solve the rational equation. check your answer. 3/ m-4 + 1/3 (m - 4) = 6/m
factorise on left hand side and cross multiply
I don't know how to factor
this my teach u how to factor: If u still need help, tag me http://www.purplemath.com/modules/solvrtnl.htm
\[\large \frac{ 3 }{ m-4 }+\frac{ 1 }{ 3(m-4) }=6m\]\[\large \frac{ 3 }{ 3 }*\frac{ 3 }{ (m-4) }+\frac{ 1 }{ 3(m-4) }=6m\] you multiply both 3 on the top and bottom of the first fraction because the LCD = 3(m-4) So you get \[\large \frac{ 3(3) }{ 3(m-4) }+\frac{ 1 }{ 3(m-4) }=6m\]Now you've got your LCD are you able to simplify the rest?
From the beginning multiply both sides by 3( m-4) in order to eliminate fractions quickly and in one
No, I don't really understand any of it
What part u didn't get?
I know my answer needs to be 9, but i don't know howto show my work.
jhanny was on the right track, what part u didn't get from it?
I understand what she has written so far I just don't know what to do after that.
so attention on the right side wann being there 6 divide by m and not 6 time m
From what I have posted,simplify the expression. \[\large \frac{ 9 }{ 3m-12 }+\frac{ 1 }{ 3m-12 }\] since we have a common denominator, we can simplify further, \[\large \frac{ 10 }{ 3m-12 }=6m\] Now we want all the m's on one side. s we can "cross multiply" \[\large \frac{ 10 }{ 3m-12 }=\frac{ 6m }{ 1}\]\[\large 10=6m(3m-12)\] can you carry on from here?
I only know how to do the first step. I can try if you tell me what I need to do next.
@Jhannybean on the right side wann being 6/m and not 6m
Thank you @jhonyy9 it was a typo :( Missed that!
was my pleasure but ATTENTION
So, what do I need to do next?
@Jhannybean help for medal please
Are you asking for one? haha
I will give a medal to whoever helps finish the problem
@jhonyy9 and @Jhannybean
so we have \[\large \frac{ 10 }{ 3m-12 }=\frac{ 6 }{ m }\]\[\large 10m=6(3m-12)\]\[\large 10m = 18m-72\]\[\large 72=18m-10m\]\[\large 72=8m\] and now what is \[\frac{ 72 }{ 8 }\]? You will have your answer.
9, Thank you very much.
No problem
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