# algebra 2 fraction and linear equations

(1/8)(3m - 5) = (6/5)(6m - 2) + 5...Type your answer as a fraction. Do not use decimals!

**i got -129/273 is this correct ?

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What's m? $\quad$ – J. M. Dec 22 '10 at 6:40
Is 129/273 correct for what? Is there supposed to be an equation? Note that 1/8(3m-5) is ambiguous. Is it $\frac{1}{8(3m-5)}$ or $\frac{1}{8}(3m-5)$? – Jonas Meyer Dec 22 '10 at 6:40
the answer... no equation you use LCD so i believe its set up as: 40/1*1/8(3m-5)=40/1*6/5(6m-2)*40/1*5 then you use the distributive property after getting rid of the fractions i think – niki Dec 22 '10 at 6:42
@niki: This doesn't make sense. You say there is no equation, but there are equations in your comments. If the first $+$ in your question is supposed to be $=$ as in your comments, then the question would make sense, but your work would be incorrect due to the second $+$ becoming multiplication and an apparent incorrect distribution of multiplication: $a(bc)\neq(ab)(ac)$. Your fractions are still ambiguous; please add parentheses or use LaTeX to clarify. Please clarify your question. – Jonas Meyer Dec 22 '10 at 6:54
@niki: You can edit your question to correct it. If you wouldn't mind, it would also be nice to use LaTeX formatting to make the math clearer. Some guidance on this can be found at meta.math.stackexchange.com/questions/107/…. Otherwise, could you please add parentheses to make your fractions clearer? E.g., 1/2(x+1) is ambiguous, but (1/2)(x+1) and 1/(2(x+1)) are not. – Jonas Meyer Dec 22 '10 at 7:10

Assuming you mean: $\frac{1}{8}(3m - 5) = \frac{6}{5}(6m - 2) + 5$
You would solve as:
$$40 * \frac{1}{8}(3m - 5) = 40 * (\frac{6}{5}(6m - 2) + 5)$$ $$5 * (3m - 5) = (48 * (6m - 2)) + 200$$ $$15m - 25 = 288m - 96 + 200$$ $$15m - 25 = 288m + 104$$ $$-129 = 273m$$ $$m = -\frac{129}{273} = -\frac{43}{91}$$

Assuming you mean $\frac{1}{8(3m - 5)} = \frac{6}{5(6m - 2)} + 5$
You would solve as:
$$\frac{1}{24m - 40} = \frac{6}{30m - 10} + 5$$ $$(30m - 10)(24m - 40) * \frac{1}{24m - 40} = (30m - 10)(24m - 40) * (\frac{6}{30m - 10} + 5)$$ $$30m - 10 = 24m - 40 + (5*(30m - 10)(24m - 40))$$ $$30m - 10 = 24m - 40 + (5 * (720m^2 - 1140m + 400))$$ $$30m - 10 = 24m - 40 + 3600m^2 - 7200m + 2000$$ $$0 = 3600m^2 - 7206m + 1970$$ $$m = \frac{1201 \pm \sqrt{654401}}{1200}$$

(I would double check the work on both of these though, I'm tired =/)

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i did it the first way and got the same answer but the program im doing it on is saying im correct so i wanted to double check and a few other questions but i think it may just be the program considering we both came to the same conclusion; thank you for your help tho.. – niki Dec 22 '10 at 7:38
@niki: Just to double check, when you are entering it, are you entering it as a negative (as your question statement has your answer as a positive). Also, you could try reducing it (as I did above) and seeing if that makes a difference. (Personally, I hate auto graders because they always seem to miss some nuance of how the correct answer could be represented). – Reese Moore Dec 22 '10 at 7:40
i think i should have reduced ; it accepted the reduced fraction thank you sooo much ! – niki Dec 22 '10 at 7:48
Which is why I usually hate auto grading mechanisms, there is nothing inherently incorrect about $m = -\frac{129}{273}$, but because it wasn't reduced the program decided it wasn't correct. Unless it is explicitly stated that the goal of the assignment is to reduce fractions, the machine should simply check to see if the solution entered is a valid solution to the problem (IMO, some would disagree with me...) – Reese Moore Dec 22 '10 at 7:51
@ Reese Moore i agree, it never told me to reduce but to only write it in fraction form. – niki Dec 22 '10 at 8:03