Find A, B, and C

[nycmathguy]

How about this beauty?

 

[MathLover1]

do you want to try first?

 

[nycmathguy]

do you want to try first?

I will try first.

 

[MathLover1]

if denominators same, numerators must be same too
also, recall that denominator cannot be equal to zero, so exclude x=1, x=2, and x=3 are restrictions

use restricted values of x to find A,B, and C

Substitute x=1

1=2A +0+0
=>A =1/2

Substitute x=2

1=0-B+0
B=-1

Substitute x=3

1=0+0+2C
C=1/2

or

 

[nycmathguy]

if denominators same, numerators must be same too
also, recall that denominator cannot be equal to zero, so exclude x=1, x=2, and x=3 are restrictions

use restricted values of x to find A,B, and C

Substitute x=1

1=2A +0+0
=>A =1/2

Substitute x=2

1=0-B+0
B=-1

Substitute x=3

1=0+0+2C
C=1/2

or

I thank you very much. However, I did promise to try solving for A, B, and C myself. Please, keep in mind that I work 40 overnight hours. Sleeping during the day is not easy at all. I look like a zombie most of the time.

The weekend hours fly by. Life often gets in the way. I am not lazy. I am committed to improving certain math skills, if not all. I just need time but time runs away from me, from all of us.

 

[Country Boy]

1/((x-1)(x-2)(x-3))= A/(x-1)+ B/(x-2)+ C/(x-3).

There are many ways to do this. The most basic is to add the fractions on the right. The common denominator is (x-1)(x-2)(x-3):
1/((x-1)(x-2)(x-3))= A(x-2)(x-3)/((x-1)(x-2)(x-3))+ B(x-1)(x-3)/((x-1)(x-2)(x-3))+ C(x-1)(x-2)/((x-1)(x-2)(x-3))
1/((x-1)(x-2)(x-3))= (Ax^2-5Ax+ 6A+ Bx^2- 4Bx+ 3B+ Cx^2- 3Cx+ 2C)/((x-1)(x-2)(x-3))= ((A+ B+ C)x^2- (5A+ 4B+3C)x+ (6A+ 3B+ 2C))/((x-1)(x-2)(x-3))

So we must have (A+ B+ C)x^2-(5A+ 4B+3C)x+ (6A+ 3B+ 2C)= 1 for all x which means we must have A+ B+ C= 0, 5A+ 4B+ 3C= 0, and 6A+ 3B+ 2C= 1. From the first equation, C= -A- B so the second equation becomes 5A+ 4B- 3A- 3B= 2A+ B= 0. B= -2A. The third equation becomes 6A- 6A- 2A+ 4A= 2A= 1 so A= 1/2. Then B= -1 and C= 1/2.

An easier way is to multiply the equation, 1/((x-1)(x-2)(x-3))= A/(x-1)+ B/(x-2)+ C/(x-3), by (x-1)(x-2)(x-3):
1= A(x-2)(x-3)+ B(x-1)(x-3)+ C(x-1)(x-2).

Now, if x= 1, 1= A(-1)(-2)= 2A so A= 1/2. If x= 2, 1= B(1)(-1)= -B so B= -1. And if x= 3, 1= C(2)(1)= 2C so C=1/2.

1/((x-1)(x-2)(x-3))= (1/2)/(x-1)- 1/(x-2)+ (1/2)/(x-3).

 

[nycmathguy]

1/((x-1)(x-2)(x-3))= A/(x-1)+ B/(x-2)+ C/(x-3).

There are many ways to do this. The most basic is to add the fractions on the right. The common denominator is (x-1)(x-2)(x-3):
1/((x-1)(x-2)(x-3))= A(x-2)(x-3)/((x-1)(x-2)(x-3))+ B(x-1)(x-3)/((x-1)(x-2)(x-3))+ C(x-1)(x-2)/((x-1)(x-2)(x-3))
1/((x-1)(x-2)(x-3))= (Ax^2-5Ax+ 6A+ Bx^2- 4Bx+ 3B+ Cx^2- 3Cx+ 2C)/((x-1)(x-2)(x-3))= ((A+ B+ C)x^2- (5A+ 4B+3C)x+ (6A+ 3B+ 2C))/((x-1)(x-2)(x-3))

So we must have (A+ B+ C)x^2-(5A+ 4B+3C)x+ (6A+ 3B+ 2C)= 1 for all x which means we must have A+ B+ C= 0, 5A+ 4B+ 3C= 0, and 6A+ 3B+ 2C= 1. From the first equation, C= -A- B so the second equation becomes 5A+ 4B- 3A- 3B= 2A+ B= 0. B= -2A. The third equation becomes 6A- 6A- 2A+ 4A= 2A= 1 so A= 1/2. Then B= -1 and C= 1/2.

An easier way is to multiply the equation, 1/((x-1)(x-2)(x-3))= A/(x-1)+ B/(x-2)+ C/(x-3), by (x-1)(x-2)(x-3):
1= A(x-2)(x-3)+ B(x-1)(x-3)+ C(x-1)(x-2).

Now, if x= 1, 1= A(-1)(-2)= 2A so A= 1/2. If x= 2, 1= B(1)(-1)= -B so B= -1. And if x= 3, 1= C(2)(1)= 2C so C=1/2.

1/((x-1)(x-2)(x-3))= (1/2)/(x-1)- 1/(x-2)+ (1/2)/(x-3).

Very nice.