+118659
FIND X,Y IF sqrt(x)+y=7 sqrt(y)+x=11
FIRST: sqrtx and sqrt y are both real numbers so x and y are both equal or greater than 0
\(\)
\(\sqrt{x}+y=7\)
Lets think about this.
x muxt be a perfect square because otherwise the sum will be irrational.
If x=0 y=7
As x gets bigger y gets smaller,
When x=49 y=0
So this is an ever decreasing curve (The gradient is always negative)
Now look at
\(\sqrt{y}+x=11\)
When y=121 x=0
as y decreases, x will increase until when x=11 y = 0
Again this is a monotonically decreasing curve.
This means that the 2 curves will only cross at one point.
With a little thought I can see that
sqrt(9)+4=7
sqrt(4)+9 =11
So the point of intersection is (9,4)
Here is a graph to confirm and help explain what I have told you :)
+129842 √x + y = 7 → √x = 7 - y square both sides x = y^2 - 14y + 49 (1)
√y + x = 11 (2)
Substitute (1) into (2)
√y + y^2 - 14y + 49 = 11
√y = 11 - (7 - y)^2 square both sides
y = 121 - 22(7 - y)^2 + (7 - y)^4 simplify
y = 121 - 22 [ y^2 - 14y + 49] + y^4-28 y^3+294 y^2-1372 y+2401
y = 121 - 22y^2 + 308y - 1078 + y^4 - 28y^3 + 294y^2 - 1372y + 2401
y^4- 28y^3+272 y^2-1065 y + 1444 = 0
The only integer solution to this is y = 4
And using (2) we have that √4 + x = 11 → 2 + x = 11 → x = 9
So.....the solution is (x,y) = (9,4)
Here's a graph of the intersection point of both equations :
https://www.desmos.com/calculator/ciw2xypsx4
