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)?
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
???
"Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.
I suspect that you violated the order of operations in giving your problem.
Last edited by JeffM; 11-08-2017 at 11:07 PM.
Not sure that I follow. Do you mean I should reduce the coefficients after factoring what could be factored?
Just so you know, you're missing some very very important grouping symbols. What you wrote literally means:
[tex]\dfrac{3y}{3y}+9 + 8y+\dfrac{16}{2y}+6 = y-\dfrac{4}{y}+3[/tex]
But what I think you meant was:
[tex]\dfrac{3y}{3y+9} + \dfrac{8y+16}{2y+6} = \dfrac{y-4}{y+3}[/tex]
Which would be type-set as: (3y)/(3y + 9) + (8y+16)/(2y + 6) = (y-4)/(y + 3)
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?"
Last edited by mmm4444bot; 11-09-2017 at 03:03 AM.
"English is the most ambiguous language in the world." ~ Yours Truly, 1969
Do you think you could reduce [tex]\frac{2}{2+3}[/tex] and get [tex]\frac{1}{3}[/tex] 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.
“... mathematics is only the art of saying the same thing in different words” - B. Russell
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