+0

# System

0
208
1

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

Jun 12, 2022

#1
+2667
0

We have:

$$x+y = 10$$                   (1)

$$x^2 + y^2 =56 + 2xy$$   (2)

In the second equation, subtract $$2xy$$ from both sides to give: $$x^2 + y^2 - 2xy = 56$$

Now, recall the identity: $$x^2 + y^2 = (x+y)^2 - 2xy$$. Substituting this in, along with the first equation, gives us: $$100 - 4xy = 56$$. Thus. $$xy = 11$$

This gives us a new system, which we can solve for directly:

$$x+y=10$$                   (1)

$$xy = 11$$                        (2)

The first equation gives us $$x = 10 - y$$, which we can substitute in the second equation.

This yields $$-y^2 + 10y = 11$$. Subtracting 11 from both sides to make a quadratic gives us: $$-y^2 +10y -11 = 0$$

Now, plugging in the quadratic formula, we get: $$y = {-10 \pm \sqrt{10^2-4 \times -1 \times -11} \over -2} = {-10 \pm \sqrt{56} \over -2} = 5 \pm \sqrt {14}$$

This means our x-coordinates are $$10 - (5 \pm \sqrt {14}) = 5 \pm \sqrt {14}$$

So, our final coordinates are $$\color{brown}\boxed{(5 + \sqrt {14} , 5- \sqrt{14}), (5 + \sqrt {14} , 5- \sqrt{14})}$$

Jun 12, 2022