Answer:
Step-by-step explanation:
As with any equation involving fractions, you can multiply the equation by the least common denominator to eliminate fractions. Then solve in the usual way.
c) 1/a +b = c
1 +ab = ac . . . . multiply by a
1 = ac -ab . . . . subtract ab
1 = a(c -b) . . . . . factor out a
1/(c -b) = a . . . . divide by the coefficient of a
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d) (a-b)/(b-a) = 1
a -b = b -a . . . . . multiply by b-a
2a = 2b . . . . . . . . add a+b
a = b . . . . . . . . . . divide by the coefficient of a
Please be aware that this makes the original equation become 0/0 = 1. This is why a=b is not an allowed condition for this equation. As written, it reduces to -1 = 1, which is false. One could say there is no solution.
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f) bc +ac = ab . . . . . multiply by abc
bc = ab -ac . . . . . . subtract ac
bc = a(b -c) . . . . . . factor out a
bc/(b -c) = a . . . . . . divide by the coefficient of a
