00:01
For this problem, to be able to solve for x, you need to use the least common denominator.
00:05
To start, i saw on the left side of the equation, there are six xes in the denominator, while on the right side there are only three xes.
00:13
So to start, i multiplied the right side of the equation times two to get the same denominators.
00:21
So this will be 7 minus 1 over 6x equals to 20.
00:36
Over 6x.
00:41
From here i decided that i wanted to see what numbers would be excluded.
00:46
Since they're both in the denominators we know that x cannot equal 0.
00:54
The next step will be to get like terms together.
00:59
So i added the 1 over 6x to the right side of the problem so i'm able to combine them.
01:06
On the left side they will cross out because they equal 0...