Write each fraction in terms of the LCM of the denominators. a) 3X/(〖2X〗^2- X-10)b) (-2X)/(〖2X〗^2-15X+25)

[cartier rick]

LCM fraction question




I just cant seem to work out these 2 questions


Write each fraction in terms of the LCM of the denominators.


a)
3X/(〖2X〗^2- X-10)


b)
(-2X)/(〖2X〗^2-15X+25)

 

[Steven G]

It seem that each fraction has the LCM already.

Are the two fractions [math]\dfrac {3x}{(2x)^2 - x - 10} \ and \ \dfrac {-2x}{(2x)^2 -15x + 25}[/math]?

Maybe you are just to compute (2x)² as 4x²?

Also please follow the posting guidelines by showing us your attempt at this problem so we know what type of help you need.

 

[cartier rick]

It seem that each fraction has the LCM already.

Are the two fractions [math]\dfrac {3x}{(2x)^2 - x - 10} \ and \ \dfrac {-2x}{(2x)^2 -15x + 25}[/math]?

Maybe you are just to compute (2x)² as 4x²?

Also please follow the posting guidelines by showing us your attempt at this problem so we know what type of help you need.

sorry yeah those are the equations for the first one with 3x I tired answering with this the two questions with red x marks on them

 

[Dr.Peterson]

LCM fraction question

I just cant seem to work out these 2 questions

Write each fraction in terms of the LCM of the denominators.
a)
3X/(〖2X〗^2- X-10)

b)
(-2X)/(〖2X〗^2-15X+25)

Seeing the actual problem helps a lot; you added parentheses (or whatever those are) where they don't belong:



For the first, you just multiplied the numerator by 4/3; why? You appear to have factored correctly and found the LCM, so what you should be doing is multiplying both numerator and denominator by the same "extra" factor.

For the second, you show the correct LCM, but you did something odd to the numerator again. You should have multiplied it by the same factor you multiplied the denominator by, namely (x+2).

Is there a reason some numbers are in red?