mathcalc: Here;s the problem:
1. 12n+7=8n+15
The second problem is:
2m-3=m
Hi Jyoti,
How much algebra have you done. This seems like an odd place to start.
Let's look at the first one.
1. 12n+7=8n+15
The idea is to get all the lots of n on one side and all the numbers without n's on the other side.
However, you have to think of it the other way around.
You say to yourself - What do I want to get rid of. (There is always more than one way that they can be done)
I have decided that I want to get rid of the 8n from the Right Hand side (RHS)
I can do that by taking it away. BUT if I take it away from the RHS then i have to take it away from the LHS as well.
12n + 7 = 8n + 15
12n + 7 - 8n = 8n +15 - 8n
12n - 8n +7 = 8n-8n+15
4n + 7 = 15
Now I want to get rid of the 7 so I will subtract it from both sides.
4n+7-7 = 15-7
4n = 8
4 * n =8 (4n means 4*n)
Now i want to get rid of the times 4 The opposite of timesing by four is dividing by 4 so, divide both sides by 4
4n/4 = 8/4
n = 2
Now the second one
2m-3=m
If you understand negative numbers then you would subtract 2m from both sides. HOWEVER, I am going to assume that you
do not understand negative numbers.
So
Start by adding 3 to both sides,
then, subtract m from both sides,
then, think about what you need to do to finish it.
It has only just occured to me that maybe you just supposed to 'guess' the answers. This is a valid way as well. I have shown you how to do it formally. When the questions get hard you will have to do them formally so it wont hurt you to try and understand now.
+118609 
