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

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Rujaxso

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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)?
 
How would you do this?


1/6 + 1/9 + 1/15

???
 
Rujaxso said:
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)?
Click to expand...
I suspect that you violated the order of operations in giving your problem.
 
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JeffM said:
I suspect that you violated the order of operations in giving your problem.
Click to expand...

Not sure that I follow. Do you mean I should reduce the coefficients after factoring what could be factored?
 
Rujaxso said:
3y/3y+9 + 8y+16/2y+6 = y-4/y+3
Click to expand...

Just so you know, you're missing some very very important grouping symbols. What you wrote literally means:

\(\displaystyle \dfrac{3y}{3y}+9 + 8y+\dfrac{16}{2y}+6 = y-\dfrac{4}{y}+3\)

But what I think you meant was:

\(\displaystyle \dfrac{3y}{3y+9} + \dfrac{8y+16}{2y+6} = \dfrac{y-4}{y+3}\)

Which would be type-set as: (3y)/(3y + 9) + (8y+16)/(2y + 6) = (y-4)/(y + 3)
 
ksdhart2 said:
Just so you know, you're missing some very very important grouping symbols. What you wrote literally means:

\(\displaystyle \dfrac{3y}{3y}+9 + 8y+\dfrac{16}{2y}+6 = y-\dfrac{4}{y}+3\)

But what I think you meant was:

\(\displaystyle \dfrac{3y}{3y+9} + \dfrac{8y+16}{2y+6} = \dfrac{y-4}{y+3}\)

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

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

I understand that you have to get an LCD …
Click to expand...
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?" :cool:
 
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mmm4444bot said:
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?" :cool:
Click to expand...

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.
 
Rujaxso said:
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.
Click to expand...

Do you think you could reduce \(\displaystyle \frac{2}{2+3}\) and get \(\displaystyle \frac{1}{3}\) 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.
 
Rujaxso said:
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.
Click to expand...
I am still concerned that something may be missing from this problem, a concern that explains my drastic editing of my first post.

Some teachers advocate "clearing fractions" before doing anything else. To do so, multiply BOTH sides of the equation by the product of ALL the denominators in the equation. That is in fact the quickest way to eliminate all fractions from your equation, but the resulting equation may be very complex. So I for one do not typically advocate that beginners do that.

Other teachers advocate finding the least common multiple of ALL the denominators in the equation and only then multiplying BOTH sides of the equation by that least common multiple. Multiplying by the least common multiple of the denominators may result in a considerably less complex equation than multiplying by the product of the denominators.

Still other teachers advocate simplifying each fraction before finding the least common multiple and multiplying. That process takes the longest to eliminate the fractions but gives the least complex equation without fractions.

ALL THREE METHODS WORK. They differ in how much preliminary work is involved before getting rid of the fractions and how complex the equation without fractions will be.

About simplifying fractions. If the numerator AS A WHOLE and the denominator AS A WHOLE share a common factor, you can factor them both and then cancel that common factor. Such cancelling does NOT apply to individual terms being added or subtracted in the numerator and denominator. To see why, experiment with some numbers.
 
JeffM said:
I am still concerned that something may be missing from this problem, a concern that explains my drastic editing of my first post.

Some teachers advocate "clearing fractions" before doing anything else. To do so, multiply BOTH sides of the equation by the product of ALL the denominators in the equation. That is in fact the quickest way to eliminate all fractions from your equation, but the resulting equation may be very complex. So I for one do not typically advocate that beginners do that.

Other teachers advocate finding the least common multiple of ALL the denominators in the equation and only then multiplying BOTH sides of the equation by that least common multiple. Multiplying by the least common multiple of the denominators may result in a considerably less complex equation than multiplying by the product of the denominators.

Still other teachers advocate simplifying each fraction before finding the least common multiple and multiplying. That process takes the longest to eliminate the fractions but gives the least complex equation without fractions.

ALL THREE METHODS WORK. They differ in how much preliminary work is involved before getting rid of the fractions and how complex the equation without fractions will be.

About simplifying fractions. If the numerator AS A WHOLE and the denominator AS A WHOLE share a common factor, you can factor them both and then cancel that common factor. Such cancelling does NOT apply to individual terms being added or subtracted in the numerator and denominator. To see why, experiment with some numbers.
Click to expand...

Door number 2 sounds like what we have been doing.
 
(3y) / (3(y+3))

So this is reducible because 3 times a/the quantity Y can cancel 3 times a/the quantity (y+3)

since both quantities are being multiplied.
 
background: I'm taking an evening 6:00 to 8:00 math class at my local CC.
My teacher isn't necessarily all bad in my opinion but he seems to rush things and since we meet only
once a week i guess its somewhat expected. He has given us formulas and straight up told us he doesn't have time to explain it right now.

So thanks for your help guys, I really want to know concepts not just memorize formulas and that sort of mentality, especially since I want to get into computer science.
 
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Rujaxso said:
(3y) / (3(y+3))

So this is reducible because 3 times a/the quantity Y can cancel 3 times a/the quantity (y+3)

since both quantities are being multiplied.
Click to expand...
It's the 3s that cancel, and, yes, they cancel one another because they are each factors. (3/3 is just 1).

You can simplify (8y+16)/(2y+6) also, by factoring and reducing.
 
mmm4444bot said:
It's the 3s that cancel, and, yes, they cancel one another because they are each factors. (3/3 is just 1).

You can simplify (8y+16)/(2y+6) also, by factoring and reducing.
Click to expand...




Getting right down in the weeds now, k.


So Y+16 can be seen as its own entity where as 8Y could bee seen as two separate entities?


Also, how does one go about writing math out on here that looks cleaner?
 
Rujaxso said:
So Y+16 can be seen as its own entity where as 8Y could bee seen as two separate entities?
Click to expand...
I'm not sure what you're thinking. In the expression 8y, 8 and y are factors. The only way to separate them, in the expression 8y+16, is to factor the entire expression.

8y + 16 = 8(y + 2)

So y + 2 can be seen as one factor, and 8 as another.

Earlier, you factored the denominator as 2(y + 3).

Now you can reduce the factor 8/2.

PS: Please don't switch back and forth between using symbols y and Y.
 
mmm4444bot said:
I'm not sure what you're thinking. In the expression 8y, 8 and y are factors. The only way to separate them, in the expression 8y+16, is to factor the entire expression.

8y + 16 = 8(y + 2)

So y + 2 can be seen as one factor, and 8 as another.

Earlier, you factored the denominator as 2(y + 3).

Now you can reduce the factor 8/2.

PS: Please don't switch back and forth between using symbols y and Y.
Click to expand...


Sorry about y and Y please don't hit me with ruler, k.

I should not have made an example out of those since I'm not referring to that specific problem.

What I meant was that if addition and subtraction are involved in a term you cant cancel unless its the same term.
If multiplication and division are involved it appears as if the numerical part is separate?


I'm gonna start working problems sets on these so should I post further questions here or in beginning algebra?
 
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