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Find all pairs (x,y) of real numbers such that x + y = 10 and x^2 + y^2 = 56 + xy.

 Feb 4, 2021
 #1
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+1

This is a system of equations. Let's try and solve it.

 

While it is possible to solve for x or y and then do direct substitution into the remaining equation, I noticed a possible shortcut that is worth knowing for future algebraic problems like this one. This may not save too much time for this particular problem, but I figured it would be to your benefit if I showed this strange substitution to you.

 

\(\fbox{1}\; x + y = 10\\ (x+y)^2 = 10^2\\ x^2 + 2xy + y^2 = 100\\ {\color{red}x^2 + y^2} = 100 - 2xy \)

 

I squared equation 1 and rearranged a few terms. To the untrained eye, this may seem to complicate matters unnecessarily, but I am setting up a creative substitution.

 

\(\fbox{2}\; {\color{red} x^2 + y^2} = 56 + xy\\ 100 - 2xy = 56 + xy\\ 3xy = 44\\ y = \frac{44}{3x}\)

 

Let's use this substitution in equation 1 and solve it.

 

\(\fbox{1}\; x+y = 10\\ x+\frac{44}{3x}=10\\ 3x^2+44=30x\\ 3x^2-30x+44=0\)

 

This quadratic is neither easy nor factorable. I will result to the quadratic formula for this particular question.

 

\(x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a};a=3, b=-30, c=44\\ x_{1,2}=\frac{30\pm\sqrt{900-528}}{6}\\ x_{1,2}=\frac{30\pm\sqrt{372}}{6}\\ x_{1,2}=\frac{30\pm2\sqrt{93}}{6}\\ x_1=\frac{15-\sqrt{93}}{3}\text{ and }x_2=\frac{15+\sqrt{93}}{3}\)

 

Let's find the corresponding y-values by substituting these values for x into \(y=\frac{44}{3x}\).

 

\(\)\(y_1=\frac{44}{3x_1}\\ y_1=\frac{44}{3*\frac{15-\sqrt{93}}{3}}\\ y_1=\frac{44}{15-\sqrt{93}}\) \(y_2=\frac{44}{3*\frac{15+\sqrt{93}}{3}}\\ y_2=\frac{44}{15+\sqrt{93}}\)

 

There are two coordinates of intersections. One is at \(\left(\frac{15-\sqrt{93}}{3},\frac{44}{15-\sqrt{93}} \right)\), and the other is at \(\left(\frac{15+\sqrt{93}}{3}, \frac{44}{15+\sqrt{93}} \right)\)

 Feb 4, 2021
 #2
avatar+118470 
+1

x +  y  =10    ⇒   y  =10 - x

 

So

 

x^2 + y^2  =   56 +  xy

 

x^2  +  (10 - x)^2  =  56  + x ( 10- x)

 

x^2  + x^2    -20x  + 100  =  56  + 10x  - x^2

 

3x^2  - 30x  + 44  =  0

 

Q Formula

 

 

x  =   30  ±  sqrt  [ 30^2  -  4 * 3 * 44 ]                 30 ± sqrt (372)

        ___________________________  =          ___________     =

                     2 *  3                                                      6

 

30 ±  2sqrt (93)

____________

         6

 

So       x  =   one  of these  solutions  and  y = the other

 

cool cool cool

 Feb 4, 2021

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