# Thread: When is it permissible to take the reciprocal of both sides of the equation

1. ## When is it permissible to take the reciprocal of both sides of the equation

Hello,

I'm studying for the GRE and am a bit confused about when it is permissible to take the reciprocal of both sides of the equations in order to solve for a variable and when it's not. For example, the study book I'm using said that if you're trying to find "1/a= 1/b + 1/c," you can't just take the reciprocal of both sides, i.e., it's not just a= b +c, but first must add "1/b" and "1/c" to get (b+c/bc)= 1/a, and then you can take the reciprocal to get a=bc/b+c. OR for (1/a-b)=5, just take the reciprocal to find a-b=1/5, so a=b+1/5. So my question is, what is the general rule about when it is ok to take the reciprocal of both sides of the equation? what are the conditions? How come in the above example you couldn't take the reciprocal with the first problem until the two fractions were added together? Thank you for your time.

2. ## Re: When is it permissible to take the reciprocal of both sides of the equation

You can always take the reciprocal of both sides, as long as it is done correctly.

Note that the reciprocal of $\displaystyle \frac{1}{b} + \frac{1}{c}$ is not $\displaystyle b+c$.

3. ## Re: When is it permissible to take the reciprocal of both sides of the equation

Two conventional rules when manipulating an equation.
1) Multiplying both sides by a form of 0 is bad. (You lose information.)
2) Dividing both sides by a form of 0 is bad.

$\displaystyle \frac{1}{a} = \frac{1}{b} + \frac{1}{c}$
Note that: $\displaystyle a,b,c \neq 0$ as a domain. This is important later.

$\displaystyle \frac{1}{a} = \frac{1}{b} + \frac{1}{c} \implies a = b+c$
This is simply incorrect algebra. Adding the fractions to the right to get $\displaystyle \frac{b+c}{bc}$ is an additional step to solve for a.

With
$\displaystyle \frac{1}{a} = \frac{b+c}{bc}$

What does it mean to 'take the reciprocal'? The quick way is to simply 'flip' both fractions (a learned heuristic). However, this ignores implicit restrictions on the domain. Operationally, taking the reciprocal on both sides (in the case above) is equivalent to multiplying both sides by
$\displaystyle \frac{a(bc)}{b+c}$

This operation is allowed because since $\displaystyle a,b,c \neq 0$ by restrictions of domain, $\displaystyle abc \neq 0$ (so you're not multiplying by a form of 0) and $\displaystyle b+c \neq 0$ (so you're not dividing by a form of 0)

Now would most questions like this require this amount of thinking? Most likely not. Flipping as a heuristic is usually pretty safe.

4. ## Re: When is it permissible to take the reciprocal of both sides of the equation

Originally Posted by rgragsda
For example, the study book I'm using said that if you're trying to find "1/a= 1/b + 1/c," you can't just take the reciprocal of both sides, i.e., it's not just a= b +c, but first must add "1/b" and "1/c" to get (b+c/bc)= 1/a, and then you can take the reciprocal to get a=bc/b+c. OR for (1/a-b)=5, just take the reciprocal to find a-b=1/5, so a=b+1/5. So my question is, what is the general rule about when it is ok to take the reciprocal of both sides of the equation? what are the conditions?
It is not a good idea to ask professional mathematicians for general rules, because such rules just don't exist.
That said, here is a natural idea that you yourself suggest. If we have an equality between two singular fractions (one unified fraction on each side) then the reciprocal rule can apply.

5. ## Re: When is it permissible to take the reciprocal of both sides of the equation

Taking reciprocals of both sides just means 1/(entire LH expression) = 1/(entire RH expression) and then you try to simply if it permits...

Also, if you are ever in doubt, just plug in any numbers into the equation and compare your answers of the original and the "transformed" equation!