Question

Can someone please help me with how to verify my solution in an equation?

Answer 1

Question: \[\frac{ 2m + 1 }{ 3 } = \frac{ 3m - 2 }{ 5 }\] (LCD = 15)

My answer: \[\frac{ 2m + 1 }{ 3 } = \frac{ 3m - 2 }{ 5 }\]\[15\left(\begin{matrix}\frac{ 2m + 1 }{ 3 } \\\end{matrix}\right) = 15\left(\begin{matrix}\frac{ 3m - 2 }{ 5 } \\\end{matrix}\right)\]\[^51̶5̶\left(\begin{matrix}\frac{ 2m + 1 }{ ^13̶ } \\\end{matrix}\right) = ^31̶5̶\left(\begin{matrix}\frac{ 3m - 2 }{ ^15̶ } \\\end{matrix}\right)\]\[5(2m + 1) = 3 (3m - 2)\]\[10m + 5 = 9m - 6\]\[10m - 9m + 5 = 9m - 9m - 6\]\[1m + 5 = -6\]\[1m + 5 - 5 = - 6 - 5\]\[m = -11\]
I'm not entirely positive that my answer is correct (I came up with a few others before but this one seemed the most accurate since I forgot to group like terms together in previous solutions I tried).
Any way, if this answer is correct -- how do I "check" my answer? As in, calculating LS = RS to prove that the solution is correct. Everything that I try doesn't end up working out, and I feel very stuck with this. All I have come up with is the first two lines: \[LS = \frac{ 2m + 1 }{ 3 }...........................RS = \frac{ 3m - 2 }{ 5 }\]\[ = \frac{ 2(-11) + 1 }{ 3 }........................... = \frac{ 3(-11) - 2 }{ 5 }\]
Everything always goes downhill from there... if anyone could help me out with this, I would really appreciate it! Thank you very much!!

Answer 2

I very appreciate your effort in typing out the question, so here's a medal :D

Answer 3

You could go like this:
\[\hspace{11pt}\mbox{L.H.S}\]\[=\frac{2m+1}3\]\[=\frac{2(-11)+1}3\]\[=\frac{-22+1}3\]\[=-\frac{21}3\]\[=-7\]
\[\hspace{11pt}\mbox{R.H.S}\]\[=\frac{3m-2}5\]\[=\frac{3(-11)-2}5\]\[=\frac{-33-2}5\]\[=-\frac{35}5\]\[=-7\]
\[\because \mbox{L.H.S.=R.H.S.}\]\[\therefore\mbox{The equation is valid.}\]

Answer 4

Ah! Wow! Thank you very much!! I guess I see where I went wrong: (grouping the (-11) with the standalone numbers right away and attempting to multiply each individually). I really thank you for clearing this up for me, it has been a very big help and I'm glad to be able to understand this now!

I will also give you a medal for this. Once again, thank you very much for all your help and for showing all the steps!!

Answer 5

I red from a website that you cannot start with a hypothetically true statement and then prove that it's right, because it's like doing the following.
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Assume that the statement \(1=2\) is valid.
\[1=2\]\[2=1\]\[1+2=2+1\]\[3=3\]\[\therefore 1=2\]

Answer 6

sadly i cannot find that website again