solve 3 equation?

You have a system of two equations and two unknowns:

x² + y² = 225 and x² + z² = 169 and y + z = 14

We can solve the last equation for z in terms of y then substitute into the second equation:

y + z = 14

z = 14 - y

x² + z² = 169

x² + (14 - y)² = 169

x² + 196 - 28y + y² = 169

x² - 28y + y² = -27

Now if we solve the first equation for x² in terms of y we can substitute that into the above equation, then solve for y:

x² + y² = 225

x² = 225 - y²

x² - 28y + y² = -27

225 - y² - 28y + y² = -27

the y² terms cancel out so you can then solve for one solution for y:

225 - 28y = -27

-28y = -252

y = 9

Now that we have y we can solve for x and z:

x² + y² = 225 and y + z = 14

x² + 9² = 225 and 9 + z = 14

x² + 81 = 225 and z = 5

x² = 144 and z = 5

x = ±12 and z = 5

There are two solutions to this system:

(-12, 9, 5) and (12, 9, 5)

x^2 + y^2 = 225 

x^2 + z^2 = 169

y + z = 14

y^2 - z^2 = 56

y - z = 4

y = 9

z = 5

x = 12

Subtract the second equation from the first giving:

y^2 - z^2 = 56

Now take the third equation, solve for z = 14 - y and substitute into the result above giving:

y^2 - (14 - y)^2 = 56

y^2 - (196 - 28y + y^2) = 56

28y - 196 = 56

28 y = 252

y = 9.

Now, z = 14 - y = 14 - 9 = 5.

Finally, x^2 + y^2 = 225

x^2 + 9^2 = 225

x^2 + 81 = 225

x^2 = 144

x = 12.  Notice that x = -12 is also a solution.