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February 7th, 2018, 07:50 PM   #1
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Question Simultaneous equations

i have look into the solving of the equation just want to confirm if my calculation is right.

Q: X + Y + Z = 1
X^2 + Y^2 + Z^2 = 35
X^3 + Y^3 + Z^3 = 97
FIND THE VALUE OF X, Y AND Z

Last edited by greg1313; February 8th, 2018 at 02:47 PM.
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February 7th, 2018, 07:56 PM   #2
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just looking at the second equation I see that

$35 = 25 + 9 + 1$

making that work for the first equation we'd get

$1 = 5 + (-3) + (-1)$

now let's check via the third equation

$(5)^3 + (-3)^3 + (-1)^3 = 125 - 27 - 1 = 97$

hurrah

$x = 5, ~y=-3,~z=-1$

or since the system is completely symmetric any permutation of those equalities.
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February 7th, 2018, 11:40 PM   #3
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Quote:
Originally Posted by romsek View Post
just looking at the second equation I see that

$35 = 25 + 9 + 1$

making that work for the first equation we'd get

$1 = 5 + (-3) + (-1)$

now let's check via the third equation

$(5)^3 + (-3)^3 + (-1)^3 = 125 - 27 - 1 = 97$

hurrah

$x = 5, ~y=-3,~z=-1$

or since the system is completely symmetric any permutation of those equalities.
How are you sure you found all the solutions?
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February 8th, 2018, 12:53 PM   #4
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Is there an underlying condition that only integer solutions be considered? If so, you are done.
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February 8th, 2018, 01:04 PM   #5
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this is the basic algebra forum people.

They aren't looking for real solutions of a non-linear system of equations.
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February 8th, 2018, 06:28 PM   #6
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Quote:
Originally Posted by romsek View Post
just looking at the second equation I see that

$35 = 25 + 9 + 1$

making that work for the first equation we'd get

$1 = 5 + (-3) + (-1)$

now let's check via the third equation

$(5)^3 + (-3)^3 + (-1)^3 = 125 - 27 - 1 = 97$

hurrah

$x = 5, ~y=-3,~z=-1$

or since the system is completely symmetric any permutation of those equalities.
well i try solving using substitution metrhod starting with equation 1 i.e
x = 1 -y - z------- eqn.1
substitute eqn. 1 for eqn. 2
x^2 + Y^2 + z^2 = 35
(1 - y - z)^2 + Y^2 + z^2 = 35
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February 8th, 2018, 10:54 PM   #7
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(1 - Y - Z)² + Y² + z² = 35 implies (Y + Z - 2)² + 2(Y + Z - 2) - 16 = (Y - 1)(Z - 1).

From X³ + Y³ + Z³ = 97, one gets (1 - Y - Z)³ + Y³ + Z³ = 97,
so 3Y² + 3Z² - 3Y²Z - 3YZ² + 6YZ - 3Y - 3Z = 96,
which implies (Y - 1)(Z - 1)(Y + Z) = -32.
Hence ((Y + Z - 2)² + 2(Y + Z - 2) - 16)(Y + Z) = -32,
which implies Y³ + Z³ + 3Y²Z + 3YZ² - 2Y² - 2Z² - 4YZ - 16Y - 16Z + 32 = 0,
which factorizes as (Y + Z + 4)(Y + Z - 4)(Y + Z - 2) = 0.
Hence Y + Z = ±4 or 2.

If Y + Z = 2, X = -1 and (Y - 1)(Z - 1) = -16.
From there, it's easy to show that (Y, Z) = (-3, 5) or (5, -3).

I'll leave you to finish from there.
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