How do inverse operations help solve equations?

2 Answers
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 #\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)#

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#
#(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#.