Question 1203631
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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)}}}
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<pre>
They want you find A, B, C from this given identity

    {{{A/(x-1)}}} + {{{B/(x-2)}}} + {{{C/(x-3)}}} = {{{(2x^2-6x+6)/(x-1)(x-2)(x-3)}}}


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

<A HREF=https://www.algebra.com/algebra/homework/expressions/expressions.faq.question.1196220.html>https://www.algebra.com/algebra/homework/expressions/expressions.faq.question.1196220.html</a>


     +--------------------------------------------------+
     |     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 + {{{B*((x-1)/(x-2))}}} + {{{C*((x-1)/(x-2))}}} = {{{(2x^2-6x+6)/((x-2)*(x-3))}}}.    (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

    {{{(2*1^2 - 6*1 + 6)/((1-2)*(1-3))}}} = {{{(2 -6 + 6)/((-1)*(-2))}}} = {{{2/2}}} = 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

    {{{A*((x-2)/(x-1))}}} + B + {{{C*((x-2)/(x-3))}}} = {{{(2x^2-6x+6)/((x-1)*(x-3))}}}.    (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*2^2 - 6*2 + 6)/((2-1)*(2-3))}}} = {{{(8 -12 + 6)/(1*(-1))}}} = {{{(-2)/(-1)}}} = -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

    {{{A*((x-3)/(x-1))}}} + {{{B*((x-3)/(x-2))}}} + C = {{{(2x^2-6x+6)/((x-1)*(x-2))}}}.    (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

    {{{(2*3^2 - 6*3 + 6)/((3-1)*(3-2))}}} = {{{(18 -18 + 6)/(2*1)}}} = {{{6/2}}} = 3.


Thus  C = 3.


<U>ANSWER</U>.  A = 1,  B = -2,  C = 3.
</pre>

Solved as a joke Math problem.



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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.