7. Solve for   x   and check.

First, clear denominators. Multiply both sides by   x + b.

x + a = c(x + b)

Remove parentheses

x + a = cx + bc

Transpose all terms that involve   x   to one side and the terms that do not to the other side.

x - cx = bc - a

Factor out the   x

x(1 - c) = bc - a

Divide by the coefficient of the unknown

c   cannot be equal to one or else you will have a zero denominator in the solution. If   c = 1,  then the top of the fraction and the bottom have to be the same. That will happen only if   a  =  b,   in which case any value of   x   will be a solution. Otherwise, there will be no value of   x   which will work

Check. Copy down the original equation except where you see an   x,  copy down the solution in parentheses.

On the left we have a compound fraction. Add the fractions on both the top and bottom in order to prepare for inverting and multiplying. The common denominator is   1 - c.

After we add the fractions

we find some like terms that cancel.

We are now ready to invert and multiply. Let us factor out the c from the t wo terns in the top of the top.

After we cancel we get

c = c

Note that in the check, in addition to the requirement that   c   is not equal to  1,  there is another occasion to get zero denominators in the check. If   a = b   there will be a zero denominator in the check. If   a = b,   then the fraction will always reduce to  1,  and so any value of  x   would work if c were 1, but no value would work if &mbsp; c   were not  1.  In either case, there is not a unique solution, and this is manifested by the problems with zero denominators.

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