# Homework Help: Difficult rational expressions thinking question

1. Mar 22, 2013

1. The problem statement, all variables and given/known data

Find the Values of B and C given this:
$$\frac{(3x-18)}{(2x+1)(7x-3)}= \frac{B}{2x+1} + \frac{C}{7x-3}$$
2. Relevant equations

The equation given:$$\frac{(3x-18)}{(2x+1)(7x-3)}= \frac{B}{2x+1} + \frac{C}{7x-3}$$

3. The attempt at a solution

My attempt:
$$\frac{(3x-18)}{(2x+1)(7x-3)}= \frac{B}{2x+1} + \frac{C}{7x-3}$$
$$\frac{(3x-18)}{(2x+1)(7x-3)}= \frac{B(2x+1)+C(7x-3)}{(2x+1)(7x-3)}$$
$$3x-18=B(2x+1)+C(7x-3)$$
At this point I'm stuck and I think something I did was wrong.
I'd appreciate any assistance that could be provided!

Last edited: Mar 22, 2013
2. Mar 22, 2013

### SammyS

Staff Emeritus
$3x-18=B(2x+1)+C(7x-3)$​
is true for all values of x, then the initial rational equation
$\displaystyle \frac{(3x-18)}{(2x+1)(7x-3)}= \frac{B}{2x+1} + \frac{C}{7x-3}$​
will be true for all x except 3/7 and -1/2 .

There are at least a couple of ways to solve the identity
$3x-18=B(2x+1)+C(7x-3)$​
for B & C.

One is to pick a couple of values for x, which will give you two equations in the two unknowns, B & C. One such value is x = -1/2 .

Another method is to equate the coefficient of x on the left with the coefficient of x on the right; also equate the constant term on the left with the constant term on the right.

3. Mar 22, 2013

Hi, Thank You so much for your response. I was just wondering if you could clarify what you meant by this:
I apologize my math vocab is weak.

4. Mar 22, 2013

### Infrared

He means that you can expand 3x−18=B(2x+1)+C(7x−3) as 3x-18=2Bx+B+7Cx-3C. Cleaning this up a bit, we get 3x-18=(2B+7C)x+(B-3C). Two polynomials are equal only if their corresponding coefficients are equal, we get 3x=2B+7C and -18=B-3C. This is a system of equations which you should be able to sove.

5. Mar 22, 2013

But don't I have 3 variables still?

6. Mar 22, 2013

### Infrared

I am sorry, there should be no x there. If 3x=(2B+7C)x, then 3=2B+7C. I apologize for not using latex, I am just beginning to learn as I am new to these forums.

7. Mar 22, 2013

### SammyS

Staff Emeritus
If you expand the right side & collect terms, you get
$\displaystyle (2B+7C)x+(B-3C)\,,\$ right?​

So, the coefficient of x on the left, which is 3 must equal the coefficient of x on the right, which is 2B+7C . Similarly, the constant term on the left must equal the constant term on the right.

These must be equal because, the equation, $\displaystyle \ 3x-18=B(2x+1)+C(7x-3)\$ is true for all possible values of x, thus this equation is an identity.

8. Mar 22, 2013

Hi! Thank you both for responding.

I'm with you guys so far until here:
$3x−18=(2B+7C)x+(B−3C){}$

I don't quite understand why $3x$ must only equal $(2B+7C)x$.

After I try and solve this is what I get. Is everything I did okay because I don't know and method of verifying if this is right (not from a textbook)?

$-18=B-3C$
$B=-18+3C$

$3x=(2B+7C)x$
$3=2B+7C$
sub in $B=-18+3C$
$3=2(-18+3C)+7C$
$3=-36+6C+7C$
$39=13C$
$C=3$

$B=-18+3C$
$B=-18+3(3)$
$B=-18+9$
$B=-9$

∴B=-9 and C=3

9. Mar 22, 2013

### Dick

Yes, you solved it right. But $$\frac{(3x-18)}{(2x+1)(7x-3)}= \frac{B}{2x+1} + \frac{C}{7x-3}$$ actually leads to the equation $$3x-18=B(7x-3)+C(2x+1)$$ Which is not what you originally posted. The B and C are backwards.

As to your first question Ax+B=Cx+D for all values of x means A=C and B=D. Can you show that?

10. Mar 22, 2013

Ah, I plugged in the values of B and C and the expressions on both sides were equal. And that quote clarified quite a lot.

Thank you so much to all of you for the help!

11. Mar 23, 2013

### SammyS

Staff Emeritus
Sorry. I didn't check the details.