Find all pairs $(x,y)$ of real numbers such that $x + y = 10$ and $x^2 + y^2 = 56$. For example, to enter the solutions $(2,4)$ and $(-3,9)$, you would enter "(2,4),(-3,9)" (without the quotation marks).
Rearrange the first equation to get y = 10 - x.
Plug this into the second equation: x^2 +(x - 10)^2 = 56
Expand this and rearrange to get a quadratic in x. Solve for the two values of x. then substitute these back into y = 10 - x to find the corresponding two values of y.
I am a bit confused. I did what you told me, and got x^2+x^2-20x+100 = 2x^2-20x+100=56. I divided my 2 to get x^2-10x+50=28, and subtracted by 25 on both sides to get x^2-10x-25=3. I factored the left side to give (x-5)(x-5) = 3. The two values of x I got from here were 8 and 6. So, the sets of x and y should be (8,2), and (6,4), but these do not satisfy x^2+y^2 = 56.......
Please help me! Thanks!
x +y =10,
x^2 +y^2 =56, solve for x, y
Use substitution to get:
x = 5 - sqrt(3) AND y = 5 + sqrt(3), OR:
x = 5 + sqrt(3) AND y = 5 - sqrt(3)