MHB No problem, glad I could contribute! Have a great day :)

[anemone]

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]

Re: Evaluate x²-2y²+z²

My solution:

Spoiler

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

$$x^2-2y^2+z^2=2$$

 

[kaliprasad]

Re: Evaluate x²-2y²+z²

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

Spoiler

We are given:

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

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

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

$$82x-82y=82$$

Which implies:

$$x=y+1$$

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

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

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

Thus, we have:

(3) $$x-y=1$$

(4) $$y-z=1$$

Adding (3) and (4) we get:

(5) $$x-z=2$$

Now, we may write:

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

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

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

Using (5), we finally conclude:

$$x^2-2y^2+z^2=2$$

 

[anemone]

Re: Evaluate x²-2y²+z²

My solution:

Spoiler

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

$$x^2-2y^2+z^2=2$$

Thanks for participating, MarkFL! I noticed sometimes solving a challenge problem by inspection can save us a lot of time and hassle! Bravo!

- - - Updated - - -

my ans agrees with markFL and solution is below

Spoiler

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 = 2now 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]

Re: Evaluate x²-2y²+z²

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

View attachment 1689​

[/sp]

 

[anemone]

Re: Evaluate x²-2y²+z²

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

View attachment 1689
​

[/sp]

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