Question

Solve the ques 17 in the attachment

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Answer 1

I skipped small calculations : hope you can figure it out . 

Steps:
1) We have two linear equations in two variables ,here α & β:
  $\alpha x^{3} - \beta y-(x-y)=0 \\ \alpha x+ \beta y^{2} -xy^{2}=0$ 

where 
[tex] a_{1} = x^{3} \:,b_{1} = -y ,c_{1} = -(x-y) \\ a_{2} = x \:,b_{2} = y^{2} ,c_{2} = -xy^{2} [/tex] 

Above Symbols have usual meanings :)

2) By Cross Multiplication Method,
 
$\frac{ \alpha }{b_{1}c_{2}-b_{2}c_{1}} = \frac{ \beta }{c_{1}a_{2}-c_{2}a_{1}} = \frac{1}{a_{1}b_{2}-a_{2}b_{1}}$ 

On substituting values , we get 
[tex] \frac{ \alpha }{xy^{3}+xy^{2}-y^{3}} = \frac{ \beta }{-x(x-y)+x^{4}y^{2}} = \frac{1}{x^{3}y^{2}+ xy} \\ \\ =\ \textgreater \ \alpha = \frac{xy^{3}+xy^{2}-y^{3}}{x^{3}y^{2}+xy}= \frac{xy^{2}+xy-y^{2}}{x^{3}y+x} \\ \\ =\ \textgreater \ \beta = \frac{-x(x-y)+x^{4}y^{2}}{x^{3}y^{2}+xy} = \frac{-x+y+x^{3}y^{2}}{x^{2}y^{2}+y} [/tex]
 
Hence we obtained values of $\alpha$ & β  in terms of x, & y . 
 
$\alpha = \frac{xy^{2}+xy-y^{2}}{x^{3}y+x} \\ \\ \beta =\frac{-x+y+x^{3}y^{2}}{x^{2}y^{2}+y}$