Question

How do inverse operations help solve equations?

Answer 1

They are helpful for solving with respect to a variable: if after some step you find yourself in a situation such as `\cos(x)=0.7`, if you want to find the value of `x` for which this statement is true, you need to apply the inverse cosine function at both sides and obtain `x=\cos^{-1}(0.7)`

Answer 2

They allow us to "undo" what has been done to the variable.

Example 1
Solve: `x+3=8`

`3` has been added to the variable, `x`. The inverse of addition is subtraction, so, by subtracting `3`, I can undo the addition.

After `3` was added, the result was equal to `8`. We undo the addition, by subtracting `3` and see that, the starting amount was `5`.. We do't need to write all of that, except when questions like the one here are asked.

Usually, we just do the inverse operation to the quantities that are equal.

`x+3=8`
`(x+3)-3=8-3`
`x=5` (Because on the left, `(x+3)-3 =x+3-3` which is just plain `x`.)
The solution is `5`.

As you gain practice, you probably won't even write that much. You understand that you're subtracting `3` to get back to `x`, so you often won't bother writing it.

Example 2 (much shorter)

Solve: `5x=35`.

Here, we have multiplied the unknown by `5`. The inverse of multiplication is division. So we'll divide both quantities (the quantities on the right and on the left) by `5`.

`(5x)/5=35/5`, so
`x=7`.

The solution is `7`.