Disclaimer: I'm new to adding fractions this way.
Does anyone understand how the author justifies the circled equation?
$\begingroup$
$\endgroup$
2
-
3$\begingroup$ This is a technique known as Partial Fraction Decomposition [1][2][3]. I particularly recommend watching PatrickJMT's video on the topic here and the later examples. $\endgroup$– JMoravitzSep 20, 2015 at 23:52
-
$\begingroup$ Note that this procedure (as JMoravitz says, "partial fraction decomposition") is not really adding fractions -- it's actually the opposite. It's analogous to answering the question "what two fractions sum to 13/12?". Similar to how multiplying and factoring are opposite procedures. $\endgroup$– Daniel R. CollinsSep 21, 2015 at 6:56
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
2
First consider (as a concrete example) the possible fractions that have $6$ as the denominator. They can all be expressed as some combination of halves or thirds:
$\frac 16 = \frac 12 - \frac 13$
$\frac 26 = \frac 13$
$\frac 36 = \frac 12$
$\frac 46 = \frac 13 + \frac 13$
$\frac 56 = \frac 12 + \frac 13$
This observation leads us to conclude that all fractions with denominator $pq$ can be expressed as a combination of $\frac 1p$ and $\frac 1q$ (provided $p$ and $q$ have no common factors).
Moving on to algebraic fractions, all fractions with denominator $(x-p)(x-q)$ can be expressed as a combination of $\frac 1{x-p}$ and $\frac 1{x-q}$.
-
-
$\begingroup$ That explains the "decomposition" term nicely, since p and q are originally "composed" $\endgroup$ Sep 21, 2015 at 11:20
$\begingroup$
$\endgroup$
When we write out an equation such as $$\frac{7x-1}{(x+2)(x-3)} = \frac A{x+2} + \frac B{x-3},$$ where $A$ and $B$ are unknown real numbers, the general idea is that we want to solve the equation for $A$ and $B$. (I assume it is stated, or at least implicit, in the preceding discussion of the "original fraction", that the equation must be true for all values of $x$.)
For a very general equation with two unknowns, it is possible there is a unique solution for the two unknowns; it is also possible that there are many solutions, or that there are none at all. You can always write an equation with two unknowns without knowing whether it actually is possible for those two unknowns to be given values that make the equation true. If that is not possible, you have an equation with no solutions.
This particular equation has properties that you (hopefully) will come to recognize (because you will likely use equations like this to perform partial-fraction decomposition of integrals if you study calculus), which guarantee it has a solution; but if this has not already been shown to you, then the justification comes later, when you actually solve for $A$ and $B$, thereby proving that a solution exists.
