How would you solve (2x+16)/(x+6)(x+7) in partial fractions?

This is a simple partial fraction to solve as the denominator has already been given to you as a factorised quadratic. 

Because the x terms are to the dgree 1 aka x1 we use the form 

(2x+16)/(x+6)(x+7) = A/(x+6) + B/(x+7)

where A and B are coefficients that have no x terms. 

You then multiply out the equation by taking the denominators up top and cancelling. Step by Step this goes:

1: move (x+6) up top to give 

(2x+16)(x+6)/(x+6)(x+7) = A + B(x+6)/(x+7)

we can see this has removed (x+6) from the A term.

2: now do the same for (x+7) to give

(2x+16)(x+6)(x+7)/(x+6)(x+7) = A(x+7) + B(x+6)

we can see on the left hand side that the brackets cancel out giving:

2x+16 = A(x+7) + B(x+6)

But how do we find A, B and x?

Work you way through the x coefficients. 

x0: 16 = 7A +6B

Then we look at x1 (this is the only x used in this example but in others you may have xetc) then:

x1: 2 = A + B

We now have 2 simultaneous equations which can be solved giving us 



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