Question

Hi! Could someone please help me out with understanding two algebraic equation examples? They don't need to be solved -- they are examples given in my book with the question and answers, but I'm having a hard time understanding a couple of lines in them (including the verifications). Sorry if the answers are obvious, I was up a little too late last night but I can't have enough time to e-mail a support teacher.
Here's the first example (I'll post it in a minute):

Answer 1

Question #1: \[\frac{ 2c }{ 3 } + 2 = \frac{ 1 }{ 2 }\] LCD = 6

Answer: \[6\left(\begin{matrix}\frac{ 2c }{ 3 } \\\end{matrix}\right) + 6(2) = 6\left(\begin{matrix}\frac{ 1 }{ 2 } \\\end{matrix}\right)\]\[^26̶\left(\begin{matrix}\frac{ 2c }{ ^13̶ } \\\end{matrix}\right) + 6(2) = ^36̶ \left(\begin{matrix}\frac{ 1 }{ ^12̶ } \\\end{matrix}\right)\]\[4c + 12 = 3\]\[4c + 12 - 12 = 3 - 12\]\[4c = -9\]\[\frac{ 4c }{ 4 } = \frac{ -9 }{ 4 }\]\[c = \frac{ -9 }{ 4 }\]
Now for the verification:
\[LS = \frac{ 2c }{ 3 } + 2 ....................RS = \frac{ 1 }{ 2 }\]\[ = \frac{ 2 }{ 3 }\left(\begin{matrix}\frac{ -9 }{ 4 } \\\end{matrix}\right) + 2\]\[ = \frac{ -18 }{ 12 } + 2\]\[ = \frac{ -3 }{ 2 } + \frac{ 4 }{ 2 }\]\[ = \frac{ 1 }{ 2 }\] So LS = RS

I'm sorry for this first question for being so long (I still will post the second example after this post) but can someone answer my confusion:

In the verification, in the third line, it states -18 ÷ 12, + 2. I understand that this is from multiplying the above line (2 x 9, and 3 x 4). But how is the next line -3 ÷ 2, + 4 ÷ 2?! How did they get from 18 ÷ 12 to 3 ÷ 2, and where did the 4 come from in the "+ 4 ÷ 2" if there was only a + 2 in the previous line, and no 4? How did they end up with 1 ÷ 2 as the answer? I truly apologize if this sounds confusing... I hope you can understand. Did they reduce the fractions? If so, why...? I don't *think* I recall seeing it in previous equations in my book...

On to the next question I'll reply with! Thanks in advance!!

Answer 2

Question #2:
\[\frac{ 3 - y }{ 6 } + \frac{ y - 1 }{ 3 } = 7\] LCD = 6

Answer:
\[6\left(\begin{matrix}\frac{ 3-y }{ 6 } \\\end{matrix}\right) + 6\left(\begin{matrix}\frac{ y - 1 }{ 3 } \\\end{matrix}\right) = 6(7)\]\[^16̶\left(\begin{matrix}\frac{ 3 - y }{ ^16̶ } \\\end{matrix}\right) + ^26̶\left(\begin{matrix}\frac{ y - 1 }{ ^13̶ } \\\end{matrix}\right) = 6(7)\]\[3 - y + 2(y - 1) = 42\]\[3 - y + 2y - 2 = 42\]\[y + 1 = 42\]\[y + 1 - 1 = 42 - 1\]\[y = 41\]
Alright, now onto the verification:
\[LS = \frac{ 3 - y }{ 6 } + \frac{ y - 1 }{ 3 }........................ RS = 7\]\[ = \frac{ 3 - 41 }{ 6 } + \frac{ 41 - 1 }{ 3 }\]\[ = \frac{ -38 }{ 6 } + \frac{ 40 }{ 3 }\]\[ = \frac{ -19 }{ 3 } + \frac{ 40 }{ 3 }\]\[= \frac{ 21 }{ 3 }\]\[ = 7\]So, LS = RS

OK: I am having trouble understanding a rather big portion of this example, but I will try to explain what I don't understand.
Near the middle of the equation, right after this line: 3 - y + 2y - 2 = 42...
It starts saying: y + 1 = 42 and continues from there to find "y". Where did the 1 come from? Wouldn't it be just "y" if you are subtracting the 3 AND the 2... not "y" PLUS 1? Did the "3 - y + 2y" become 3y, and subtracting the 2, it becomes y + 1? (I thought if an answer becomes 1 it would just stay as the variable without having to add a number?)

One more problem... (Sorry this is so long) In the verification portion, similar to my first question, how did the line: -38 ÷ 6 become: -19 ÷ 3? Why were the numbers divided by 2? I don't believe the my first example I posted had to be divided... if this is the case, how will I know when I need to divide by 2 in any equations in the future?

Anyway: Thank you very much in advance for any help! Once again, I apologize if this happens to be a painfully obvious answer and if I sound out of it -- I'm not sure why I can't figure out some of these parts. Thank you for reading and sorry for writing such a ridiculously long post.

Answer 3

they r right