I have two equations: 4 = 0.5x*(5-y)^2 and 1 = 0.5x*(3-y)^2 Can you solve this for x and y, and if yes, how?
4 = 0.5x*(5-y)^2 and 1 = 0.5x*(3-y)^2
Rearanging both equations, we have
x = 8/(5- y)^2 and x = 2/(3-y)^2
This implies that
8/(5- y)^2 = 2/(3-y)^2 divide both sides by 2
4/(5 - y)^2 = 1/(3-y)^2 cross-multiply and simplify
4(y^2 - 6y + 9) = y^2 - 10y + 25
4y^2 - 24y + 36 = y^2 - 10y + 25
3y^2 - 14y + 11 = 0 factor
(3y -11) (y - 1) = 0 set each factor to 0
y = 11/3 and y = 1
And when y = 11/3, x = 9/2
And when y = 1, x = 1/2
Here's the graph of both x = 8/(5- y)^2 and x = 2/(3-y)^2 and the points of intersection
https://www.desmos.com/calculator/ky62ryvrhs
P.S.- The graphs look as though they might intersect below the y axis at some point.....in fact..they don't !!!
You can start by doing this:
4 = 0.5x(5 - y)²
---> 8 = x(5 - y)² ---> 8/x = (5 - y)² ---> ± √(8)/√(x) = 5 - y
---> y = 5 ± √(8)/√(x) ---> y = 5 ± 2√(2)/√(x)
similarly:
1 = 0.5x(3 - y)²
---> 2 = x(3 - y)² ---> 2/x = (3 - y)² ---> ± √(2)/√(x) = 3 - y
---> y = 3 ± √(2)/√(x)
Choosing one from each group and setting them equal to each other:
5 - 2√(2)/√(x) = 3 - √(2)/√(x)
---> 2 = 2√(2)/√(x) - √(2)/√(x) ---> 2 = √(2)/√(x) ---> 2√(x) = √(2) ---> √(x) = √(2)/2
---> x = 2/4 ---> x = 1/2
When x = 1/2, y = 3 - √(2)/√(1/2) ---> y = 3 - √(4) ---> y = 3 - 2 = 1
So, one answer is (0.5, 1).
But, I choose only one possibility from each. Now, it's up to you to choose the other 3 possibilities (not all may work).
This is rather sketchy; if you want further help, please ask.
Thank you very much! I tried to isolate x, but that wasn't very helpful, ended up with a long division.
You're my hero for today! :)
4 = 0.5x*(5-y)^2 and 1 = 0.5x*(3-y)^2
Rearanging both equations, we have
x = 8/(5- y)^2 and x = 2/(3-y)^2
This implies that
8/(5- y)^2 = 2/(3-y)^2 divide both sides by 2
4/(5 - y)^2 = 1/(3-y)^2 cross-multiply and simplify
4(y^2 - 6y + 9) = y^2 - 10y + 25
4y^2 - 24y + 36 = y^2 - 10y + 25
3y^2 - 14y + 11 = 0 factor
(3y -11) (y - 1) = 0 set each factor to 0
y = 11/3 and y = 1
And when y = 11/3, x = 9/2
And when y = 1, x = 1/2
Here's the graph of both x = 8/(5- y)^2 and x = 2/(3-y)^2 and the points of intersection
https://www.desmos.com/calculator/ky62ryvrhs
P.S.- The graphs look as though they might intersect below the y axis at some point.....in fact..they don't !!!