# Math Help - where is the error in algebra?

1. ## where is the error in algebra?

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?

2. Originally Posted by blertta
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 $y \neq 3x$ for all values of x. Notice that
$(y + x)^2 = 4(y - x)^2$
in conjunction with $(y + x)^2 = 4y^2$ and $(y - x)^2 = 16y^2$ leads to
$4y^2 = 64y^2 \implies y = 0$.

You can see that this is the only solution by what you derived:
$y = 3x$
$3y = x$
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 $z = (y + x)^2 = 4(x - y)^2$
which can be written as the system of equations
$z = (y + x)^2$
and
$z = 4(y - x)^2$
which only has one intersection point: (x, y, z) = (0, 0, 0).

-Dan

3. Originally Posted by blertta
Find the error:
x and y are positive numbers.
(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?
You forgot the absolute value: $\left( {y + x} \right)^2 = 4\left( {y - x} \right)^2 \, \Rightarrow \,\left| {y + x} \right| = 2\left| {y - x} \right|$

4. Originally Posted by blertta
Find the error:
x and y are positive numbers.
1) x=3*y
2) (y-x)2=(y-3y)=(-2y)2=4y2
3) (y+x)2=(y+3y)2=(4y)2=16y2
4) So, (y+x)2=4(y-x)2
5) or y+x=2(y-x)
6) → 3*x=y
But x=3*y, So, 9*y=y,→ 9=1!!!!
Where is the error?
In step 4 you have $(y+x)^2=4(y-x)^2$

Then, in Step 5 you took the square root of both sides to get $y+x=2(y-x)$

You should've gotten a positive root and a negative root: $y+x=\pm 2(y-x)$

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 $y+x=-2(y-x)\Longrightarrow y+x=-2y+2x \Longrightarrow \boxed {3y=x}$ which agrees with your premise: $\boxed{x=3y}$.