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Evaluate x² - 2y² + z²

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,756
Let $x, y, z$ be real numbers such that $9x-10y+z=8$ and $x+8y-9z=10$.

Evaluate $x^2-2y^2+z^2$.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: Evaluate x²-2y²+z²

My solution:

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

\(\displaystyle x^2-2y^2+z^2=2\)
 

kaliprasad

Well-known member
Mar 31, 2013
1,322
Re: Evaluate x²-2y²+z²

My answer agrees with that of MarkFL and my solution is below:

We are given:

(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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anemone

MHB POTW Director
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Feb 14, 2012
3,756
Re: Evaluate x²-2y²+z²

My solution:

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

\(\displaystyle x^2-2y^2+z^2=2\)
Thanks for participating, MarkFL! I noticed sometimes solving a challenge problem by inspection can save us a lot of time and hassle! Bravo!:cool:

- - - Updated - - -

my 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
Hey kaliprasad, your method works well too! Well done!
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Re: Evaluate x²-2y²+z²

[Not a solution, just a comment on the previous solutions.]
Geometrically, the equations $9x-10y+z=8$ and $x+8y-9z=10$ represent planes. Their intersection is the line with parametric equation $(x,y,z) = (0,1,2) + t(1,1,1).$ The whole of this line lies on the conic $x^2-2y^2+z^2 = 2.$ The conic is a ruled surface (in fact, a circular hyperboloid of one sheet), which can be entirely generated by a family of straight lines, as in the figure.

220px-Ruled_hyperboloid.jpg
 
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anemone

MHB POTW Director
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Feb 14, 2012
3,756
Re: Evaluate x²-2y²+z²

[Not a solution, just a comment on the previous solutions.]
Geometrically, the equations $9x-10y+z=8$ and $x+8y-9z=10$ represent planes. Their intersection is the line with parametric equation $(x,y,z) = (0,1,2) + t(1,1,1).$ The whole of this line lies on the conic $x^2-2y^2+z^2 = 2.$ The conic is a ruled surface (in fact, a circular hyperboloid of one sheet), which can be entirely generated by a family of straight lines, as in the figure.

Hi Opalg, thank you so much for the insight and I find it interesting!:)