Verified Answer :
\(\Large \frac { x }{ y }\) \(\begin{matrix} \rightarrow numerator \\ \rightarrow denominator \end{matrix}\)
If two (or more) fractions have the same number as their denominator, you can just go ahead and add their numerators, without changing the denominator. For example:
\(\frac { 1 }{ 2 }\) and \(\frac { 4 }{ 2 }\) has the same number (2) as their denominator. So we will add their numerators, without changing the denominator.
\(\frac { 1 }{ 2 } +\frac { 4 }{ 2 } \quad =\quad \frac { 1+4 }{ 2 } \quad =\quad \frac { 5 }{ 2 }\)
So we added the numerators (1 + 4 = 5) and kept the denominator same (2) and we got (\(\frac{5}{2}\)
This was the case where the fractions' denominator was same. What if it's not ?
If their denominator isn't equal, we're gonna have to equalize them ourself. How to do that ?
Let's show it with an example :
\(\frac { 1 }{ 2 }\) and \(\frac { 3 }{ 4 }\) do no have the same number as their denominator. To be able to add them, we have to equalize their denominators.
\(\frac { 1 }{ 2 }\) 's denominator is 2 and \(\frac { 3 }{ 4 }\) 's is 4
What can we do to equalize them ? Well, 4 is two times 2 ( \(4 = 2\cdot 2\) ) So we cane multiply the denominator of \(\frac { 1 }{ 2 }\) (which is 2) with 2 , to equal it to \(\frac { 3 }{ 4 }\) 's denominator (which is 4).
But, there is a catch here. When multiplying a fraction's denominator before adding it to another, you should make sure that you're preserving its ratio. What does that mean ?
Let's take the number \(\frac { 4 }{ 6 }\)
(1) if we multiply only its numerator with a number (let it be 3)
\(\frac { 3\cdot 4 }{ 6 } \quad =\quad \frac { 12 }{ 6 } \quad =\quad 2\)
You got a new fraction with a different ratio than \(\frac{4}{6}\). And it is also equal to 2, but \(\frac{4}{6}\) isn't equal to 2.
(2) if we multiply only its denominator with a number (let it be 3 again)
\(\frac { 4 }{ 3\cdot 6 } \quad =\quad \frac { 4 }{ 18 }\)
You got a new fraction again, with a different ration than \(\frac{4}{6}\) .
How can we know that ? Well, if you simplify these two numbers to the simplest number, you'll get a different fraction or integer. Let's do so.
\(\frac { 4 }{ 6 } \quad =\quad \frac { 2\cdot 2 }{ 2\cdot 3 } \quad =\quad \frac { 2 }{ 2 } \cdot \frac { 2 }{ 3 } \quad =\quad 1\cdot \frac { 2 }{ 3 } \quad =\quad \frac { 2 }{ 3 }\)
So the simplified form of \(\frac{4}{6}\) in the fraction form is \(\frac{2}{3}\)
\(\frac { 4 }{ 18 } \quad =\quad \frac { 2\cdot 2 }{ 2\cdot 9 } \quad =\quad \frac { 2 }{ 2 } \cdot \frac { 2 }{ 9 } \quad =\quad 1\cdot \frac { 2 }{ 9 } \quad =\quad \frac { 2 }{ 9 }\)
And the simplest form of \(\frac{4}{18}\) as a fraction is \(\frac{2}{9}\) , which is not equal to \(\frac{2}{3}\)
\(\frac { 2 }{ 3 } \quad
eq \quad \frac { 2 }{ 9 }\)
So what to do, to preserve the ratio ? Simle. We'll multiply also the numerator with the same number we're going to multiply the denominator with.
Let's get back to our example.
Adding \(\frac { 1 }{ 2 }\) and \(\frac { 3 }{ 4 }\)
We were going to multiply \(\frac{1}{2}\) 's denominator with 2. Now that we know, the ratio must not change, we'll also multiply the numerator with 2.
\(\frac { 2\cdot 1 }{ 2\cdot 2 } \quad =\quad \frac { 2 }{ 4 }\)
Now we've got a number which has the same denominator as \(\frac { 3 }{ 4 }\)
We can add them now,
\(\frac { 2 }{ 4 } \quad +\quad \frac { 3 }{ 4 } \quad =\quad \frac { 2+3 }{ 4 } \quad =\quad \frac { 5 }{ 4 }\)
\(\boxed { \frac { 1 }{ 2 } \quad +\quad \frac { 3 }{ 4 } \quad =\quad \frac { 5 }{ 4 } }\)
I hope this was clear, if not please ask and I'll try to explain.