Difficult rational expressions thinking question

[AvocadosNumber]

Homework Statement

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

Homework Equations

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

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!

 

[SammyS]

Homework Statement

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

Homework Equations

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

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!

If your last equation

$3x-18=B(2x+1)+C(7x-3)$​

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.

 

[AvocadosNumber]

Hi, Thank You so much for your response. I was just wondering if you could clarify what you meant by this:

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.

I apologize my math vocab is weak.

 

[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.

 

[AvocadosNumber]

we get 3x=2B+7C and -18=B-3C. This is a system of equations which you should be able to solve.

But don't I have 3 variables still?

 

[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.

 

[SammyS]

Hi, Thank You so much for your response. I was just wondering if you could clarify what you meant by this:

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.

I apologize my math vocab is weak.

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.

 

[AvocadosNumber]

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

 

[Dick]

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

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?

 

[AvocadosNumber]

Ax+B=Cx+D for all values of x means A=C and B=D.

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!

 

[SammyS]

.., 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.
...

Sorry. I didn't check the details.