√x+y=11
x+√y=7
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since √y = 7-x,
√x + (7-x)^2 = 11
√x = -x^2+14x-38
x = x^4-28x^3+272x^2-1064x+1444
x^4-28x^3+272x^2-1065x+1444 = 0
(x-4)(x^3-24x^2+176x-361) = 0
good luck with the cubic. But, if x=4, y=9
To find the values for x and y, we can solve the system of equations:
Equation 1: √x + y = 11
Equation 2: x + √y = 7
To begin, let's isolate the square roots in each equation.
In Equation 1, isolate the square root by subtracting y from both sides:
√x = 11 - y
In Equation 2, isolate the square root by subtracting x from both sides:
√y = 7 - x
Now, square both sides of Equation 1 to eliminate the square root:
(√x)^2 = (11 - y)^2
x = (11 - y)^2
Similarly, square both sides of Equation 2 to eliminate the square root:
(√y)^2 = (7 - x)^2
y = (7 - x)^2
Now substitute the expressions for x and y into the other equation:
x + √y = 7
(11 - y)^2 + (√y) = 7
Simplify the equation and solve for y:
(11 - y)^2 + √y = 7
(121 - 22y + y^2) + √y = 7
y^2 - 22y + 121 + √y = 7
y^2 - 22y + (√y - 114) = 0
We now have a quadratic equation in terms of y. To solve for y, you can use factoring, completing the square, or the quadratic formula.
Once you find the value(s) of y, substitute it back into one of the original equations to find the corresponding value(s) of x.
Keep in mind that this quadratic equation may have two real solutions, one real solution, or no real solutions, depending on the values of x and y.