how do you solve these equations?

Question

z=(x+y)/2 z=(x^2+y)/5 z=(x^3+y)/14

Solve for x y and z

Answer 1

eliminate the denominators from the 3 given equations:

Equation 1: 2z = x+y
Equation 2: 5z = x^2 + y
Equation 3: 14z = x^3 + y

Subtract eq.1 from eq.2, the result is:
equation 4: 3z = x^2 - x

Subtract eq. 2 from eq.3, the result is:
equation 5: 9z = x^3 - x^2 = x * ( x^2 - x )

Divide eq. 5 by eq. 4, the result is:
equation 6: 3 = x

Substitute 3 for x in eq.4, the result is;
3z = 9 - 3 = 6
z = 2

Substitute 3 for x, and 2 for z, in eq. 1, the result is:
4 = 3 + y
y = 1

Therefore, the answers are:
x = 3
y = 1
z = 2

Answer 2

(x+y)/2 = z = (x² + y)/5
5x + 5y = 2x² + 2y
3y = 2x² - 5x
y = 2x²/3 - 5x/3 .......... and
(x+y)/2 = z = (x³ + y)/14
7x + 7y = x³ + y
6y = x³ - 7x
y = x³/6 - 7x/6
so
x³/6 - 7x/6 = y = 2x²/3 - 5x/3
x³ - 7x = 4x² - 10x
x³ - 4x² + 3x = 0
x(x² - 4x + 3) = 0
x(x-3)(x-1) = 0

x = 0, y = 0, z = 0 .... solution 1

x = 3,
y = [2(3)² - 5(3)]/3 = 1
z = (3+1)/2 = 2 ..........solution 2

x = 1
y = [2(1)² - 5(1)]/3 = -3
z = (1-3)/2 = -1 ..........solution 3

graphing the 2 y equations gives you a parabola and a cubic with 3 intersection points.

Answer 3

since z is equal to the three equations, set two of them equal and solve for x and y. Then that should be your (x, y). then check your answers by plugging in those two values, x and y, into all of the equations and the z values should be the same.

Answer 4

use matrices.