Solve-simultaneously-x-y-1-y-x-1-5-3-i-x-2-y-2-2-ii-

Question Number 9236 by tawakalitu last updated on 24/Nov/16

Solve simultaneously

\frac{x}{y + 1_{}} + \frac{y}{x + 1} = \frac{5}{3} \dots \dots \dots \dots . (i)

x^{2} + y^{2} = 2 \dots \dots \dots . . (ii)

Answered by RasheedSoomro last updated on 25/Nov/16

\frac{x}{y + 1_{}} + \frac{y}{x + 1} = \frac{5}{3} \dots \dots \dots \dots . (i)

x^{2} + y^{2} = 2 \dots \dots \dots . . (ii)

\frac{x^{2} + y^{2} + x + y}{(x + 1) (y + 1)} = \frac{5}{3}

\frac{2 + x + y}{(x + 1) (y + 1)} = \frac{5}{3}

6 + 3 x + 3 y = 5 xy + 5 x + 5 y + 5

5 xy + 2 x + 2 y - 1 = 0

x (5 y + 2) = - 2 y + 1

x = \frac{- 2 y + 1}{5 y + 2}

(ii) \Rightarrow {(\frac{- 2 y + 1}{5 y + 2})}^{2} + y^{2} = 2

{(- 2 y + 1)}^{2} + y^{2} (5 y + 2) = 2 (5 y + 2)

{4 y}^{2} - 4 y + 1 + {5 y}^{3} + {2 y}^{2} = 10 y + 4

{5 y}^{3} + {6 y}^{2} - 14 y - 3 = 0

Continue

Commented by tawakalitu last updated on 25/Nov/16

Thank you sir .

Answered by RasheedSoomro last updated on 26/Nov/16

\frac{x}{y + 1} + \frac{y}{x + 1} = \frac{5}{3} [x \neq - 1, y \neq - 1]

x^{2} + y^{2} = 2

- - - - - - - - -

Let x + 1 = X, y + 1 = Y

(i) \Rightarrow \frac{X - 1}{Y} + \frac{Y - 1}{X} = \frac{5}{3} [X \neq 0, Y \neq 0]

X^{2} - X + Y^{2} - Y = \frac{5}{3} XY

X^{2} + Y^{2} - (X + Y) = \frac{5}{3} XY

(ii) \Rightarrow {(X - 1)}^{2} + {(Y - 1)}^{2} = 2

X^{2} - 2 X + 1 + Y^{2} - 2 Y + 1 = 2

X^{2} + Y^{2} - 2 (X + Y) = 0

X + Y = \frac{X^{2} + Y^{2}}{2}

X^{2} + Y^{2} - (X + Y) = \frac{5}{3} XY

\Rightarrow X^{2} + Y^{2} - \frac{X^{2} + Y^{2}}{2} = \frac{5}{3} XY

{6 X}^{2} + {6 Y}^{2} - {3 X}^{2} - {3 Y}^{2} = 10 XY

{3 X}^{2} - 10 XY + {3 Y}^{2} = 0

{3 X}^{2} - XY - 9 XY + {3 Y}^{2} = 0

X (3 X - Y) - 3 Y (3 X - Y) = 0

(3 X - Y) (X - 3 Y) = 0

3 X = Y ∣ X = 3 Y

Case - 1 Y = 3 X

X + Y = \frac{X^{2} + Y^{2}}{2} \Rightarrow X + 3 X = \frac{X^{2} + {(3 X)}^{2}}{2}

8 X = {10 X}^{2}

4 X = {5 X}^{2}

4 X - {5 X}^{2} = 0 [Dividing by X, as X \neq 0]

4 - 5 X = 0

X = \frac{4}{5}

X + Y = \frac{X^{2} + Y^{2}}{2}

\Rightarrow \frac{4}{5} + Y = \frac{{(4 / 5)}^{2} + Y^{2}}{2}

\Rightarrow 8 + 10 Y = 5 (\frac{16}{25} + Y^{2}) = \frac{16}{5} + {5 Y}^{2}

\Rightarrow 40 + 50 Y = 16 + {25 Y}^{2}

\Rightarrow {25 Y}^{2} - 50 Y - 24 = 0

\Rightarrow Y = \frac{50 \pm \sqrt{50^{2} - 4 (25) (- 24)}}{2 (25)} = \frac{50 \pm 70}{50}

\Rightarrow Y = 2 \frac{2}{5} ∣ Y = - \frac{2}{5}

Solution set for (X, Y) : {(\frac{4}{5}, 2 \frac{2}{5}), (\frac{4}{5}, - \frac{2}{5})}

As x = X - 1, y = Y - 1, so

Solution set for (x, y) : {(- \frac{1}{5}, 1 \frac{2}{5}), (- \frac{1}{5}, - \frac{7}{5})}

\boldsymbol BUT (- \frac{1}{5}, - \frac{7}{5}) {doesn}^{'} t satisfy one of the given

equations . S {it}^{'} s an \boldsymbol extraneous \boldsymbol root . Hence the

solution set is {(- \frac{1}{5}, 1 \frac{2}{5})}

By Case - 2 it can be seen that (1 \frac{2}{5}, - \frac{1}{5}) is also

a solution .

Finally {(- \frac{1}{5}, 1 \frac{2}{5}), (1 \frac{2}{5}, - \frac{1}{5})} is solution set .

Commented by tawakalitu last updated on 26/Nov/16

I really appreciate sir . God bless you .

I really appreciate sir . God bless you .