# Thread: Solving Rational Equations: 3y/3y+9 + 8y+16/2y+6 = y-4/y+3

1. ## Solving Rational Equations: 3y/3y+9 + 8y+16/2y+6 = y-4/y+3

3y/3y+9 + 8y+16/2y+6 = y-4/y+3

I understand that you have to get an LCD
so when I factor the denominators I get

3(y+3) 2(y+3) and (y+3) making the LCD: 6(y+3)

at this point do you multiply 3y by 6(y+3)?

2. How would you do this?

1/6 + 1/9 + 1/15

???

3. Originally Posted by Rujaxso
3y/3y+9 + 8y+16/2y+6 = y-4/y+3

I understand that you have to get an LCD
so when I factor the denominators I get

3(y+3) 2(y+3) and (y+3) making the LCD: 6(y+3)

at this point do you multiply 3y by 6(y+3)?
I suspect that you violated the order of operations in giving your problem.

4. Originally Posted by tkhunny
How would you do this?

1/6 + 1/9 + 1/15

???
Is this step one of a walk-through?

5. Originally Posted by JeffM
I suspect that you violated the order of operations in giving your problem.
Not sure that I follow. Do you mean I should reduce the coefficients after factoring what could be factored?

6. Originally Posted by Rujaxso
3y/3y+9 + 8y+16/2y+6 = y-4/y+3
Just so you know, you're missing some very very important grouping symbols. What you wrote literally means:

$\dfrac{3y}{3y}+9 + 8y+\dfrac{16}{2y}+6 = y-\dfrac{4}{y}+3$

But what I think you meant was:

$\dfrac{3y}{3y+9} + \dfrac{8y+16}{2y+6} = \dfrac{y-4}{y+3}$

Which would be type-set as: (3y)/(3y + 9) + (8y+16)/(2y + 6) = (y-4)/(y + 3)

7. Originally Posted by ksdhart2
Just so you know, you're missing some very very important grouping symbols. What you wrote literally means:

$\dfrac{3y}{3y}+9 + 8y+\dfrac{16}{2y}+6 = y-\dfrac{4}{y}+3$

But what I think you meant was:

$\dfrac{3y}{3y+9} + \dfrac{8y+16}{2y+6} = \dfrac{y-4}{y+3}$

Which would be type-set as: (3y)/(3y + 9) + (8y+16)/(2y + 6) = (y-4)/(y + 3)
Your assumption is correct. I think Jeff wrote it out correctly too (before he edited his post that I was about to reply to) .

8. Originally Posted by Rujaxso
3y/(3y+9) + (8y+16)/(2y+6) = (y-4)/(y+3)

I understand that you have to get an LCD …
Actually, in this exercise, you already have a common denominator. You just don't see it, yet, because you have factored only the denominators.

Factor everything that you can (numerators and denominators), and then look for cancellations of common factors.

For example:

(17x - 34)/(68y + 17) reduces to (x - 2)/(4y + 1)

because we can cancel a common factor. Factor both the numerator and denominator, to see it:

[17(x - 2)]/[17(4y + 1)]

The common factor 17 cancels.

Now, in the same way, reduce the two algebraic ratios on the left-hand side of your given equation. You can then immediately add the reduced ratios, yes?

Once you have a single, algebraic ratio on each side of the equation, ask yourself, "If the denominators are the same on each side, what does that say about the numerators?"

9. Originally Posted by mmm4444bot
Actually, in this exercise, you already have a common denominator. You just don't see it, yet, because you have factored only the denominators.

Factor everything that you can (numerators and denominators), and then look for cancellations of common factors.

For example:

(17x - 34)/(68y + 17) reduces to (x - 2)/(4y + 1)

because we can cancel a common factor. Factor both the numerator and denominator, to see it:

[17(x - 2)]/[17(4y + 1)]

The common factor 17 cancels.

Now, in the same way, reduce the two algebraic ratios on the left-hand side of your given equation. You can then immediately add the reduced ratios, yes?

Once you have a single, algebraic ratio on each side of the equation, ask yourself, "If the denominators are the same on each side, what does that say about the numerators?"
Okay, so I get (y+3) as the common denom.

Whats the general rules for reducing after you factor? I think this is where I'm foggy on concepts.
for instance why cant Y cancel out the Y in (Y+3) ---> (Y)/(Y+3) ?

p.s thanks for the replies so far.

10. Originally Posted by Rujaxso
Okay, so I get (y+3) as the common denom.

Whats the general rules for reducing after you factor? I think this is where I'm foggy on concepts.
for instance

why cant Y cancel out the Y in (Y+3) ---> (Y)/(Y+3) ?

p.s thanks for the replies so far.
Do you think you could reduce $\frac{2}{2+3}$ and get $\frac{1}{3}$ by "cancelling" out the 2s in the numerator and denominator.

The answer is "No" and I hope you "No" it!!

The reason for not "cancelling" out Y in that expression is same.

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