Solving the HARD Equation: Finding Values for A, B, and C | Step-by-Step Guide

  • Thread starter Thread starter Sirsh
  • Start date Start date
  • Tags Tags
    Hard
AI Thread Summary
To solve the equation \(\frac{A}{(x-1)} + \frac{B}{(x+1)} + \frac{C}{x} = \frac{5x^2+2x+1}{x^3-x}\), it's essential to combine the left-hand side into a single fraction and equate the numerators after multiplying both sides by the common denominator \(x(x-1)(x+1)\). By substituting specific values for \(x\), such as 0, 1, and -1, you can derive a system of equations to find the values of A, B, and C. The correct approach involves isolating like terms and solving the resulting equations, leading to the values C = -1, A = 4, and B = 2. Verifying these values in the original equation confirms their accuracy.
Sirsh
Messages
262
Reaction score
10
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.

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

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.

Thanks
 
Physics news on Phys.org
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.
 
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:
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
 
you leave x as an unknown. Instead you will need equate the coefficients of x^{0},x^{1},x^{2}

this will give you enough equations to solve for A,B and C
 
apologies, I said to equate the denominators in an earlier post. I meant to say "equate numerators"
 
Sirsh said:
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
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).
 
I am confused, I know that you can solve for three variables by doing an eliminiation process like a silmutaneous solving method, but i don't understand how to get it to the three equations..
 
Starting from this equation:
\frac{A}{x - 1} + \frac{B}{(x+1)} + \frac{C}{x} = \frac{5x^2 + 2x + 1}{x(x - 1)(x + 1)}

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.
 
  • #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
 
  • #11
Both sides have the variable x right? So you don't 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.
 
  • #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
 
  • #13
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.
 
Back
Top