Mathematics 101 |
---|

When solving an equation involving fractions, eliminate the fractions by multiplying both sides of the equation by the least common multiple of the denominators.

Example 1: x/3 +5 = 2x +7.

Multiply both sides by 3 to obtain the equivalent equation x+15 = 6x + 21.

Example 2: x + 2/3 = x/4 - 1.

Multiply both sides by 12, *since 12 = 3*4 is the least common multiple of 3
and 4,* to obtain the equivalent equation 12x + 8 = 3x - 12.

Example 3: 3x/4 +5 = x - 1/6.

Multiply both sides by 12 to obtain the equivalent equation 9x + 60 = 12x -
2. *Since 4 = 2 ^{2} and 6 = 2*3, the least common multiple of
4 and 6 is 2^{2}*3 = 12.*

**Warning:** Remember that there is a difference between an equation (which
has two sides connected by an equal symbol) and an algebraic expression
(which only has one side). One obviously *cannot* multiply both sides of
an algebraic expression--if only because it only has one side.

Remember, when solving an equation or inequality involving absolute values, there is very little algebraic manipulation one can do with absolute values and, for the most part, they need to be treated in the same way variables are treated.

For example, one starts to solve |x-3|+4<15 the same way one would solve x+4<15, by subtracting 4 from both sides. In the case of the inequality x+4<15, one obtains x<11 and is essentially finished, needing only to make sure that one's answer is expressed clearly.

In the case of the inequality |x-3|+4<15, one obtains |x-3|<11 and then needs to ask oneself how that is possible. Since the numbers which have an absolute value less than 11 are the numbers between -11 and 11, exclusively, the answer to that question is the numbers for which -11<x-3<11.

In other words, x is a solution if the two inequalities -11<x-3 and x-3<11 are both satisfied. Since the former is satisfied when -8<x and the latter when x<14, the solution to the question is {x | -8<x<14}. In interval notation, the solution may also be written as (-8,14).

After you get an equation or inequality into the form |x-a| R b, where R is either =, <, <=, > or >=, the key is to figure out how that can possibly be. That can be done by answering the question:

**A. How can the absolute value of x-a be R b?**

If you can't answer that immediately, ask yourself the question:

**B. How can the absolute value of any number be R b?**

If you can't answer that immediately, ask yourself the question:

**C. What numbers have an absolute value that is R b?**

The answer to C leads immediately to the answer to B which leads immediately to the answer to A. Once the answer to A is known, one merely has to figure out the values for x which fit in with that condition.

For example, suppose you tried to solve the inequality |x-3|+4=15 and got to the point where you were left with solving |x-3|=11. In terms of the notation used above, a is 3, b is 11 and R is =.

Question C becomes: What numbers have an absolute value that is = 11?

The answer (hopefully obviously) is 11 and -11.

Question B becomes: How can the absolute value of any number be = 11?

Once C has been answered, the answer (again hopefully obviously) is "if the number is either 11 or -11."

Question A becomes: How can the absolute value of x-3 be = 11? (The phrasing here is somewhat awkward, but the meaning should be clear.)

Once B has been answered, the answer (again hopefully obviously) is "if x-3 is either 11 or -11."

At this point, you merely have to figure out the values of x for which x-3 is either 11 or -11. This is easily done, since x-3=11 precisely when x=14 and x-3=-11 precisely when x=-8. Thus the solution set of the original equation is {14,-8}.

*Warning: When the solution set of an equation or inequality contains more
than one number, it's important to use proper set notation in order to make
your answer clear.*