WHAT d'hell are we "seeing" in your attachment?
A partial solution?
YOU started at top, and simplified (or complicated!)
step by step to the bottom line, and you're stuck there
unable to continue?
And what does solve by "pencil" mean?
Let k = |x-2| + |x-4] - 2 ; then the whole thing STARTS with:
[(y - x/2 - 1)^2 + k]^2 * [(y - 3)^2 +k]
Is that correct?
If so, why don't you simplify that as much as you can,
then substitute back in as the grand finale?
Anyway, like Romsek, I think you're on drugs...
if not, your teacher certainly is
$y = 5\ and\ x = 30$ is one solution. There are an infinite number of other answers in integers of course.
The solution given above is easy enough to prove using paper and pencil although you can use a calculator or computer to check your work if you have one handy.
$y = 5\ and\ x = 30$ is one solution. There are an infinite number of other answers in integers of course.
The solution given above is easy enough to prove using paper and pencil although you can use a calculator or computer to check your work if you have one handy.
Yes we are certainly on drugs. Unfortunately substitution isn't working because through the 23 lines pod the formula there aren't consistent values in each line. Meaning I'd have to substitute several different letters.
This problem is weird.
First, you do not have a formula, you have an equation of the form $f(x,\ y) = 0.$
Second, the function is in three dimensions and cannot be graphed.
Third, my answer given above is wrong. Sorry for that. I misread the problem.
Fourth, I do not have time to spend solving this, but this looks like a promising approach.
$(y^2 - 6y + 8)^2 = (y^2 - 6y + 8)(y^2 - 6y + 8) =$
$y^4 - 6y^3 + 8y^2 - 6y^3 + 36y^2 - 48y + 8y^2 - 48y + 64 = y^4 - 12y^3 + 52y^2 - 96y + 64.$
So $(y^2 - 6y + 8 + \sqrt{y^4 - 12y^3 + 52y^2 - 96y + 64})^2 = \{2(y^2 - 6y + 8)\}^2 = 4(y^2 - 6y + 8)^2.$
And so we have that the long product $= -\ 4\{(y - 4)(y - 2)\}^2 \le 0\ if\ y \in \mathbb R.$
If y is real, then $4((y^2 - 6y + 8)^2$ is non-negative. But that means that if y is real, the long, complex product must be non-positive.
But that product appears to contain all squares so if x and y are both real, y = 2 or y = 4.
I think that is all OK so far, but I am VERY pressed for time. Please check my logic.
For 3D plotting wouldn't I need 3 variables? Because there are only 2 in the original equation.
Your logic flows smooth. My first step I took everything to the right of the "+ sign", or the 23rd line of the equation, and subtracted to move it to the other side of the "= sign". So now I think the equation would look like
..... = - (y^2-6y+8+sqrt(y^4-12y^3+52y^2-96y+64))^2
Since every line of equation has a ^2 at the end of each open and closed parenthesis then can I slap a square root over the whole thing? So line 1 would simplify from...
sqrt{((y-1/2x-1)^2+abs(2x-6)-2)^2((y-3)^2+abs(2x-6)-2)^2}
simplified to...
((y-1/2x-1)^2+abs(2x-6)-2)((y-3)^2+abs(2x-6)-2)
as you may see I also simplified line 1 where abs(x-2)+abs(x-4) to = abs(2x-6)
then the right side of the equation I think would look like...
= sqrt(-2(y^2-6y+8))^2
so would that now simplify to...
= 2(y^2-6y+8)i