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How to do mix numbers
- Thread starter Honey
- Start date
I would like to know morer about fraction and mixed numbers like this 25/3 + 36/15=
You need to get a common denominator before you can add or subtract fractions. In this case, 15 would work. Multiply the fraction 25/3 by 5/5 to makes its denominator equal 15:
(25/3)(5/5) = 125/15
Now you can add the fractions. Add the numerators and keep the same denominator:
125/15 + 36/15 = 161/15
Hello, Honey!
When adding or subtracting mixed numbers,
. . I prefer to keep the whole numbers and fraction separated.
For example: \(\displaystyle \,3\frac{1}{3}\,+\,4\frac{1}{3}\)
This is: \(\displaystyle \:\begin{array}{ccc}3\frac{1}{3} \\ 4\frac{1}{3} \\ --- \end{array}\)
And you can see the answer: \(\displaystyle \,7\frac{2}{3}\;\) . . . right?
A harder example: \(\displaystyle \,2\frac{2}{3}\,+\,5\frac{1}{2}\)
This is: \(\displaystyle \:\begin{array}{ccc}2\frac{2}{3} \\ 5\frac{1}{2} \\ ---\end{array}\)
We can see: \(\displaystyle \,2\,+\,5\:=\:7\)
And we must add the fractions:
. . \(\displaystyle \frac{2}{3}\,+\,\frac{1}{2}\;=\;\frac{4}{6}\,+\,\frac{3}{6}\;=\;\frac{7}{6}\;=\;1\frac{1}{6}\)
So we have: \(\displaystyle \,7\,+\,1\frac{1}{6}\;=\;8\frac{1}{6}\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Many teachers would have you change everything to "improper" fractions.
So your boss says, "There's \(\displaystyle 10\frac{1}{4}\) pounds of coffee in one bin
. . and \(\displaystyle 20\frac{1}{4}\) pounds in another. .How much coffee do we have?"
You grab a pencil and your calculator and start working:
. . \(\displaystyle 10\frac{1}{4}\,+\,20\frac{1}{4}\:=\:\frac{41}{4}\,+\,\frac{81}{4}\:=\:\frac{122}{4}\:=\:\cdots\)
And the guy emptying wastebaskets says, "Thirty and a half."
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Subtraction gets trickier because "borrowing" is often required.
I'll let someone else explain it.
When adding or subtracting mixed numbers,
. . I prefer to keep the whole numbers and fraction separated.
For example: \(\displaystyle \,3\frac{1}{3}\,+\,4\frac{1}{3}\)
This is: \(\displaystyle \:\begin{array}{ccc}3\frac{1}{3} \\ 4\frac{1}{3} \\ --- \end{array}\)
And you can see the answer: \(\displaystyle \,7\frac{2}{3}\;\) . . . right?
A harder example: \(\displaystyle \,2\frac{2}{3}\,+\,5\frac{1}{2}\)
This is: \(\displaystyle \:\begin{array}{ccc}2\frac{2}{3} \\ 5\frac{1}{2} \\ ---\end{array}\)
We can see: \(\displaystyle \,2\,+\,5\:=\:7\)
And we must add the fractions:
. . \(\displaystyle \frac{2}{3}\,+\,\frac{1}{2}\;=\;\frac{4}{6}\,+\,\frac{3}{6}\;=\;\frac{7}{6}\;=\;1\frac{1}{6}\)
So we have: \(\displaystyle \,7\,+\,1\frac{1}{6}\;=\;8\frac{1}{6}\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Many teachers would have you change everything to "improper" fractions.
So your boss says, "There's \(\displaystyle 10\frac{1}{4}\) pounds of coffee in one bin
. . and \(\displaystyle 20\frac{1}{4}\) pounds in another. .How much coffee do we have?"
You grab a pencil and your calculator and start working:
. . \(\displaystyle 10\frac{1}{4}\,+\,20\frac{1}{4}\:=\:\frac{41}{4}\,+\,\frac{81}{4}\:=\:\frac{122}{4}\:=\:\cdots\)
And the guy emptying wastebaskets says, "Thirty and a half."
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Subtraction gets trickier because "borrowing" is often required.
I'll let someone else explain it.