if x*x +y =21 and y*y+x =29 then find the value of x and y.?

Question

the answer is x=4 and y=5 but i want complete logical solution.

Answer 1

x^2 + y = 21
y^2 + x = 29

Rewrite the first equation as an expression for y.

y = -x^2 + 21

Substitute in the expression for y in the second equation, expand, and combine like terms.

(-x^2 + 21)^2 + x = 29
x^4 - 42x^2 + 441 + x = 29
x^4 - 42x^2 + x + 412 = 0

This has zeroes at four values: -5.18136, -3.90686, 4, and 5.0882, which I found using a graphic calculator. Taking the one integer zero, I used polynomial long division to divde (x - 4) from the fourth-order polynomial, giving a quotient of x^3 + 4x^2 - 26x - 103, which unfortunately isn't exactly easy to solve for exact values instead of decimals.

Anyway, we need to apply the formula for y to each of the four values of x.

y = -(-5.18136)^2 + 21 = -5.84649
y = -(-3.90686)^2 + 21 = 5.73644
y = -(4)^2 + 21 = 5
y = -(5.0882)^2 + 21 = -4.88978

The ordered pairs are therefore

(-5.18, -5.85)
(-3.91, 5.74)
(4, 5)
(5.09, -4.89)

If you check them, you will find that all four work and are therefore solutions.

Answer 2

x² + y = 21
y² + x = 29

Since this is involving squares, it also involves square roots. This can be done mentally You'll get:

x² + y = 21 = 16 + 5 = ?
y² + x = 29 = 25 + 4 = ?

Or solve it as a system:

(y + x² = 21)-² = -y² + -x^4 =
y² + x = 29

Answer 3

y = 21 - x^2

(21 - x^2)^2 + x = 29
441 - 42x^2 + x^4 + x = 29
x^4 - 42x^2 + x + 412 = 0
(x^4 - 16x^2) + (-26x^2 + x + 412) = 0
x^2(x^2 - 16) + (26x + 103)(-x + 4) = 0
x^2(x - 4)(x + 4) + (26x + 103)(4 - x) = 0
(x - 4)[x^2(x + 4) - (26x + 103)] = 0
(x - 4)[x^3 + 4x^2 - 26x + 103] = 0

x - 4 = 0
x = 4

x^3 + 4x^2 - 26x + 103 = 0
x = -8.490820277
This is an extraneous solution.

So x = 4
4^2 + y = 21
16 + y = 21
y = 5

Answer 4

x^2 + y = 21
y = 21 - x^2 --- (1)
y^2 + x = 29 --- (2)

Sub (1) into (2)
(21 - x^2)^2 + x = 29
(21)^2 - 2(21)(x^2) + (x^2)^2 + x = 29
x^4 - 42x^2 + x + 412 = 0
x = 4.000, x = 5.088, x = -3.907, x = -5.181

when x = 4, y = 21 - (4)^2
y = 5

when x = 5.09, y = 21 - (5.088)^2
y = 4.89

when x = -3.907, y = 21 - (-3.907)^2
y = 5.74

when x = -5.181, y = 21 - (-5.181)^2
y = -5.84

Therefore, there are four solution sets to the equations,
they are (4, 5), (-3.91, 5.74), (5.09, -4.89) and (-5.18, -5.85).