2-43^2)x=24(67+43)$

$(67+43)(67-43)x=24(110)$

$110(24)x=24(110)$

$x=1$

This implies,

$43(1)+67y=-24$

$67y=-24-43$

$67y=-67$

$y=-1$

Therefore,

$x=1$

$y=-1$

(vi) $\frac{x}{a}+\frac{y}{b}=a+b$..........(i)

$\frac{x}{a^{2}}+\frac{y}{b^{2}}=2$.........(ii)

Multiplying (i) by $\frac{1}{a}$ and then subtracting from (ii), we get,

$[\frac{x}{a^{2}}+\frac{y}{b^{2}}]-[\frac{x}{a^{2}}+\frac{y}{a b}]=2-(1+\frac{b}{a})$

$y(\frac{1}{b^{2}}-\frac{1}{a b})=2-1-\frac{b}{a}$

$y(\frac{a-b}{a b^{2}})=1-\frac{b}{a}$

$=(\frac{a-b}{a})$

$y=\frac{a b^{2}}{a}$

$y=b^{2}$

This implies,

$\frac{x}{a^{2}}+\frac{b^{2}}{b^{2}}=2$

$\frac{x}{a^{2}}=2-1$

$\frac{x}{a^{2}}=1$

$x=a^{2}$

Hence, the required values of $x$ and $y$ are $a^{2}$ and $b^{2}$ respectively.

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

This implies,

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

$\frac{x}{x y}+\frac{y}{x y}=\frac{2\times2}{3}$

$\frac{1}{y}+\frac{1}{x}=\frac{4}{3}$.......(i)

$\frac{x y}{2 x-y}=\frac{-3}{10}$

$\frac{2 x-y}{x y}=\frac{-10}{3}$

$\frac{2 x}{x y}-\frac{y}{x y}=\frac{-10}{3}$

$\frac{2}{y}-\frac{1}{x}=\frac{-10}{3}$...........(ii)

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

This implies,

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

$2 v-u=\frac{-10}{3}$........(iv)

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

$3 v=\frac{4}{3}-\frac{10}{3}$

$3v=\frac{4-10}{3}$

$3v=\frac{-6}{3}$

$3 v=-2$

$v=\frac{-2}{3}$

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

$\frac{-2}{3}+u =\frac{4}{3}$

$u=\frac{4}{3}+\frac{2}{3}$

$u=\frac{6}{3}$

$u=2$

This implies,

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

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

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

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Updated on: 10-Oct-2022

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