x-y-y-x-13-6-x-y-5-find-the-solution-

Question Number 104062 by bramlex last updated on 19/Jul/20

{\begin{cases} \frac{x}{y} + \frac{y}{x} = \frac{13}{6} \\ x + y = 5 \end{cases}

f i n d t h e s o l u t i o n

Commented by Rasheed.Sindhi last updated on 19/Jul/20

A n O t h e r W a y_{}^{}

\frac{x^{2} + y^{2}}{x y} = \frac{13}{6}

6 x^{2} - 13 x y + 6 y^{2} = 0

(3 x - 2 y) (2 x - 3 y) = 0

({\begin{cases} 3 x - 2 y = 0 \\ x + y = 5 \end{cases}) \lor ({\begin{cases} 2 x - 3 y = 0 \\ x + y = 5 \end{cases})

3 x - 2 (5 - x) = 0 \lor 2 x - 3 (5 - x) = 0

x = 2, y = 3 \lor x = 3, y = 2

Commented by bemath last updated on 19/Jul/20

c o o l l

Commented by bramlex last updated on 19/Jul/20

t h a n k y o u a l l

Commented by Dwaipayan Shikari last updated on 19/Jul/20

Another way

\frac{x}{y} + 1 = \frac{5}{y} \Rightarrow \frac{x}{y} = \frac{5}{y} - 1 and \frac{y}{x} = {(\frac{5}{y} - 1)}^{- 1}

\frac{x}{y} + \frac{y}{x} = \frac{13}{6}

\frac{5 - y}{y} + \frac{y}{5 - y} = \frac{13}{6} \Rightarrow \frac{25 + {2 y}^{2} - 10 y}{5 y - y^{2}} = \frac{13}{6}

\frac{25 + {2 y}^{2} - 10 y}{10 y - {2 y}^{2}} + 1 = \frac{13}{12} + 1 \Rightarrow \frac{25}{10 y - {2 y}^{2}} = \frac{25}{12} \Rightarrow {2 y}^{2} - 10 y + 12 = 0 \to (a)

So y = 3, 2 (from (a))

\frac{x}{3} + 1 = \frac{5}{3} or \frac{x}{2} + 1 = \frac{5}{2}

x = 2 or 3

{\begin{cases} x = 2, 3 \\ y = 3, 2 \end{cases}

Answered by Dwaipayan Shikari last updated on 19/Jul/20

\frac{x^{2} + y^{2}}{2 x y} = \frac{13}{12}

\frac{x^{2} + y^{2} + 2 x y}{x^{2} + y^{2} - 2 x y} = \frac{25}{1}

{(\frac{x + y}{x - y})}^{2} = \frac{25}{1}

\frac{25}{{(x - y)}^{2}} = \frac{25}{1}

x - y = 1 o r y - x = 1

x + y = 5

{\begin{cases} x = 3, 2 \\ y = 2, 3 \end{cases}

Commented by Rasheed.Sindhi last updated on 19/Jul/20

E v e r y b o d y i s n o t a s i n t e l l i g e n t

a s y o u a r e! :)

S o p l i n s e r t s o m e s t e p s b e t w e e n

1 s t & 2 n d l i n e s .

Commented by Dwaipayan Shikari last updated on 19/Jul/20

S o r r y s i r f o r y o u r i n c o n v e n i e n c e

Commented by Rasheed.Sindhi last updated on 19/Jul/20

\frac{x^{2} + y^{2} + 2 x y}{x^{2} + y^{2} - 2 x y} = \frac{13 + 12}{13 - 12}

T h i s w a s n e c e s s a r y s t e p t o

i n s e r t . W i t h o u t t h i s t h e a n s w e r

s e e m e d a r i d d l e!

P l {d o n}^{'} t m i n d!

Commented by Rasheed.Sindhi last updated on 19/Jul/20

N i c e S o l u t i o n!

Commented by 1549442205PVT last updated on 19/Jul/20

applying the property : \frac{a}{m} = \frac{b}{n} = \frac{a + b}{m + n} = \frac{a - b}{m - n}

\frac{x^{2} + y^{2}}{13} = \frac{2 xy}{12} = \frac{x^{2} + y^{2} + 2 xy}{25} = \frac{x^{2} + y^{2} - 2 xy}{1}

\Rightarrow \frac{x^{2} + 2 xy + y^{2}}{x^{2} - 2 xy + y^{2}} = \frac{25}{1}

Answered by bobhans last updated on 19/Jul/20

s e t x + y = u, x y = v

\frac{x^{2} + y^{2}}{x y} = \frac{13}{6} \Rightarrow \frac{{(x + y)}^{2} - 2 x y}{x y} = \frac{13}{6}

25 - 2 v = \frac{13 v}{6} \Rightarrow 150 = 25 v

{\begin{cases} v = 6 \to y = \frac{6}{x} \\ u = 5 \to x + \frac{6}{x} = 5; x^{2} - 5 x + 6 = 0 \end{cases}

{\begin{cases} x = 3 \to y = 2 \\ x = 2 \to y = 3 \end{cases} s o l u t i o n s e t {(3, 2); (2, 3)}

Answered by Rasheed.Sindhi last updated on 19/Jul/20

Using formula :

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

{\begin{cases} \frac{x}{y} + \frac{y}{x} = \frac{13}{6} \dots . . (i) \\ x + y = 5 \dots \dots \dots \dots . . (i i) \end{cases}

(i) \Rightarrow \frac{x}{y} + 1 + \frac{y}{x} + 1 = \frac{13}{6} + 2

\frac{x + y}{y} + \frac{x + y}{x} = \frac{25}{6}

\frac{5}{y} + \frac{5}{x} = \frac{25}{6}

\frac{1}{y} + \frac{1}{x} = \frac{5}{6} \dots \dots \dots \dots \dots . (i i i)

(i i) \Rightarrow \frac{x}{x y} + \frac{y}{x y} = \frac{5}{x y}

\frac{1}{y} + \frac{1}{x} = 5 (\frac{1}{x}) (\frac{1}{y}) \dots \dots . (i v)

(i i i) & (i v) :

(\frac{1}{x}) (\frac{1}{y}) = \frac{1}{6} \Rightarrow x y = 6

x + y = 5 \land x y = 6

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

{(5)}^{2} - {(x - y)}^{2} = 4 (6)

x - y = \pm 1

{\begin{cases} x + y = 5 \\ x - y = 1 \end{cases} \lor {\begin{cases} x + y = 5 \\ x - y = - 1 \end{cases}

x = 3, y = 2 \lor x = 2, y = 3