((x^3 +y^3 )/(x+y))=7 ((x^3 −y^3 )/(x−y))=19. find x and y


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Question Number 171728 by Mikenice last updated on 20/Jun/22

$\frac{x^{3} + y^{3}}{x + y} = 7$ $\frac{x^{3} - y^{3}}{x - y} = 19 . f i n d x a n d y$
	Answered by Ar Brandon last updated on 20/Jun/22
	$((x^3 +y^3 )/(x+y))=(((x+y)(x^2 −xy+y^2 ))/(x+y))=x^2 −xy+y^2 =7 ...eqn(i) ((x^3 −y^3 )/(x−y))=(((x−y)(x^2 +xy+y^2 ))/(x−y))=x^2 +xy+y^2 =19 ...eqn(ii) eqn(ii)− eqn(i) (x^2 +xy+y^2 )−(x^2 −xy+y^2 )=19−7 ⇒2xy = 12 ⇒xy=6 (At this point we can deduce (3, 2), (2, 3) are the solutions. But we can also get the result by solving.) ⇒y=(6/x) ...eqn(iii) Replacing eqn(iii) in eqn(i):- x^2 −x((6/x))+((6/x))^2 =7 ⇒x^2 −6+((36)/x^2 )=7 ⇒x^4 −13x^2 +36=0, (x^2 −9)(x^2 −4)=0 ⇒(x−3)(x+3)(x−2)(x+2)=0 ⇒x=3; x=−3; x=2; x=−2. Replacing the values of x in eqn(iii) to get corresponding values of y ⇒y=2; y=−2; y=3; y=−3. S={(x, y)∈R∣(3, 2); (−3, −2); (2, 3); (−2, −3)}$
	$\frac{x^{3} + y^{3}}{x + y} = \frac{(x + y) (x^{2} - x y + y^{2})}{x + y} = x^{2} - x y + y^{2} = 7 . . . eqn (i)$ $\frac{x^{3} - y^{3}}{x - y} = \frac{(x - y) (x^{2} + x y + y^{2})}{x - y} = x^{2} + x y + y^{2} = 19 . . . eqn (i i)$ $eqn (i i) - eqn (i)$ $(x^{2} + x y + y^{2}) - (x^{2} - x y + y^{2}) = 19 - 7$ $\Rightarrow 2 x y = 12 \Rightarrow x y = 6 (At this point we can deduce (3, 2),$ $(2, 3) are the solutions . But we can also get the result by solving .)$ $\Rightarrow y = \frac{6}{x} . . . eqn (i i i)$ $Replacing eqn (i i i) in eqn (i) : -$ $x^{2} - x (\frac{6}{x}) + {(\frac{6}{x})}^{2} = 7 \Rightarrow x^{2} - 6 + \frac{36}{x^{2}} = 7$ $\Rightarrow x^{4} - 13 x^{2} + 36 = 0, (x^{2} - 9) (x^{2} - 4) = 0$ $\Rightarrow (x - 3) (x + 3) (x - 2) (x + 2) = 0$ $\Rightarrow x = 3; x = - 3; x = 2; x = - 2 .$ $Replacing the values of x in eqn (i i i) to get corresponding$ $values of y$ $\Rightarrow y = 2; y = - 2; y = 3; y = - 3 .$ $S = {(x, y) \in R ∣ (3, 2); (- 3, - 2); (2, 3); (- 2, - 3)}$
	Commented by Mikenice last updated on 20/Jun/22

	$t h a n k s s i r$
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