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Solve system of equation $x + y + xy = 11$ and $x^{2} y + x y^{2} = 30$ and find the solution set.

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{1, 4}; {5, 1}; {4, 3} and {3, 1}.

{2, 5}; {5, 0}; {2, 3} and {2, 2}.

{1,4}; {5,2}; {6,3} and {3, 3}.

{1, 5}; {5, 1}; {2,3} and {3,2}.

answer is D.

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Detailed Solution

Given the equation

x + xy + y = 11

……….(1)

x^{2} y + x y^{2} = 30

……….(2)
From the equation (2) taking common of

xy

xy (x + y) = 30

………….(3)
Equation (1) can be written as

xy = 11 - (x + y)

…….(4)
Put value of

xy

in equation (3)

11 - (x + y) (x + y) = 30

…….(5)
Now, suppose that

x + y = z

(11 - z) z = 30

[From equation (5)]

11 z - z^{2} = 30

\Rightarrow Z^{2} - 11 z + 30 = 0

We obtain, a quadratic equation

z^{2} - 11 z + 30 = 0

Comparing the general quadratic equation

a x^{2} + bx + c = 0

We got value of coefficients

a = 1, b = - 11, c = 30

This quadratic being it has 2 roots.

\Rightarrow x = \frac{(- b + \sqrt{b^{2} - 4 ac})}{2 a} \frac{(- b - \sqrt{b^{2} - 4 ac})}{2 a}

Substitute the value of coefficients, a, b, c in equation (1) then

\Rightarrow Z = \frac{(- (- 11) + \sqrt{(- 11)^{2} - 4 \times (1) \times (30)})}{2 \times 1} \frac{(- (- 11) - \sqrt{(- 11)^{2} - 4 \times (1) \times (30)})}{2 \times 1}

\Rightarrow Z = \frac{(11 + \sqrt{1})}{2} \frac{(11 - \sqrt{1})}{2}

\Rightarrow Z = \frac{(11 + 1)}{2} \frac{(11 - 1)}{2}

\Rightarrow Z = 6 5

Here

D = \sqrt{(- 11)^{2} - 4 \times 1 \times (30)} = \sqrt{1} ⩾ 0

Two distinct real roots are

z_{1} = 6 z_{2} = 5

Therefore,

x + y

= to 5 or 6

11 - (x + y) = xy

Case 1st, when

x + y = 5

Then

xy = 6

[By hit and trial method]
= {2, 3} and {3, 2}
Case 2nd when

x + y = 6

Then

xy = 5

= {5, 1} and {1, 5}.
Therefore we obtain
{1,5}; {5,1}; {2,3} and {3,2}.
Correct option is 4.

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