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Homework Help: HARD equation.

  1. Jun 10, 2010 #1
    Hi, i'm just wondering if anyone can help me with this equation: I need to determine the values of A,B and C from the given equation.

    [tex]\frac{A}{(x-1)}[/tex] + [tex]\frac{B}{(x+1)}[/tex] + [tex]\frac{C}{x}[/tex] = [tex]\frac{5x^2+2x+1}{x^3-x}[/tex]

    I've tried to simplify it by factorising the denominator on the LHS to x^3 - x, and i am suspicious that the values of A,B and C are the 3 parts of the numerator of the RHS respectively to 5x^2,2x and 1.

  2. jcsd
  3. Jun 10, 2010 #2
    No, A, B, and C are numbers. You need to combine terms on the LHS and then compare the numerators to get a system of equations for A, B, and C.
  4. Jun 10, 2010 #3
    You need to make the three "fractions" you have on the LHS into just one "fraction" (google "addition and subtraction of algebraic fractions" as a hint).

    once you have that you can then equate the [STRIKE]denominators[/STRIKE] numerators in a separate equation. Then you can start to solve for A, B and C
    Last edited: Jun 10, 2010
  5. Jun 10, 2010 #4
    How is that possible, i'd have to solve for 4 unknowns A,B,C and x?

    (Ax^2+Ax+x^2B+x^2C-xB-C)/(x^3-x) is the LHS
  6. Jun 10, 2010 #5
    you leave x as an unknown. Instead you will need equate the coefficients of [tex]x^{0},x^{1},x^{2}[/tex]

    this will give you enough equations to solve for A,B and C
  7. Jun 10, 2010 #6
    apologies, I said to equate the denominators in an earlier post. I meant to say "equate numerators"
  8. Jun 10, 2010 #7


    Staff: Mentor

    As already noted, the idea is not to solve for x. The original equation is supposed to be true for all values of x other than those that make the denominators zero (i.e., x = 0, 1, or -1).
  9. Jun 10, 2010 #8
    I am confused, I know that you can solve for three variables by doing an eliminiation process like a silmutaneous solving method, but i dont understand how to get it to the three equations..
  10. Jun 10, 2010 #9


    Staff: Mentor

    Starting from this equation:
    [tex]\frac{A}{x - 1} + \frac{B}{(x+1)} + \frac{C}{x} = \frac{5x^2 + 2x + 1}{x(x - 1)(x + 1)}[/tex]

    Multiply both sides of the equation by x(x - 1)(x + 1).
    Solve for A, B, and C. The simplest way to get the three equations is to substitute specific values for x. I would use x = 0 to get one equation, x = 1 to get another, and x = -1 to get the third equation.
  11. Jun 10, 2010 #10
    Whether you choose to combine them into 1 fraction, and compare numerators, or multiply both sides by x(x-1)(x+1), it would be easier on you if you do not expand the brackets

    eg. instead of writing "Ax2 + Ax", leave it as "A(x+1)(x)"

    Since the values of A, B and C are the same for all values of x, you yourself can choose any value of x to substitute into the equation to obtain expressions for A, B and C

    Choosing the right numbers to substitute is the key to making this simple

    eg.: for the term "A(x+1)(x)", if I substitute the values x = -1 or x = 0, that term becomes 0, which means I eliminate the term containing A completely
  12. Jun 11, 2010 #11
    Both sides have the variable x right? So you dont want to solve for x. What is missing is the A B C on the right hand side. You want to expand out the left hand side (through cross multiplication) and then say all x^2 terms must have some coefficient (in this case 5 on the right side) and so any terms on the left with x^2 must sum to give 5x^2. Similarly with 2x and the constant 1.
  13. Jun 11, 2010 #12
    When I attempted the problem I managed to get C = -1, A = 4 and B = 2...

    any idea if this is right?

    I multiplied through to get a common denominator and then set the numerators equal to one another.

    A(x+1)(x) + B(x-1)(x) + C(x2-1) = 5x2+2x+1

    I distributed and found...

    Ax2 + Ax + Bx2-Bx+Cx2-C = 5x2+2x+1

    Then I simply isolated like terms, found C = -1; A - B = -2 and A + B + C = 5

    Solving the system of equations yielded C = -1, A = 4, and B = 2

    Might have messed up the arithmetic somewhere but i think that is the general premise to the solution
  14. Jun 11, 2010 #13


    Staff: Mentor

    These kinds of problems are easy to check. Put the numbers in the equation in the OP and see if you get a true statement.
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