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- Feb 14, 2012

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Let $x, y, z$ be real numbers such that $9x-10y+z=8$ and $x+8y-9z=10$.

Evaluate $x^2-2y^2+z^2$.

Evaluate $x^2-2y^2+z^2$.

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Let $x, y, z$ be real numbers such that $9x-10y+z=8$ and $x+8y-9z=10$.

Evaluate $x^2-2y^2+z^2$.

Evaluate $x^2-2y^2+z^2$.

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- Mar 31, 2013

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My answer agrees with that of

(1) \(\displaystyle 9x-10y+z=8\)

(2) \(\displaystyle x+8y-9z=10\)

To eliminate $z$, multiply (1) by 9 and add to (2) to get:

\(\displaystyle 82x-82y=82\)

Which implies:

\(\displaystyle x=y+1\)

Substituting for $x$ in (1), we get:

\(\displaystyle 9(y+1)-10y+z=8\) or \(\displaystyle z-y=-1\) so \(\displaystyle y-z=1\)

putting this in (2), we find no contradiction.

Thus, we have:

(3) \(\displaystyle x-y=1\)

(4) \(\displaystyle y-z=1\)

Adding (3) and (4) we get:

(5) \(\displaystyle x-z=2\)

Now, we may write:

\(\displaystyle x^2-2y^2+z^2=\left(x^2-y^2 \right)+\left(z^2-y^2 \right)=(x+y)(x-y)+(z-y)(z+ y)\)

Using (3) and (4), this becomes:

\(\displaystyle x^2-2y^2+z^2=(x+y)-(z+y)=x-z\)

Using (5), we finally conclude:

\(\displaystyle x^2-2y^2+z^2=2\)

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- Feb 14, 2012

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Thanks for participating,My solution:

By inspection, we see $(x,y,z)=(2,1,0)$ is a solution to the two given equations. Hence:

\(\displaystyle x^2-2y^2+z^2=2\)

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Heymy ans agrees with markFL and solution is below

given

9x−10y+z=8 ...(1)

x+8y−9z=10.... (2)

to eliminate z multiply (1) by 9 and add to (2)

82x - 82 y = 82

or x = y + 1

put it in (1) to get

9(y+1) - 10 y + z = 8 or z -y = - 1 so y -z = 1

putting in (2) it satisfies so no contradicyion

so x - y = 1 ..(3)

and y- z = 1 ...(4)

adding we get x- z = 2

now x^2 - 2y^2 + z^2

= (x^2- y^2) + (z^2 -y^2)

= (x+y)(x-y) + (z-y)(z+ y)

= (x+y) - (z + y) as x - y = 1 and z - y = -1

= x - z

= 2

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- Feb 7, 2012

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[Not a solution, just a comment on the previous solutions.]

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Hi[Not a solution, just a comment on the previous solutions.]