# How do inverse operations help solve equations?

Jan 21, 2015

They are helpful for solving with respect to a variable: if after some step you find yourself in a situation such as $\setminus \cos \left(x\right) = 0.7$, 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 $x = \setminus {\cos}^{- 1} \left(0.7\right)$

Mar 9, 2015

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.

$x + 3 = 8$
$\left(x + 3\right) - 3 = 8 - 3$
$x = 5$ (Because on the left, $\left(x + 3\right) - 3 = x + 3 - 3$ 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: $5 x = 35$.

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 $5$.

$\frac{5 x}{5} = \frac{35}{5}$, so
$x = 7$.

The solution is $7$.