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Question

Solve the following equations :
$$\dfrac { x }{ a+\lambda  } +\dfrac { y }{ b+\lambda  } +\dfrac { z }{ c+\lambda  } =1,\dfrac { x }{ a+\mu  } +\dfrac { y }{ b+\mu  } +\dfrac { z }{ c+\mu  } =1,\\ \dfrac { x }{ a+\nu  } +\dfrac { y }{ b+\nu  } +\dfrac { z }{ c+\nu  } =1.$$


Solution

Consider the following equation in $$t$$,
$$\dfrac { x }{ a+t } +\dfrac { y }{ b+t } +\dfrac { z }{ c+t } =1-\dfrac { \left( t-\lambda  \right) \left( t-\mu  \right) \left( t-\gamma  \right)  }{ \left( t+a \right) \left( t+b \right) \left( tc \right)  }$$
$$x,y,z$$ being for the present regarded as know quantities.
This equation when cleared of fractions is of second degree in $$t$$, and is satisfied for three values   $$t=\lambda, t=\mu$$ and $$t=v$$ by virtue of the given equations; hence it must be an identity.To find the value of $$x,$$ we multiply $$(1)$$ by $$a+t$$, and then put $$t=-a$$. Thus
$$x=\dfrac { \left(a+\lambda  \right) \left( a+\mu  \right) \left( a+v \right)  }{ \left( a-b \right) \left( a-c \right)  }$$ 
or $$x=\dfrac { \left( -a-\lambda  \right) \left( -a-\mu  \right) \left( -a-v \right)  }{ \left( -a+b \right) \left( -a+c \right)  }$$. 
By reason of symmetry, we have
$$y=\dfrac { \left(b+\lambda  \right) \left( b+\mu  \right) \left( b+v \right)  }{ \left( b-c \right) \left( b-a \right)  }$$
and $$z=\dfrac { \left(c+\lambda  \right) \left( c+\mu  \right) \left( c+v \right)  }{ \left( c-a \right) \left(c-b\right)  }$$
We know that if $$\alpha ,\beta$$ are the roots of an eqution, then this equation is
$${ x }^{ 2 }-(\alpha +\beta )x+\alpha \beta =0$$ 
or $${ x }^{ 2 }-{ S }_{ 1 }x+{ S }_{ 2 }=0$$ 
where $${S}_{1}$$ is sum of roots taken one at a time and $${S}_{2}$$ is sum of products of roots taken two at a time.Similarly if there is an equation whose roots are $$\alpha \beta \gamma$$ then $${ S }_{ 1 }=\alpha +\beta +\gamma ,{ S }_{ 2 }=\alpha \beta +\beta \gamma +\gamma \alpha ,{ S }_{ 3 }=\alpha \beta \gamma$$ and the corresponding equation will be $${ t }^{ 3 }-{ S }_{ 1 }{ t }^{ 2 }+{ S }_{ 2 }t-{ S }_{ 3 }=0.$$
Again we know that if $$a$$ is a root of the equation $$f(x)=0$$ then $$a$$ will satisfy the equation, $$i.e.f(a)=0.$$
Alternative Solution of Q.22 (i).
Consider the equation $$z+ty+{t}^{2}x+{t}^{3}=0$$
It is clear that the above equation is satisfied by $$t=a,b,c$$ by virtue of the above three given equations.which is true by $$1st$$ equation and so on. Thus $$a,b,c$$ are the roots of the equation.
$$\because z+ay+{a}^{2}x+{a}^{3}=0.$$
which is ture by $$1st$$ equation and so on. Thus $$a,b,c$$ are the roots of the equation.
$${t}^{3}+{t}^{2}x+ty+z=0$$
$$\therefore { S }_{ 1 }=a+b+c=-x,$$
$${ S }_{ 2 }=ab+bc+ca=y,{S}^{3}=abc=-z$$
$$\therefore x=-(a+b+c),y=\sum { ab },z=-abc.$$

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