Question

Solve the following equations $x + y = 7 + \sqrt {xy} $, ${x^2} + {y^2} = 133 - xy$
$
  A)x = 9,y = 4 \\
  B)x = 4,y = 8 \\
  C)X = 2,Y = 7 \\
  D)X = 5,Y = 12 \\
 $

Answer 1

Hint: Here to proceed with a solution we need to know the basic factorization of quadratic equations.

Complete step-by-step answer:

And we should also use a very basic algebra formula which will help us to solve the equation.
Let us consider

 $x + y = 7 + \sqrt {xy} - - - - - - - - > (1)$

${x^2} + {y^2} = 133 - xy$

Now let us rewrite above equation

${x^2} + {y^2} + xy = 133 - - - - - - - > (2)$

By using the formula ${(a + b)^2} = {a^2} + {b^2} + 2ab$ we can rewrite the equation (2) as
${(x + y)^2} - xy = 133 - - - - - - - > (3)$

From equation (1) we know that $x + y = 7 + \sqrt {xy} $

Now on substituting (1) in (3) we get

$

  {(7 + \sqrt {xy} )^2} - xy = 133 \\

   \Rightarrow 49 + xy + 14\sqrt {xy} - xy = 133 \\

   \Rightarrow 14\sqrt {xy} = 133 - 49 \\

   \Rightarrow 14\sqrt {xy} = 84 \\
   \Rightarrow \sqrt {xy} = \dfrac{{84}}{{14}} \\

   \Rightarrow \sqrt {xy} = 6 \\

   \Rightarrow xy = 36 \\

   \Rightarrow y = \dfrac{{36}}{x} \\

 $


Therefor here we got $y = \dfrac{{36}}{x}$


Let us expand equation (3) using the formula ${(a + b)^2} = {a^2} + {b^2} + 2ab$ and on further simplification we get it as

$

   \Rightarrow {x^2} + {y^2} + 2xy - xy = 133 \\

   \Rightarrow {x^2} + {y^2} + xy = 133 - - - - - - - (4) \\

 $


On substituting y value in equation (4) we get

$ \Rightarrow {x^2} + {\left( {\dfrac{{36}}{x}} \right)^2} + x\left( {\dfrac{{36}}{x}} \right) = 133 - - - - - - > (5)$


Now again substituting y value in equation (2) we get

$ \Rightarrow {x^2} + {y^2} + xy = 133$

$ \Rightarrow {x^2} + {\left( {\dfrac{{36}}{x}} \right)^2} + x\left( {\dfrac{{36}}{x}} \right) =

133$ (Since x get cancelled in third term of equation)

$

   \Rightarrow {x^2} + \dfrac{{1296}}{{{x^2}}} = 133 - 36 \\

   \Rightarrow {x^2} + \dfrac{{1296}}{{{x^2}}} = 97 \\

$

Let us simplify it further

$

   \Rightarrow {x^4} - 97{x^2} + 1296 = 0 \\

   \Rightarrow {x^4} - 16{x^2} - 81{x^2} + 1296 = 0 \\

   \Rightarrow {x^2}({x^2} - 16) - 81({x^2} - 16) = 0 \\

   \Rightarrow ({x^2} - 16)({x^2} - 18) = 0 \\

   \Rightarrow {x^2} = 16,81 \\

   \Rightarrow x = \pm 4, \pm 9 \\

 $

Here negative values of x are not true for equation (1)

So, x values are 4, 9

Now we know that

            $y = \dfrac{{36}}{x}$

If $x = 4 \Rightarrow y = \dfrac{{36}}{4} = 9$

If $x = 9 \Rightarrow y = \dfrac{{36}}{9} = 4$

Therefore we can say that for values x=4, 9 the corresponding y values are y=9, 4

Hence the values from the options are x=9 and y=4.

Option A is the correct answer.

NOTE: Here we have used basic algebra formula to convert the equation (2) into proper format. Later using equation (1) we have rewritten the equation (3) and we got y value on further simplification. Again on further substitution we get the x value.