Converting Mixed Numbers to Improper Fractions

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To convert mixed numbers to improper fractions, ensure both fractions have the same denominator. For example, 1/2 + 1 can be expressed as 1/2 + 2/2, resulting in 3/2. When adding fractions with different denominators, find the least common denominator (LCD), which for 1/2 and 1/3 is 6. Rewrite each fraction using the LCD, then add the numerators. This method effectively combines fractions into a single improper fraction.
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Homework Statement



If i have 1/2 + 1 = 1,5 (on the calculator), how do i make this into a fraction?

I know this answer would be 1/2 + 1 = 3/2

But how excactly is this done?


Homework Equations





The Attempt at a Solution

 
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To add fractions, both fractions have to have the same denominator.

For your problem, 1/2 + 1 = 1/2 + 2/2 = 3/2.

Here's a slightly harder problem: 1/2 + 1/3

The least common denominator is the least common multiple of 2 and 3, which is 6. So the common denominator will be 6. In a fraction, I can get the common denominator by multiplying by 1 in a suitable form. After the fractions have been rewritten using the common denominator, add the numerators and use the common denominator.

\frac{1}{2} + \frac{1}{3} = \frac{1}{2} * \frac{3}{3} + \frac{1}{3} * \frac{2}{2}
=\frac{3}{6} + \frac{2}{6} = \frac{5}{6}
 
Thank you ! :)
 
Well, you have one fraction (1/2) and a whole number 1 i.e. (2/2).
To make this into one fraction you just have to add the numerators together because the denominator is the same on both sides, 1/2 + 2/2 = (1+2)/2 or 3/2. hope that helps.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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