Solution 2.2:3b

From Förberedande kurs i matematik 1

First, we move all the terms over to the left-hand side:

4x4x−7−12x−3−1=0

Then, we multiply the top and bottom of all three terms by appropriate factors so that they have the same common denominator, in the following way,

4x4x−72x−32x−3−12x−34x−74x−7−2x−34x−72x−34x−7=0 

and so that we can rewrite the left-hand side giving

2x−34x−74x2x−3−4x−7−2x−34x−7=0 

We expand the numerator

2x−34x−78x2−12x−4x−7−8x2−14x−12x+21=0 

and simplify

10x−142x−34x−7=0 

This equation is satisfied when the numerator is zero (provided the denominator is not also zero) and this happens when

10x−14=0

which gives x=75.

It can easily happen that we calculate incorrectly, so we must check that the answer x=75 satisfies the equation:

LHS    =    457457−7−1257−3 = multiply top and bottom by 5    = 457457−755−1257−355=4747−75−527−35=44−5−514−15=−4−−5=1    = RHS