# Solving 4 equations in 4 variables - how?

Discussion in 'Physics & Math' started by Confused2, Sep 17, 2016.

1. ### Confused2Registered Senior Member

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de-TEXted but otherwise hopefully straight:-
Just interested (flummoxed) - how do you get 4 nice looking equations AND a (fairly) nice solution? I can't see how to start to do that.

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3. ### mathmanValued Senior Member

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Possible approach: The first equation can be substituted into the other three to make them linear equations in x, y, and z. Solving this system will give you x, y, z as functions of t. These can be substituted into the first equation to give you an equation in t, which may or may not be easy to solve.

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5. ### Confused2Registered Senior Member

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From the original context we're looking at an expanding 'sphere' of light in different frames. Having chosen the point of origin the 4 equations should (?) be easy (?) - apart from the trick of choosing the origin to get nice integers in all four equations.
Edit... knowing the point of origin in the past should give the point of coincidence in the future. Er... maybe.

Last edited: Sep 17, 2016
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7. ### Confused2Registered Senior Member

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Edit... or not different frames... does it make any difference?

8. ### Confused2Registered Senior Member

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Last edited: Sep 18, 2016
9. ### Confused2Registered Senior Member

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This looks more context sensitive than I thought - here is the original post
A four dimensional universe
I'll try to read it a few times before proceeding further.

10. ### rpennerFully WiredRegistered Senior Member

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Easiest substitution for t, assuming t > 5 (the future)
x^2+y^2+z^2−(t−5)^2=0 t = 5 + √(x^2+y^2+z^2)

This makes each quadratic equation reduce to a linear equation in x, y, z, and t
(x−4)^2+(y−3)^2+(z−8)^2−(t−7)^2=0
(x−4)^2+(y−3)^2+(z−8)^2−(–2 + √(x^2+y^2+z^2))^2 = 0
x^2 – 8 x +16 + y^2 – 6 y + 9 + z^2 – 16 z + 64 – 4 + 4 √(x^2+y^2+z^2) – x^2 – y^2 – z^2 = 0
–8 x – 6 y – 16 z + 4 t = –65

This makes each quadratic equation reduce to a linear equation in x, y, z, and t
(x+4)^2+(y−7)^2+(z+8)^2−(t−3)^2=0
(x+4)^2+(y−7)^2+(z+8)^2−(2 + √(x^2+y^2+z^2))^2 = 0
x^2 + 8 x + 16 + y^2 – 14 y + 49 + z^2 + 16 z + 64 – 4 – 4 √(x^2+y^2+z^2) – x^2 – y^2 – z^2 = 0
8 x – 14 y + 16 z – 4 t = -145

Combining these two, we have an explicit formula for y.
–8 x – 6 y – 16 z + 4 t + 8 x – 14 y + 16 z – 4 t = –65 + -145
y = 210/20 = 189/18 = 21/2

Substitute in y = 21/2
t = 5 + √(x^2+y^2+z^2)
t = 5 + √(x^2+z^2+441/4)

Substitute in y = 21/2
–8 x – 6 y – 16 z + 4 t = –65
–8 x – 16 z + 4 t = –2

Substitute in y = 21/2
8 x – 14 y + 16 z – 4 t = -145
8 x – 14 y + 16 z – 4 t = 2 (no surprise here)

Substitute in y = 21/2 AND t = 5 + √(x^2+z^2+441/4)
(x−7)^2+(y−12)^2+(z+4)^2−(t−4)^2=0
(x−7)^2+9/4+(z+4)^2−(1 + √(x^2+z^2+441/4))^2 = 0
x^2 –14 x + 49 + 9/4 + z^2 + 8 z + 16 – 1 – 2 √(x^2+z^2+441/4) + 10 – 10 – x^2 – z^2 – 441/4 = 0
– 14 x + 8 z – 2 t = 34
– 28 x + 16 z – 4 t = 68

Combining two, we have an explicit formula for x.
–8 x – 16 z + 4 t – 28 x + 16 z – 4 t = –2 + 68
x = –66/36 = –33/18 = –11/6

So now our four equations are:
x^2+y^2+z^2−(t−5)^2=0 t = 5 + √(z^2 + 2045/18), x = –11/6, y = 21/2
And any one of:
(x−4)^2+(y−3)^2+(z−8)^2−(t−7)^2=0 – 16 z + 4 t = –50/3
(x+4)^2+(y−7)^2+(z+8)^2−(t−3)^2=0 16 z – 4 t = 50/3
(x−7)^2+(y−12)^2+(z+4)^2−(t−4)^2=0 16 z – 4 t = 50/3

By substituting in our expression for t into one of these, squaring both sides, we get a quadratic equation in z.
16 z – 4 t = 50/3
4 z – t = 25/6
4 z – 25/6 = 5 + √(z^2 + 2045/18)
4 z – 55/6 = √(z^2 + 2045/18)
16 z^2 – (440/6) z + 3025/36 = z^2 + 4090/36
15 z^2 – (440/6) z – 1065/36 = 0
z = (44 ± 5 √103)/18

By substituting in our (indefinite) expression for z into one of the linear equations, we get an indefinite expression for t, only one value of which satisfies t>5
4 z – t = 25/6
t = 4 z – 25/6
t = (101 ± 20 sqrt(103))/18
t = (101 + 20 sqrt(103))/18
z = (44 + 5 √103)/18

Done.

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11. ### Confused2Registered Senior Member

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501
Just amazing. Thank you. More questions may follow when I can do it without having to peek at rpenner's exemplary solution.