Partial Fractions: Solving Simple Linear Equations

In summary, the conversation is about figuring out how to integrate using partial fractions. The main problem is splitting up expressions such as x^2-8/x+3 and x^2+1/x-1 and making them equal to a simpler form. The solution involves using long division or recognizing a difference of two squares and factoring the numerator. Working backwards and checking the result can also be helpful.
  • #1
Help1212
9
0

Homework Statement


Hello I'm trying to figure out how to integrate with the use of partial fractions but I'm a bit stuck on something that should probably be simple.. but I can't see it clearly

How do you split up x2-8/x+3 and make it equal to (x-3) + 1/x+3

or for instance x2 +1/x-1 split up into x+1 +2/x-1

Homework Statement


Homework Equations


The Attempt at a Solution


They seem simple.. but i think I'm missing something here because I only see one linear denominator so I don't know how to apply the sequence.. or I'm overlooking something.. a bitof hellp please?
 
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  • #2
It is basically "long division". To divide [itex]x^2- 8[/itex] by x+ 3, note that x divides into [itex]x^2[/itex] x times. Taking x times x+3, we get [itex]x^2+ 3x[/itex] and subtract to get [itex]x^2- 8 - (x^2+ 3x)= -3x- 8[/itex]. That is, x+ 3 divides into [itex]x^2- 8[/itex] x times with remainder -3x- 8. x divides into -3x -3 times. Taking -3 times x+ 3 we get [itex]-3x- 9[/itex] and subtracting that from -3x- 8 gives a remainder of +1. That tells you that
[tex]\frac{x^2- 8}{x+ 3}= x- 3+ \frac{1}{x+ 3}[/tex].

Another way to see that, simpler but not "rote", is to note that
[tex]\frac{x^2- 8}{x+ 3}= \frac{x^2- 9+ 1}{x+ 3}= \frac{(x- 3)(x+ 3)+ 1}{x+ 3}[/tex]
[tex]= x- 3+ \frac{1}{x+ 3}[/tex]
again.
 
  • #3
Help1212 said:

Homework Statement


Hello I'm trying to figure out how to integrate with the use of partial fractions but I'm a bit stuck on something that should probably be simple.. but I can't see it clearly

How do you split up x2-8/x+3 and make it equal to (x-3) + 1/x+3

or for instance x2 +1/x-1 split up into x+1 +2/x-1

Knowing what to look for it is easy enough even when you have left out the brackets.

One useful thing for you would be to work backwards and check it that way - may give some insights.

One thing to look for is express your numerator as a [difference of two squares (check the formula for that) that has your denominator as factor] + something.
 
Last edited:

1. What are partial fractions?

Partial fractions are fractions that can be broken down into simpler fractions. They are useful in solving equations with complex fractions.

2. How do you solve simple linear equations with partial fractions?

To solve simple linear equations with partial fractions, you need to first decompose the given fraction into partial fractions. Then, set the numerators of the partial fractions equal to each other and solve for the variables.

3. What is the purpose of using partial fractions in solving equations?

The purpose of using partial fractions is to simplify complex fractions and make equations easier to solve. It also allows us to solve equations that would otherwise be difficult or impossible to solve.

4. Can you give an example of solving a simple linear equation using partial fractions?

Yes, for example, let's say we have the equation (x+3)/(x+2) = 5/3. By decomposing the fraction into partial fractions, we get (x+3)/(x+2) = 1/3 + 1/3x. Setting the numerators equal to each other, we get 1/3 + 1/3x = 5/3. Solving for x, we get x = 2.

5. Are there any special cases when solving equations with partial fractions?

Yes, there are special cases such as repeated factors, non-distinct linear factors, and quadratic factors. These cases require different methods of decomposition and solving, but the general process remains the same.

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