If x^2-yz=10, y^2+xz=10, z^2 +xy=10, and z not=y, what is the value of x^2+y^2+z^2
10
15
20
25
30
C
is there an easy way to do this problem? shortcut?
3 unknowns
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x^2 - yz = 10 (eqn a)
y^2 + xz = 10 (eqn b)
z^2 + xy = 10 (eqn c)
x^2 + y^2 + z^2 = ?
(eqn a) - (eqn b):
x^2 - yz - y^2 - xz = 0
x^2 - y^2 = xz + yz
(x + y)(x - y) = z(x + y)
x - y = z (eqn 1)
Add equations a, b and c:
x^2 + y^2 + z^2 - yz + xz + xy = 30
x^2 + y^2 + z^2 - yz + xz + (10 - z^2) = 30 (xy substituted with equation c)
x^2 + y^2 + z^2 - yz + xz - z^2 = 20
x^2 + y^2 - yz + xz = 20
x^2 + y^2 + z(x - y) = 20
x^2 + y^2 + z(z) = 20 (insert equation 1)
x^2 + y^2 + z^2 = 20
Choose C
-BM-
y^2 + xz = 10 (eqn b)
z^2 + xy = 10 (eqn c)
x^2 + y^2 + z^2 = ?
(eqn a) - (eqn b):
x^2 - yz - y^2 - xz = 0
x^2 - y^2 = xz + yz
(x + y)(x - y) = z(x + y)
x - y = z (eqn 1)
Add equations a, b and c:
x^2 + y^2 + z^2 - yz + xz + xy = 30
x^2 + y^2 + z^2 - yz + xz + (10 - z^2) = 30 (xy substituted with equation c)
x^2 + y^2 + z^2 - yz + xz - z^2 = 20
x^2 + y^2 - yz + xz = 20
x^2 + y^2 + z(x - y) = 20
x^2 + y^2 + z(z) = 20 (insert equation 1)
x^2 + y^2 + z^2 = 20
Choose C
-BM-
- sanju09
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How do we know that (x + y) ≠ 0, BM?
The mind is everything. What you think you become. -Lord Buddha
Sanjeev K Saxena
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Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
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You are right, i didnt consider (x+y) = 0 in my solution.sanju09 wrote:How do we know that (x + y) ≠ 0, BM?
However, with (x+y) = 0, can equations a, b and c all be satisfied? I cannot find a solution for x, y and z under this condition.
How would you solve this problem, sanju?
-BM-