given 12a-3(a+1)=a+1, what is the value of a?

Then Dr Khan advised me to go with pen and paper and lev also agreed and Otis too. That is what I reposted it again. I still do not understand why Symbolab came with a-3 on both sides. I still don't, but I do not need it because when I did it on paper it panned out clearly.
The only thing I did not understand was why the use of a-3 to both sides. On that lev was right. Why put it in a solver if I was not getting the way the solver used. Now, I totally see what lev meant. And he is right.
 
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eddy2017 said:
Then Dr Khan advised me to go with pen and paper and lev also agreed and Otis too. That is what I reposted it again. I still do not understand why Symbolab came with a-3 on both sides. I still don't, but I do not need it because when I did it on paper it panned out clearly.
The only thing I did not understand was why the use of a-3 to both sides. On that lev was right. Why put it in a solver if I was not getting the way the solver used. Now, I totally see what lev meant. And he is right.
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We have: 9a-3=a+1
To solve it we want to combine terms with the variable on the left and constant terms on the right. To do this we add 3 to both sides AND we subtract a from both sides: 3-a. That's where it comes from.
 
Thank you so much, Mr lev. It is always clear as a bell when explained. I failed to see it, though. That happens to me a lot, and, then, knowing you are out there makes the difference for me. Thank you.
I'll take Dr Khan's advice. I'll always do it on paper first, and then, if in doubt put it into a solver.
 
The only thing I WANT YOU ALL TO KNOW IS THAT I want to learn. Nothing I said or write is to hurt or bad-mouth or mislead anyone. If you see I am not following something because I ask please, please, be patient. I love this site. there are amazing tutors here with a depth of knowledge that is really unfathomable. The thirst for knowledge I have is unquenchable. But I need to study on my own. too late for me to go to school and for a Math career. I can't. I only have you and some other resources I am studying and plenty of books now I have bought.
 
[imath]\begin{gathered} 12a - 3(a + 1) = (a + 1) \\ 12a - 4(a + 1) = 0 \\ 12a - 4a - 4 = 0 \\ 8a - 4 = 0 \\ a = \frac{1}{2} \\ \end{gathered} [/imath]
 
Thank you so much, pka. It is so good to see your post!!. Thank you. Neat as always!
 
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