ab-c=ae+f
Solve for a
Please solve and explain this. I can't understand the concept!
Hi Curry
sorry, your question got flagged automatically, I am not sure why.
Probably the program thought those letters in your equation looked too random and it was flagged as a potential rubbish question. These questions viewed later by a moderator and shown.
ab-c=ae+f solve for a
The idea is to get a by itself in fron to the equal sign.
First get every term WITH an a on the left and everything eles on the right.
\(ab-c=ae+f\\ \text{subtract ae from both sides}\\ ab-c-ae=ae+f-ae\\ ab-c-ae=f\\ \text{Now add c to both sides}\\ ab-c-ae+c=f+c\\ ab-ae=f+c\\ \text{Now factor out the a on the left side}\\ a(b-e)=f+c\\ \text{Now divide both sides by (b-e)}\\ \frac{a(b-e}{b-e}=\frac{f+c}{b-e}\\ a=\frac{f+c}{b-e}\\ \\~\\ \text{You cannot divide by 0 so you should put }b-e\ne0\\\ \text{which is the same as } b\ne e\)
Hi Curry
sorry, your question got flagged automatically, I am not sure why.
Probably the program thought those letters in your equation looked too random and it was flagged as a potential rubbish question. These questions viewed later by a moderator and shown.
ab-c=ae+f solve for a
The idea is to get a by itself in fron to the equal sign.
First get every term WITH an a on the left and everything eles on the right.
\(ab-c=ae+f\\ \text{subtract ae from both sides}\\ ab-c-ae=ae+f-ae\\ ab-c-ae=f\\ \text{Now add c to both sides}\\ ab-c-ae+c=f+c\\ ab-ae=f+c\\ \text{Now factor out the a on the left side}\\ a(b-e)=f+c\\ \text{Now divide both sides by (b-e)}\\ \frac{a(b-e}{b-e}=\frac{f+c}{b-e}\\ a=\frac{f+c}{b-e}\\ \\~\\ \text{You cannot divide by 0 so you should put }b-e\ne0\\\ \text{which is the same as } b\ne e\)
Sorry I'm saying this a little late, but thank you. This really helped, and now I understand the concept