So you substitute in different values of x to cancel out one of the unknowns and find the values of A and B. x is a variable, you can let it equal whatever you want.

(edited 5 years ago)

Reply 4

Maths1210

Original post by Sinnoh

Moved to Maths.

Start by treating it like an equation.

\dfrac{1}{(x-2)(x+1)} = \dfrac{A}{x-2} + \dfrac{B}{x+1}

.

When you multiply both sides by

(x - 2)(x + 1)

, you get

1 = A(x + 1) + B(x - 2)

So you substitute in different values of x to cancel out one of the unknowns and find the values of A and B. x is a variable, you can let it equal whatever you want.

What do you mean by cancel out one of the unknowns?

Reply 5

Sinnoh

Original post by Maths1210

What do you mean by cancel out one of the unknowns?

Supposing you let x = 2. What happens then?

Reply 6

Maths1210

Original post by Sinnoh

Supposing you let x = 2. What happens then?

Oh right so it would get rid of the bracket next to B. Wouldnt that answer be okay though since it is no longer in a fraction?

Reply 7

Sinnoh

Original post by Maths1210

Oh right so it would get rid of the bracket next to B. Wouldnt that answer be okay though since it is no longer in a fraction?

Wait until you've found the values of A and B, then put those values into the format they're asking for.

Reply 8

Maths1210

Original post by Sinnoh

Wait until you've found the values of A and B, then put those values into the format they're asking for.

Okay well I put 2 into it and 1= A(3) + B?

Reply 9

Sinnoh

Original post by Maths1210

Okay well I put 2 into it and 1= A(3) + B?

It would be 3A + 0B, because you have B(2 - 2). So 3A = 1, A = 1/3, so one part of your fraction is

\dfrac{1}{3(x-2)}

. Now substitute -1 for x to find B

Reply 10

Maths1210

Original post by Sinnoh

It would be 3A + 0B, because you have B(2 - 2). So 3A = 1, A = 1/3, so one part of your fraction is

\dfrac{1}{3(x-2)}

. Now substitute -1 for x to find B

would B be -1/3?

Reply 11

Sinnoh

Original post by Maths1210

would B be -1/3?

Yep

Reply 12

Maths1210

Original post by Sinnoh

Yep

But then how would I not get it into a fraction?

Reply 13

Maths1210

Original post by Maths1210

But then how would I not get it into a fraction?

Just subtract them?

Reply 14

Sinnoh

Original post by Maths1210

But then how would I not get it into a fraction?

Remember we're trying to get it so that you've got two fractions being added together, each with a linear denominator - as opposed to the original one, which had a quadratic in the denominator.

Since

Unparseable latex formula:

\dfrac{A}{x-2} \equiv A*\dfac{1}{x-2}

, we can substitute our values of A and B that we found to get our linear partial fractions:

\dfrac{1}{3} * \dfrac{1}{x-2} - \dfrac{1}{3} * \dfrac{1}{x+1}

So just multiply them together and you have your partial fractions.

Back in 15-20

Reply 15

Maths1210

Original post by Sinnoh

Unparseable latex formula:

\dfrac{A}{x-2} \equiv A*\dfac{1}{x-2}

, we can substitute our values of A and B that we found to get our linear partial fractions:

\dfrac{1}{3} * \dfrac{1}{x-2} - \dfrac{1}{3} * \dfrac{1}{x+1}

So just multiply them together and you have your partial fractions.

Back in 15-20

what does the star mean?

Reply 16

Sinnoh

Original post by Maths1210

what does the star mean?

Multiplied

Reply 17

ghostwalker

Given that the original post was simply to "write the equation without a fraction", I do wonder if it was not just a case of multiplying through by the two denominator factors and perhaps rearranging, and leaving it at that (especially as A and B are to remain in the answer); rather than doing the full blown partial fractions. I suspect the question is leading into partial fractions, but that's not yet been covered.