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Solution 2.2:3b

From Förberedande kurs i matematik 1

Revision as of 07:40, 24 September 2008 by Tek (Talk | contribs)
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First, we move all the terms over to the left-hand side,

4x4x712x31=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,

4x4x72x32x312x34x74x7(2x3)(4x7)(2x3)(4x7)=0

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

(2x3)(4x7)4x(2x3)(4x7)(2x3)(4x7)=0.

We expand the numerator

(2x3)(4x7)8x212x(4x7)(8x214x12x+21)=0

and simplify

10x14(2x3)(4x7)=0.

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

10x14=0

which gives x=75.

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

LHS =457457712573=multiply top and bottom by 5=4574577551257355=47477552735=44551415=4(5)=1=RHS.