Search 75,714 tutors
FIND TUTORS
Ask a question
0 0

X-4/X^2-251+1/X-5

Tutors, please sign in to answer this question.

2 Answers

I was not sure exactly how it was set up w/o parethesis so I did it the most complicated way I could think of.  Hopefully I'll touch on something that will help.

(X-4)/(X^2-251) + 1/(X-5), then find LCD

(X-4)(X-5)/(X^2-251)(X-5) + (X^2-251)/(X-5)(X^2-251), multiply each side by the other's denomiator to find the LCD, then combine fractions

((X-4)(X-5) + (X^2-251)) / (X^2-251)(X-5) , then factor out

X^2 -9X + 20 + X^2 - 251 / (X^2-251)(X-5) , combine terms and get ready for partial fractions

(2X^2 -9X -231) / (X^2-251)(X-5), partial fractions

(2X^2 -9X -231) / (X^2-251)(X-5) = A/(X-5) + (B+CX)/(X^2-251), now solve for A, B, and C.

(2X^2 -9X -231)= A(X^2-251) + (BX+C)(X-5), expand out

AX^2-A251+BX^2-B5+CX-C251, it gets tricky here, I always put the x terms together like so...

X^2: 2=B+A

X: -9=C

1:-231= -251A-251C

Now remove the X variables and solve... C=-9,A=(2028/251),B=(1429/753)

Judging by the answers I either did it wrong or you don't need to do a paritial fraction.  Let me know if you needed to solve it a different way.

 

Hi Jonathan,

I'm also having a little trouble understanding your question, but I can give a generalized complex fraction explanation.  Hopefully, it might help you a little.  Most complex fraction problems are comprised of one fraction over (divided by) another fraction.  I'll start with a somewhat simple one...numbers only.

6/25   /    3/5    (vertically 6 over 25 over 3 over 5)

I usually start by setting up the probem horizontally as you see it above.  Division of fraction requires that we multiply the 1st fraction by the reciprocal of the second. This would become:

6/25   *    5/3

WE then need to reduce like factors in the numerators/denominators and multiply across.  This would become:

2/5   *     1/1   =   2/5

With algebra involved the reducing often requires factoring....example....

x+4/x^2-25    /      1/x-5

multiply first fraction by reciprocal of 2nd

x+4/x^2-25    *   x-5/1

factor numerators and deominators if possible,

(x+4)/(x+5)(x-5)   *   (x-5)/1

reduce like factors n the numerators/denominators  ...will reduce the (x-5)s  ... then multiply across

x+4/x+5               *     1/1     =    x+4/x+5

I know it's not your original problem, but I hope this helps