What does it mean to equate the coefficients of like terms when solving for A and B in partial fractions?

Question

I'm trying to step myself through solving partial fractions in a year 10 book by Cambridge. This is a concept they're introducing early for students who want to challenge themselves and it's pretty light on the explanation.

For example: 7/( x+2 ) ( 2x-3 ) = A/2x-3 + B/x+2. I understand how to work this to the point where I reach 7=x(A+2B)+2A−3B. From there I've read that I need to do something called "equating coefficients. The coefficients near the like terms should be equal, so the following system is obtained: A+2B=0 2A−3B=7.

But I don't understand WHY or how it is valid that we set these parts of the equation to these values. Why not A+2B=7 2A−3B=0 for instance. I've tried looking at YouTube and asking friends, but I can't seem to get my head around it.

I can do it and I can solve for A and B using this method. But I'm really struggling to understand what it is I'm doing at that point in the process. The phrase that keeps popping up when I look into this is "we can equate the coefficients of like terms". For example on the wikipedia page on Fraction Decomposition it says "Equating the coefficients of x and the constant (with respect to x) coefficients of both sides of this equation...". Second example: it says "The coefficients near the like terms should be equal, so the following system is obtained: " on the emathhelp page when I enter the equation 7/( x+2 ) ( 2x-3 ).

Answer 1

Assume you were working with

$$ax+b=3x+2.$$

We mean that this holds for any $x$. So in particular, we could write

$$x=0\to b=2,\\x=1\to a+b=5,\\ x=-1\to -a+b=-1,\\ x=2\to 2a+b=8,\ \ \ \ \ \ \ \ \ \ \ \ \\ x=5000\to 5000a+b=1502,\\\cdots$$

This is a system of two unknowns and infinitely many equations. But it turns out that if you solve for a minimum number of equations (with the first two, $a=3, b=2$), the solution is valid for all equations, because the symbolic expressions are fully equivalent.

The same holds for rational fractions or any kind of identification.

Answer 2

I think you're a little confused about the steps in this problem. Note that, after you've multiplied both sides by the denominator, you need to try and solve the resulting equation, in this case $$7 = x(A+2B) + 2A - 3B.$$ Like terms are the coefficients of identical powers of $x$. Observe that $7 = 0x + 7$. Can you see the similarity now? Had $(A+2B)$ been anything but $0$, you'd have a non-zero $ax$ term on the left-hand side of the equation above. The same logic applies to $(2A-3B)$.

So really you end up with $$A+2B = 0 \\ 2A - 3B = 7$$ which, when solved simultaneously, gives $A= 2$, $B = -1$.