Incorrect in solving a b − a c = 1 for cI have this: a b − a c = 1Solve for c. Then, a b − a c = 1 ⋅ b c = a c − a b = b c = a ( c − b ) = b c c = b c a + bThis is my final result.But the correct result is: c = a b a − b What I development wrong in this equation ?

Question

Incorrect in solving       a    b    −      a    c    =  1 for cI have this:      a    b    −      a    c    =  1Solve for c. Then,      a    b    −      a    c    =  1  ⋅  b  c  =  a  c  −  a  b  =  b  c  =  a  (  c  −  b  )  =  b  c  c  =            b      c        a    +  bThis is my final result.But the correct result is:  c  =            a      b              a      −      b      What I development wrong in this equation ?

Andrejkoxg · Accepted Answer

To isolate c we can proceed as follow      a    b    −      a    c    =  1    ⟺        a    c    =      a    b    −  1  =            a      −      b        b      ⟺        c    a    =      b          a      −      b          ⟺    c  =            a      b              a      −      b

Paul Reilly · Accepted Answer

You have expressed the solution to c in terms of c, where you need to have it in terms of a and b. This is what gives you a different(although correct) answer.From   a  c  −  a  b  =  b  c, you should have taken the variables with   c to one side, like so:  a  c  −  b  c  =  a  b    ⟹    c  (  a  −  b  )  =  a  bCan you continue from here?

Incorrect in solving a/b−a/c=1 for c

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