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

## Main Question or Discussion Point

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?

## Answers and Replies

pbuk
Gold Member
"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).

"Splitting an equation in two" is not in general a meaningful mathematical operation.
Why not?

(clue: a = b = 0 is not the only solution to a + b = 0).
Hmmm... you're right. It's weird that my textbook teaches this technique.

What if it's an identity?

Last edited:
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?
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 $a_0=a_1=...=a_n=0$.

pbuk
Gold Member
Why not?
Because it does not in general yield anything useful.

Hmmm... you're right. It's weird that my textbook teaches this technique.
Perhaps the textbook is looking for solutions which are valid for all values of $x$?

What if it's an identity?
I don't understand what you mean.

Integral
Staff Emeritus