Find the error:
x and y are positive numbers.
x=3*y
(y-x)2=(y-3y)=(-2y)2=4y2
(y+x)2=(y+3y)2=(4y)2=16y2
So, (y+x)2=4(y-x)2
or y+x=2(y-x)
→ 3*x=y
But x=3*y, So, 9*y=y,→ 9=1!!!!
Where is the error?
The answer is that for all values of x. Notice that
in conjunction with and leads to
.
You can see that this is the only solution by what you derived:
have only one simultaneous solution: y = x = 0. So your solution is not incorrect, you simply didn't take it far enough. It is interesting to note that if you take any line y = mx you will get a second equation by your method: y = nx where m is not equal to m.
If it helps you to see what's going on here, let
which can be written as the system of equations
and
which only has one intersection point: (x, y, z) = (0, 0, 0).
-Dan
In step 4 you have
Then, in Step 5 you took the square root of both sides to get
You should've gotten a positive root and a negative root:
After that, you only followed the positive square root solution (Step 6) to arrive at an erroneous result. We call that an "extraneous root".
If you followed the negative root, you would've received which agrees with your premise: .