# What other ways can you split an equation into two?

1. Oct 16, 2013

### ainster31

Here are two ways:

$$(x-{ x }_{ 1 })({ x }-{ x }_{ 2 })=0\\ x-{ x }_{ 1 }=0\quad \quad \quad \quad x-{ x }_{ 2 }=0\\ \\$$$$\\ { e }^{ x }({ c }_{ 1 }-3{ c }_{ 2 })+{ e }^{ -32x }({ c }_{ 5 }-{ c }_{ 4 })=0\\ { c }_{ 1 }-3{ c }_{ 2 }=0\quad \quad \quad \quad \quad { c }_{ 5 }-{ c }_{ 4 }=0$$

Any other ways?

2. Oct 16, 2013

### MrAnchovy

"Splitting an equation in two" is not in general a meaningful mathematical operation. Your second example is incorrect: can you see why? (clue: a = b = 0 is not the only solution to a + b = 0).

3. Oct 16, 2013

### ainster31

Why not?

Hmmm... you're right. It's weird that my textbook teaches this technique.

What if it's an identity?

Last edited: Oct 16, 2013
4. Oct 16, 2013

### gopher_p

The first situation relies on the fact that if the product of two real numbers is zero, then one of those numbers must be zero.

The second situation is a statement about the linear independence of $e^x$ and $e^{-32x}$; if $ae^x+be^{-32x}=0$ for all $x$ (i.e. $ae^x+be^{-32x}$ is the zero function), then $a=b=0$. It's similar to the statement that if $a_0+a_1 x+...+a_n x^n=0$ for all $x$, then $a_0=a_1=...=a_n=0$.

5. Oct 16, 2013

### MrAnchovy

Because it does not in general yield anything useful.

Perhaps the textbook is looking for solutions which are valid for all values of $x$?

I don't understand what you mean.

6. Oct 16, 2013

### Integral

Staff Emeritus
Really this is not splitting an equation into two.

It is a way to find zeros of the original equation. Always keep in mind your goals, what are you trying to do. You are not splitting and equation you are looking for zeros.