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avatar+46 

ab-c=ae+f

 

Solve for a

 

Please solve and explain this. I can't understand the concept!

CurryCrafterMaster  Sep 28, 2018

Best Answer 

 #2
avatar+94118 
+3

Hi Curry   laugh

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.

 

 

\(ab-c=ae+f\\ \text{subtract ae from both sides}\\ ab-c-ae=ae+f-ae\\ ab-c-ae=f\\ \text{Now add c to both sides}\\ ab-c-ae+c=f+c\\ ab-ae=f+c\\ \text{Now factor out the a on the left side}\\ a(b-e)=f+c\\ \text{Now divide both sides by (b-e)}\\ \frac{a(b-e}{b-e}=\frac{f+c}{b-e}\\ a=\frac{f+c}{b-e}\\ \\~\\ \text{You cannot divide by 0 so you should put }b-e\ne0\\\ \text{which is the same as } b\ne e\)

Melody  Sep 28, 2018
 #1
avatar+46 
+1

why is this flagged?

CurryCrafterMaster  Sep 28, 2018
 #2
avatar+94118 
+3
Best Answer

Hi Curry   laugh

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.

 

 

\(ab-c=ae+f\\ \text{subtract ae from both sides}\\ ab-c-ae=ae+f-ae\\ ab-c-ae=f\\ \text{Now add c to both sides}\\ ab-c-ae+c=f+c\\ ab-ae=f+c\\ \text{Now factor out the a on the left side}\\ a(b-e)=f+c\\ \text{Now divide both sides by (b-e)}\\ \frac{a(b-e}{b-e}=\frac{f+c}{b-e}\\ a=\frac{f+c}{b-e}\\ \\~\\ \text{You cannot divide by 0 so you should put }b-e\ne0\\\ \text{which is the same as } b\ne e\)

Melody  Sep 28, 2018
 #3
avatar+46 
0

Sorry I'm saying this a little late, but thank you. This really helped, and now I understand the concept

CurryCrafterMaster  Oct 2, 2018

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