Question

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Answer 1

Answer

The required solutions are (2 , 8) and (8 ,2)

Given

The equations are :

• x + y + √xy = 6
• x² + y² + xy = 84

To Find

• The value of x and y

Solution

We are given ,

$\sf x + y - \sqrt{xy} = 6 \longrightarrow (1)$

And

$\sf x^{2}+y^{2} + xy = 84 \\\\ \sf\implies x^{2} + y^{2} + 2xy - xy = 84 \\\\ \sf\implies (x+y)^{2} - (\sqrt{xy})^{2}= 84 \\\\ \sf\implies (x+y-\sqrt{xy})(x +y + \sqrt{xy}) = 84 \\\\ \sf\implies 6(x + y + \sqrt{xy}) = 84 \\\\ \sf\implies x + y + \sqrt{xy} =14 \longrightarrow (2)$

Adding (1) and (2) we have :

$\sf\implies x + y + \sqrt{xy} + x + y - \sqrt{xy} = 14 + 6 \\\\ \sf\implies 2x + 2y = 20 \\\\ \sf\implies x + y = 10 \longrightarrow(3)$

Putting the value from (3) in (2) we have

$\sf\implies 10 + \sqrt{xy}= 14 \\\\ \sf\implies \sqrt{xy} = 4 \\\\ \sf\implies xy = 16 \\\\ \sf\implies x = \dfrac{16}{y} \longrightarrow(4)$

Using the value of x from (4) in (3) we have :

$\sf\implies \dfrac{16}{y}+ y = 10 \\\\ \sf\implies \dfrac{16+ y^{2}}{y} = 10 \\\\ \sf\implies y^{2} - 10y + 16 \\\\ \sf\implies y^{2} - 2y - 8y + 16 = 0 \\\\ \sf\implies y(y-2)-8(y-2)=0 \\\\ \sf\implies (y-8)(y-2)= 0$

Therefore ,

$\sf \implies y - 8 = 0 \: \: or \: \implies y - 2 = 0 \\\\ \sf\implies y = 8 \: \: or \: \implies y=2$

Now from (4) putting the value pf y

$\sf\implies x = \dfrac{16}{8} \: \: or \: \implies x = \dfrac{16}{2} \\\\ \sf\implies x = 2 \: \: or \: \implies x = 8$

Thus when x = 2 , y = 8

and when x = 8 , y = 2

Answer 2

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