I’m starting to feel like the Rocky movie series – will I get to part 7?
Here are more examples of adding fractions with different denominators:
- \[
\frac{4}{15}\hspace{0.33em}{+}\hspace{0.33em}\frac{3}{5}
\]
So two different denominators. We have to find equivalent fractions that have the same denominator before we can add these. As mentioned in my last post, first check to see if one of the denominators evenly divides into the other. In this case, 5 divides into 15 three times so 15 can be our common denominator. We just have to convert the second fraction. We do this by multiplying top and bottom by 3 because multiplying the 5 by 3 will give us the 15 we want:
\[\frac{3}{5}\hspace{0.33em}{=}\hspace{0.33em}\frac{{3}\hspace{0.33em}\times\hspace{0.33em}{3}}{{5}\hspace{0.33em}\times\hspace{0.33em}{3}}\hspace{0.33em}{=}\hspace{0.33em}\frac{9}{15}
\]
So now,
\[\frac{4}{15}\hspace{0.33em}{+}\hspace{0.33em}\frac{3}{5}\hspace{0.33em}{=}\hspace{0.33em}\frac{4}{15}\hspace{0.33em}{+}\hspace{0.33em}\frac{9}{15}\hspace{0.33em}{=}\hspace{0.33em}\frac{13}{15}
\]
Now I’ve been concentrating on adding fraction but subtracting them is just as easy:
2. \[
\frac{7}{10}\hspace{0.33em}{-}\hspace{0.33em}\frac{3}{14}
\]
Now 10 does not divide into 14 so we will have to find the LCD of these two numbers:
10 = 2 × 5
14 = 2 × 7
LCD = 2 × 5 × 7 = 70
So now let’s convert the two fractions to equivalent fractions with 70 as the denominator. For the first fraction, I want to multiply the top and bottom by 7 and the second fraction I want to use 5:
\[\begin{array}{l}
{\frac{7}{10}\hspace{0.33em}{=}\hspace{0.33em}\frac{{7}\hspace{0.33em}\times\hspace{0.33em}{7}}{{10}\hspace{0.33em}\times\hspace{0.33em}{7}}\hspace{0.33em}{=}\hspace{0.33em}\frac{49}{70}}\\
{\frac{3}{14}\hspace{0.33em}{=}\hspace{0.33em}\frac{{3}\hspace{0.33em}\times\hspace{0.33em}{5}}{{14}\hspace{0.33em}\times\hspace{0.33em}{5}}\hspace{0.33em}{=}\hspace{0.33em}\frac{15}{70}}
\end{array}
\]
So now,
\[\frac{7}{10}\hspace{0.33em}{-}\hspace{0.33em}\frac{3}{14}\hspace{0.33em}{=}\hspace{0.33em}\frac{49}{70}\hspace{0.33em}{-}\hspace{0.33em}\frac{15}{70}\hspace{0.33em}{=}\hspace{0.33em}\frac{{49}\hspace{0.33em}{-}\hspace{0.33em}{15}}{70}\hspace{0.33em}{=}\hspace{0.33em}\frac{34}{70}
\]
So subtracting is just as easy since you just need to subtract the numerators. But we’re not quite done here. The numerator and denominator are both even numbers so at least a 2 is a common factor:
\[\frac{34}{70}\hspace{0.33em}{=}\hspace{0.33em}\frac{{17}\hspace{0.33em}\times\hspace{0.33em}\rlap{/}{2}}{{35}\hspace{0.33em}\times\hspace{0.33em}\rlap{/}{2}}\hspace{0.33em}{=}\hspace{0.33em}\frac{17}{35}
\]
And this is as simple as it can get. One more example:
3. \[
\frac{10}{21}\hspace{0.33em}{+}\hspace{0.33em}\frac{9}{24}
\]
Again, we will need the LCD to get the required denominator as 21 does not divide into 24:
21 = 3 × 7
24 = 2 × 2 × 2 × 3
LCD = 2 × 2 × 2 × 3 × 7 = 168
So now let’s convert the two fractions to equivalent fractions with 168 as the denominator. For the first fraction, I want to multiply the top and bottom by 8 and the second fraction I want to use 7:
\[\begin{array}{l}
{\frac{10}{21}\hspace{0.33em}{=}\hspace{0.33em}\frac{{10}\hspace{0.33em}\times\hspace{0.33em}{8}}{{21}\hspace{0.33em}\times\hspace{0.33em}{8}}\hspace{0.33em}{=}\hspace{0.33em}\frac{80}{168}}\\
{\frac{9}{24}\hspace{0.33em}{=}\hspace{0.33em}\frac{{9}\hspace{0.33em}\times\hspace{0.33em}{7}}{{24}\hspace{0.33em}\times\hspace{0.33em}{7}}\hspace{0.33em}{=}\hspace{0.33em}\frac{63}{168}}
\end{array}
\]
So now,
\[\frac{10}{21}\hspace{0.33em}{+}\hspace{0.33em}\frac{9}{24}\hspace{0.33em}{=}\hspace{0.33em}\frac{80}{168}\hspace{0.33em}{+}\hspace{0.33em}\frac{63}{168}\hspace{0.33em}{=}\hspace{0.33em}\frac{143}{168}
\]
So it’s not immediately obvious that there may be some simplification required. In this case, this fraction cannot be simplified but 143 is not a prime number. Can you find the factors of 143?
In my next post, we will continue with fractions (catching up to Rocky) and extend the knowledge we have so far.