Processing Math: Done

From Förberedande kurs i matematik 1

(The exercise is taken from an actual exam from Spring Term 1944!)

Start by rewriting the terms on the left-hand side as one term having a common denominator:

1x−1−1x+1=1x−1x+1x+1−1x+1x−1x−1=x+1x−1x+1−x−1x−1x+1=x−1x+1x+1−x−1=2x−1x+1

If we also write 3x−3=3x−1 , the equation can be rewritten as

2x−1x+1x2+21=6x−13x−1 

Because x=1 cannot be a solution to the equation, the factor x−1 can be removed from the denominator of both sides (i.e. actually, we multiply both sides by x−1 and then eliminate it).

2x+1x2+21=36x−1 

Then, both sides are multiplied by 3 and x+1, so that we get an equation without any denominators.

6x2+21=6x−1x+1 

Expanding both sides

6x2+3=6x2+5x−1

the x2 terms cancel each other out and we obtain a first-order equation,

3=5x−1

which has the solution

x=54

We check whether we have calculated correctly by substituting x=4/5 into the original equation:

LHS=154−1−154+1542+21=1−51−1592516+21=−5−95225162+25=−9505057=−957=−319

RHS=354−3654−1=524−55512−515=5124−55112−15=19−3=−319