Partial equation

[bingsmallbang]

Hello,
I 'm trying an equation question. If anyone would show me how to get the answer and most importantly the method to the answer for it. I appreciate very much. The equation is looking for the value of E, F, G, H.

Thanks.

[Guest]

I believe I saw this answered on another forum but I will answer it here.

This is a problem in "partial fractions", not a "partial equation".

There are a number of ways to do this, one being to get the "common denominator" on the right (which would be the denominator on the left) then add and compare numerators.
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In my opinion the simplest method is to eliminate the denominators by multiplying both sides by [tex](x+ 6)^2(x^2+ 6)[/tex]:[tex]x^2+ 30= E(x+ 6)(x^2+ 6)+ F(x^2+ 6)+ G(x+ 6)^2[/tex]. And there are a number of ways to determine those constants. One would be to multiply out the right side and compare coefficients of the same powers of x. But basically there are three numbers to be determined so we need three equations. This is to be true for all values of x so we can just choose three values of x. Seeing "x+ 6" in there, taking x= -6 simplifies a lot.

If x= -6, we have 36+ 30= 66= E(0)+ F(42)+ G(0) so F= 66/42= 33/21= 11/7. There is no other value that will simplify that much but I would just take x= 0 and x= 1. If x= 0, we have 30= 36E+ 6F+ 36G and if x= 1, 31= 49E+ 7F+ 49G. Set F= 11/7 in those and solve the two equations for E and G.

[bingsmallbang]

Hi Thanks for the reply.
I found F and I used x-6 cause it easiest to remove the E portion as multiply by 0.
Can I know how do you get 49G assuming x=1
I am stuck at 31=49E+77/7+(G(1)+H) (1+6)^2

Thanks.

[HallsofIvy]

As I said, you need two equations to solve for E and G. Originally you had three equation in E, F, and G.
Two of them were 30= 36E+ 6F+ 36G and 31= 49E+ 7F+ 49G. Having found that F= 11/7, replacing F in those equations with 11/7 we have 30= 36E+ 66/7+ 36G and 31= 49E+ 11+ 36G.

Subtracting 66/7 from both sides in the first equation,
30- 66/7= 210/7- 66/7= 144/7= 36E+ 36G. Dividing by 36, 4= E+ G.

Subtracting 11 from both sides in the second equation,
20= 49E+ 36G. Since E= 4- G, from above, 20= 49(4- G)+ 36= 196- 13G. 13G= 176.

[Guest]

Hi Thanks for the reply.
I found F and I used x-6 cause it easiest to remove the E portion as multiply by 0.
Can I know how do you get 49G assuming x=1
I am stuck at 31=49E+77/7+(G(1)+H) (1+6)^2

Thanks.

1+ 6= 7 and 7^2= 49.

[Guest]

And 77/7= 11!