Reducing fractions
Test your ability in reducing fractions!
Test your ability in reducing fractions!
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Flashcards
Before I reduce a fraction...
If I want to add up \(\frac{1}{4} + \frac{1}{5}\) , I have to obtain a common denominator. How do I do that?
Can we reduce the fraction
\(\frac{2-x-x^2}{x-1}\)
and if yes, what is the result?
In the numerator, an expression can be factored out. To see that, it helps to reorder the terms in the more common order (starting with the highest power of x). The signs have to be considered carefully.
\(2-x-x^2 = -x^2-x+2 = (-x+1)(x+2)\)
Therefore, we may continue to simplify:
\(\frac{(-x+1)(x+2)}{x-1} = \frac{(-1)(x-1)(x+2)}{x-1} = (-1)(x+2) = -x-2\)
Can we reduce the fraction
\(\frac{x^2-b^2}{x+b}\)
and if yes, what is the result?
\(\frac{x^2-b^2}{x+b} = \frac{(x-b)(x+b)}{x+b} = x-b\)
Yes! The numerator is a binomial formula of the third kind. Therefore, it can be factorized and thereafter the fraction can be reduced.
Can we reduce the fraction
\(\frac{x^2+b^2}{x+b}\)
and if yes, what is the result?
\(\frac{x^2+b^2}{x+b}\) cannot be reduced! The numerator is no binomial formula, and nothing can be factored out. Therefore, this fraction of sums cannot be reduced.
Right or wrong?
The fraction \(\frac{4}{4+y}\) cannot be reduced.
Right!
There is a sum in the denominator. Let us assume y=3, for example (as y represents a number that we can choose). Then the fraction becomes \(\frac{4}{4+3}\)
Here, it is obvious that the fraction cannot be reduced and the numbers in the denominator must simply be added up.
This is the only thing we can do; however, we do not know the value of y, therefore we must leave the denominator 4+y as it stands.
Right or wrong?
\(\frac{4a+3}{4a}=\frac{1+3}{1}=4\)
Terribly wrong!
There is a sum in the numerator, from which no parts can be reduced.
To show that, let us assume a=1 (a could be any number, so a could be 1). Then, our claim becomes:
\(\frac{4+3}{4}=\frac{1+3}{1}=4\)
which is obviously false.
There is no way around it: both numbers of the sum in the numerator must be divided by the denomiator.
Can I reduce this fraction?
\(\frac{ab^3+c^2}{ac^3}\)
No!
There is a sum in the numerator, from which nothing can be factored out.
Can I reduce this fraction?
\(\frac{ab^3c^2}{ac^3}\)
Yes, both numerator and denominator are only multiplications. The solution is:
\(\frac{b^3}{c}\)
