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

[Sirsh]

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

 

[Tedjn]

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.

 

[qualal]

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

 

[Sirsh]

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

 

[qualal]

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

 

[qualal]

apologies, I said to equate the denominators in an earlier post. I meant to say "equate numerators"

 

[Mark44]

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).

 

[Sirsh]

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..

 

[Mark44]

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.

 

[Daft]

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

 

[Shaybay92]

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.

 

[gpax42]

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

 

[Mark44]

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.