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Solution 2.1:1d

From Förberedande kurs i matematik 1

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Revision as of 08:31, 9 September 2008

After x3y2 are multiplied inside the bracket, we can eliminate factors which occur in both the numerator and denominator.

x3y2y1−1xy+1=x3y2y1−x3y21xy+x3y21=yx3y2−xyx3y2+x3y2=x3y−x2y+x3y2

where we have used

yx3y2=yx3yy=x3y,

xyx3y2=xyxxxyy=xxy=x2y

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