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Partial Fraction


Chikis
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I need help with this:

I need some explanation when it comes to dealing with partial fractions. [math]\frac{7x+19}{(x+1)(x+5)}[/math] into simpler fractions.[math]\frac{7x+19}{(x+1)(x+5)}=\frac{A}{(x+1)}+\frac{B}{(x+5)}[/math]

Why did[math]\frac{7x+19}{(x+1)(x+5)}[/math] become equal to [math]\frac{A}{(x+1)}+\frac{B}{(x+5)}[/math]?

Edited by Chikis
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It's not so much that it has become equal but rather that you are supposing it is equal. Having supposed they are equal the task is then to find some A and B which satisfy the equality. It's not necessarily the case that there are solutions for A and B that make the equations equal, in which case the supposition would be mistaken.

Edited by Prometheus
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Well when you are adding two fractions with different denominators you multiply each fraction top and bottom by the denominator of the other fraction so

 

[latex]\frac{a}{b}+\frac{b}{c} = \frac{ac}{bc}+\frac{bb}{bc} = \frac{ac+bc}{bc}[/latex]

 

If you are trying to work the equation back to A+B it isn't really possible. For instance there is two possible values for 7x being

 

[latex]\frac{7-\frac{7}{x+5}}{x+1} [/latex] and [latex]\frac{7-\frac{7}{x+1}}{x+1}[/latex]

Edited by fiveworlds
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Ok, solving

[math]\frac{7x+19}{(x+1)(x+5)}=\frac{A}{(x+1)}+\frac{B}{(x+5)}[/math]

[math]\rightarrow[/math]

[math]\frac{A(x+5)+B(x+1)}{(x+1)(x+5)}=\frac{7x+19}{1}[/math]

How does this give rise to[math]7x+19=A(x+5)+B(x+1)[/math]

Edited by Chikis
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x+1 and x+ 5 are common denominators of the fraction. That allows you to easily manipulate the equation

 

You want to rewrite the right-hand side as [math]A(x+5)/(x+1)(x+5) + B(x+1)/(x+1)(x+5)[/math]

 

All you have done is multiply each term by 1

 

Now the denominator for the whole thing is the same, so it cancels and you are left with the equation in question

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[math]\frac{A(x+5)+B(x+1)}{(x+1)(x+5)}=\frac{7x+19}{(x+5)(x+1)}[/math]

Ok! Multiplying the left and right hand side by the denominator, we have

[math]A(x+5)+B(x+1)=7x+19[/math] What happens next?

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Well, A and B are constants, but x is a variable - so we can vary it as we like. Can you think of any particular value(s) of x that might be of help in solving this equation?

 

 

No, you solve for A and B, so that the original equation is true for any x (except where it diverges). You have 2 equations and 2 unknowns.

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Well, A and B are constants, but x is a variable - so we can vary it as we like. Can you think of any particular value(s) of x that might be of help in solving this equation?

Based on the examples I have seen so far, to find A we take that [math]x=-1[/math]

So [math]7x-1+19=A(-1+5)+B(-1+1)[/math]

[math]\rightarrow[/math] [math]12=4A+0[/math]

[math]\rightarrow[/math][math]A=3[/math]

To find B, we take that [math]x=-5[/math]

So [math]7x-5+19=A(-5+5)+B(-5+1)[/math]

[math]\rightarrow[/math] [math]-35+19=0-4B[/math]

[math]\rightarrow[/math] [math]-16=-4B[/math]

[math]\rightarrow[/math] [math]B=+4[/math]

Thus,

[math]\frac{7x+19}{(x+1)(x+5)}=\frac{3}{(x+1}+\frac{4}{(x+5)}[/math]

Is my work correct?

Edited by Chikis
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Based on the examples I have seen so far, ....

[math]\frac{7x+19}{(x+1)(x+5)}=\frac{3}{(x+1}+\frac{4}{(x+5)}[/math]

Is my work correct?

 

Always but always multiply back out to check whether it is correct

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[math]\frac{3}{x+1}+\frac{4}{x+5}[/math]

[math]=\frac{3(x+5)+4(x+1)}{(x+1)(x+5)}[/math]

[math]=\frac{3x+15+4x+4}{(x+1)(x+5)}[/math]

[math]=\frac{7x+19}{(x+1)(x+5)}[/math] If the above is what you meant, I have done it and seen that it came out correctly. Thank you.

Edited by Chikis
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