SOLUTION: Find A, B and C if {{{A/(x-1)}}} + {{{B/(x-2)}}} + {{{C/(x-3)}}} = {{{(2x^2-6x+6)/(x-1)(x-2)(x-3)}}}

Question 1203631: Find A, B and C if + + =
Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3816) (Show Source): You can put this solution on YOUR website!

Multiply both sides by the LCD to clear out fractions

Equate corresponding terms
(A+B+C)x^2 = 2x^2 leads to A+B+C = 2
(-5A-4B-3C)x = -6x leads to -5A-4B-3C = -6
6A+3B+2C = 6

The task is to solve this system of equations
A+B+C = 2
-5A-4B-3C = -6
6A+3B+2C = 6

Use either elimination, substitution, or matrix row reduction.

I'll skip the steps for this part, but the answers should be:
A = 1
B = -2
C = 3

Therefore,

updates to

It is an identity that is true for nearly all real numbers x but and and (to avoid division by zero errors).

WolframAlpha can be used to confirm the answer.
GeoGebra's CAS feature can also be used.
There are many other online calculators that offer the same features.

A different approach can be found here:
https://www.algebra.com/algebra/homework/expressions/expressions.faq.question.1196220.html

Answer by ikleyn(52777) (Show Source): You can put this solution on YOUR website!
.
Find A, B and C if + + =
~~~~~~~~~~~~~~~

They want you find A, B, C from this given identity + + = This problem is, in some sense, a joking and entertainment Math problem. Indeed, there are different ways/methods to solve it. One way, shown by the other tutor, is direct and straightforward, but requires a lot of computations, such as reducing the problem to a system of 3 linear equations and solving this system. This way is like a torture and can only bring tears. But there are other ways, so beautiful that you will smile learning them. One of such method is my solution under this link https://www.algebra.com/algebra/homework/expressions/expressions.faq.question.1196220.html +--------------------------------------------------+ | The other method I will show right here. | +--------------------------------------------------+ Multiply the given identity by (x-1). Then from the given identity (1), you will get this identity A + + = . (2) In this identity, place x= 1. In its left side, you will get the second and the third addends zeroed due to presence of the factor (x-1) in the numerator. Right side of (2) at x= 1 is = = = 1. Thus A = 1. Let's find B in similar way. Multiply the given identity by (x-2). Then from the given identity (1), you will get this identity + B + = . (3) In this identity, place x= 2. In its left side, you will get the first and the third addends zeroed due to presence of the factor (x-2) in the numerator. Right side of (3) at x= 2 is = = = -2. Thus B = -2. Find C in similar way. Multiply the given identity by (x-3). Then from the given identity (1), you will get this identity + + C = . (4) In this identity, place x= 3. In its left side, you will get the first and the second addends zeroed due to presence of the factor (x-3) in the numerator. Right side of (4) at x= 3 is = = = 3. Thus C = 3. ANSWER. A = 1, B = -2, C = 3.

Solved as a joke Math problem.

------------------

Next time, if you are given a similar functional identity of 10 addends with 10 unknown literal coefficients,
you do not need to reduce it to the system of 10 equations; you also do not need solve this system.

Simply apply this method and find the coefficients one after another, independently one from another,
spending one line for each coefficient.

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