Questions on Algebra: Systems of Linear Equations answered by real tutors!

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Question 151438: Could you please help me solve this equation for A, B, and C.
11x^2-x+8/(x+1)(X^2+4)=A/(x+1)+Bx+C/X^2+4: Could you please help me solve this equation for A, B, and C.
11x^2-x+8/(x+1)(X^2+4)=A/(x+1)+Bx+C/X^2+4
Answer by jim_thompson5910(9368) About Me  (Show Source):
You can put this solution on YOUR website!
(11x^2-x+8)/((x+1)(x^2+4))=A/(x+1)+(Bx+C)/(x^2+4) Start with the given equation.


(cross((x+1)(x^2+4)))((11x^2-x+8)/(cross((x+1)(x^2+4))))=(cross((x+1))(x^2+4))(A/cross((x+1)))+((x+1)cross((x^2+4)))((Bx+C)/cross((x^2+4))) Multiply every term by the LCD (x+1)(x^2+4) to clear the fractions.


11x^2-x+8=(x^2+4)(A)+(x+1)(Bx+C) Simplify


11x^2-x+8=A(x^2+4)+(x+1)(Bx+C) Rearrange the terms


11x^2-x+8=A(x^2+4)+Bx^2+Cx+Bx+C FOIL


11x^2-x+8=Ax^2+4A+Bx^2+Cx+Bx+C Distribute





Notice how the x^2 terms on the right side are Ax^2 and Bx^2. So this means that Ax^2+Bx^2=11x^2 and A+B=11


Also, notice how the x terms on the right side are Bx and Cx. So this means that Bx+Cx=-x and B+C=-1

Finally, the constant terms on the right side are 4A and C. So this means that 4A+C=8



A+B=11 Start with the first equation.


B=11-A Subtract A from both sides.


B+C=-1 Move onto the second equation


11-A+C=-1 Plug in B=11-A


-A+C=-12 Subtract 11 from both sides.


-4A+4C=-48 Multiply both sides by 4.


Add this equation to equation #3

-4A+4C=-48
+4A+C=8
--------------
0A+5C=-40


5C=-40 Simplify


C=-8 Divide both sides by 5.


4A+C=8 Go back to the third equation


4A-8=8 Plug in C=-8


4A=16 Add 8 to both sides.


A=4 Divide both sides by 4.


B=11-A Move onto the first isolated equation


B=11-4 Plug in A=4


B=7 Subtract


====================================================

So the value of the constants are

A=4, B=7, and C=-8


So this means that the equation is (11x^2-x+8)/((x+1)(x^2+4))=(4)/(x+1)+(7x-8)/(x^2+4)