1-x-1-y-3-4-x-2-y-y-2-x-9-find-the-value-of-x-and-y-

Question Number 58195 by salaw2000 last updated on 19/Apr/19

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

\frac{x^{2}}{y} + \frac{y^{2}}{x} = 9

find the value of x and y

Answered by Kunal12588 last updated on 19/Apr/19

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

\Rightarrow \frac{x + y}{x y} = \frac{3}{4} (1)

\Rightarrow x + y = \frac{3 x y}{4} (2)

\frac{x^{2}}{y} + \frac{y^{2}}{x} = 9

\Rightarrow \frac{x + y}{x y} (x^{2} - x y + y^{2}) = 9

f r o m (1)

\Rightarrow x^{2} - x y + y^{2} = 12

\Rightarrow {(x + y)}^{2} - 3 x y = 12 (3)

\Rightarrow {(x - y)}^{2} + x y = 12 (4)

f r o m (3)

x + y = \pm \sqrt{3 x y + 12}

b u t f r o m (2)

\frac{3 x y}{4} = \sqrt{3 x y + 12}

\Rightarrow 3 \cdot 3 x^{2} y^{2} = 16 [3 (x y + 4)]

\Rightarrow 3 x^{2} y^{2} - 16 x y - 64 = 0

\Rightarrow x y = \frac{16 \pm \sqrt{16^{2} - 4 \cdot 3 \cdot (- 64)}}{2 \cdot 3}

\Rightarrow x y = \frac{48}{6}, \frac{- 16}{6}

\Rightarrow x y = 8, \frac{- 8}{3}

(i) x y = 8

p u t t i n g x y i n (2)

x + y = 6 (5)

p u t t i n g x y i n (4)

{(x - y)}^{2} = 12 - 8

\Rightarrow x - y = \pm \sqrt{4}

\Rightarrow x - y = 2 o r x - y = - 2

(a) t a k i n g x - y = 2

a n d c o m p a r i n g w i t h (5)

1^{s t} a n s w e r x = 4, y = 2

(b) t a k i n g x - y = - 2

a n d c o m p a r i n g w i t h (5)

2^{n d} a n s w e r x = 2, y = 4

(i i) x y = \frac{- 8}{3}

p u t t i n g x y i n (2)

x + y = - 2 (6)

p u t t i n g x y i n (4)

{(x - y)}^{2} + x y = 12

3 {(x - y)}^{2} - 8 = 36

x - y = \pm \sqrt{\frac{44}{3}}

(a) t a k i n g x - y = \sqrt{\frac{44}{3}}

a n d c o m p a r i n g w i t h (6)

3^{r d} a n s w e r x = - 1 + \sqrt{\frac{11}{3}}, y = - 1 - \sqrt{\frac{11}{3}}

(b) t a k i n g x - y = - \sqrt{\frac{44}{3}}

a n d c o m p a r i n g w i t h (6)

4^{t h} a n s w e r x = - 1 - \sqrt{\frac{11}{3}}, y = - 1 + \sqrt{\frac{11}{3}}

Commented by salaw2000 last updated on 19/Apr/19

thank you very much

Answered by MJS last updated on 19/Apr/19

(1) y = \frac{4 x}{3 x - 4}

(2) x^{3} + y^{3} - 9 x y = 0

(1) in (2) \Rightarrow

x^{4} - 4 x^{3} - \frac{20}{3} x^{2} + 32 x - \frac{64}{3} = 0

trying factors of \pm \frac{64}{3} we find

x_{1} = 2; x_{2} = 4

\Rightarrow

(x - 2) (x - 4) (x^{2} + 2 x - \frac{8}{3}) = 0

\Rightarrow x_{3} = - 1 - \frac{\sqrt{33}}{3}; x_{4} = - 1 + \frac{\sqrt{33}}{3}

\Rightarrow (\begin{matrix} x \\ y \end{matrix}) \in {(\begin{matrix} 2 \\ 4 \end{matrix}), (\begin{matrix} 4 \\ 2 \end{matrix}), (\begin{matrix} - 1 - \frac{\sqrt{33}}{3} \\ - 1 + \frac{\sqrt{33}}{3} \end{matrix}), (\begin{matrix} - 1 - \frac{\sqrt{33}}{3} \\ - 1 + \frac{\sqrt{33}}{3} \end{matrix})}

Commented by Kunal12588 last updated on 19/Apr/19

Sir you are making me think why I did that ��

Commented by MJS last updated on 19/Apr/19

different paths lead to the same solutions -

{that}^{'} s a good thing!

Commented by salaw2000 last updated on 19/Apr/19

a very big thanks to you