Find all pairs (x,y) of real numbers such that x + y = 10 and x^2 + y^2 = 56 + xy.
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 \(\left(\frac{15+\sqrt{93}}{3}, \frac{44}{15+\sqrt{93}} \right)\)
+130099 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)
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6
So x = one of these solutions and y = the other
