# Partial Fraction

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I need help with this:

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

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

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

$\frac{a}{b}+\frac{b}{c} = \frac{ac}{bc}+\frac{bb}{bc} = \frac{ac+bc}{bc}$

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

$\frac{7-\frac{7}{x+5}}{x+1}$ and $\frac{7-\frac{7}{x+1}}{x+1}$

Edited by fiveworlds
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Ok, solving

$\frac{7x+19}{(x+1)(x+5)}=\frac{A}{(x+1)}+\frac{B}{(x+5)}$

$\rightarrow$

$\frac{A(x+5)+B(x+1)}{(x+1)(x+5)}=\frac{7x+19}{1}$

How does this give rise to$7x+19=A(x+5)+B(x+1)$

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 $A(x+5)/(x+1)(x+5) + B(x+1)/(x+1)(x+5)$

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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The reason you cannot see how it gives rise to your third line (which is correct) is that your second line has a mistake (check the denominator on the RHS). Correct this mistake and you'll see the equality in the third.

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$\frac{A(x+5)+B(x+1)}{(x+1)(x+5)}=\frac{7x+19}{(x+5)(x+1)}$ Could this one be right? Is the right hand side correct now?

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Yep. You might be able to just see the equality of the numerators now, or you might want to multiply both sides by the denominator to make it explicit.

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$\frac{A(x+5)+B(x+1)}{(x+1)(x+5)}=\frac{7x+19}{(x+5)(x+1)}$

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

$A(x+5)+B(x+1)=7x+19$ 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?

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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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Yes, but there are particular values of x (i.e. the roots - of which there are two here) for which it is often easier to see the values of A and B.

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Yes, but there are particular values of x (i.e. the roots - of which there are two here) for which it is often easier to see the values of A and B.

At those values the equation doesn't work.

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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 $x=-1$

So $7x-1+19=A(-1+5)+B(-1+1)$

$\rightarrow$ $12=4A+0$

$\rightarrow$$A=3$

To find B, we take that $x=-5$

So $7x-5+19=A(-5+5)+B(-5+1)$

$\rightarrow$ $-35+19=0-4B$

$\rightarrow$ $-16=-4B$

$\rightarrow$ $B=+4$

Thus,

$\frac{7x+19}{(x+1)(x+5)}=\frac{3}{(x+1}+\frac{4}{(x+5)}$

Is my work correct?

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

$\frac{7x+19}{(x+1)(x+5)}=\frac{3}{(x+1}+\frac{4}{(x+5)}$

Is my work correct?

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

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Multiplying back? How?

Edited by Chikis
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Do the reverse - you have just used simultaneous equations to figure out two numbers. Put the numbers back in where your A and B were and check that it comes out correctly

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$\frac{3}{x+1}+\frac{4}{x+5}$

$=\frac{3(x+5)+4(x+1)}{(x+1)(x+5)}$

$=\frac{3x+15+4x+4}{(x+1)(x+5)}$

$=\frac{7x+19}{(x+1)(x+5)}$ 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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