Hi people, I have a question, ive looked through forums no luck, teacher has explained how to solve it but it just doesn't click into my mind. Here it is 45.26-2(Idc*0.1)=20Idc Just got to solve for Idc. I have an exam tommorow ive studied everything just stuck on this. Help much appreciated. Best regards, Anthony
There is something missing here. Idc is something like "x"? 1st degree equation is Ax+B and result must have some "B". Your result has no "B". This can be done only if B comes to zero somehow. But in your equation B =45.26 and A=0.2 (2*0.1). If your equation is correct then it must be: 45.26-2(Idc*0.1)=20Idc => 45.26-2(Idc*0.1)-20Idc = 0 => 45.26-(2*Idc*0.1)-20Idc =0 => 45.26- Idc(0.2+20)=0 => 45.26- 20.2*Idc=0 => 45.26=20.2*Idc So, in order this equation to be true, it must be Idc=45.26/20.2 otherwise something is missing
There is nothing missing. The variable in an equation doesn't have to be x. In this case, Idc is the variable, and probably represents DC current (I is normally used for current, and DC is probably a subscript in the original problem). No. A first degree equation, like all equations, shows that an expression on one side of the equals sign is equal to the expression on the other side. Ax + B is only an expression, not an equation. Ax + B = 0 is an equation. This makes no sense. The work above is correct, but the numerical answer could be gotten with fewer steps. 45.26-2(Idc*0.1)=20Idc => 45.26 -0.2*Idc = 20*Idc => 45.26 = 20.2*Idc => Idc = 45.26/20.2 2.24
Hi guys thankyou very much for that. I got the same answer however the teacher and some students had something else. They had an answer in Milli amps. I do apologise I realised I posted in the wrong forum. Regards, Anthony
You didn't give any units in the problem you posted, so without them, it's not possible to provide them with the numerical answer. This looked like a homework problem, so this was probably the right forum. Mark
Mark, i am working with some 3d lines right now and i was taken over. Of course i can solve a 1st degree equation and ofcourse i know that ax+b=0. I did not take Idc as dc current. The whole form reminded me of something that i am working right now. That's why i sid all these stuff before. And at the end, i told him that if this is the corect equation, i gave him the solution for 1st degree. I gave it in more simple way (with more steps) but i can give it in faster way. I supposed that if he wants the solution for 1st degree, he must be new to mathematics thus a simple solution is better. So next time, prior to your answer, do not haste to say "NO" and "NO" and "NO". We need to know the units of Idc. Is Idc in mAmps?
??? It's clear to me that you can solve a 1st degree equation. But ax + b is not an equation, as you said, and as I pointed out. And another of your statements made no sense to me, namely that "Your result has no "B". This can be done only if B comes to zero somehow." Your answer was correct, although not simplified (i.e., you ended up 45.26/20.2). As new to mathematics as the OP seems to be, you should have given the answer in its simplest form. If there is anything I said that is incorrect, please let me know. Otherwise I stand behind everything I said. Mark
Get some more knowledge about lines and 3d and you will see what i mean. Untilt then you can... ...stand behind