solve 2/3 + 4/6
Let a = 2, b = 3, c = 4, and d = 6.
Plug in your numbers to the formula and try it.
Before I proceed with the explanation of why it works, here are some words of advice and encouragement. (I will write the rest of this reply assuming that either you are studying elementary arithmetic. Likewise, if you are of elementary age, you should take this with your parents and work it out on paper.)
Please note that there are other ways of solving this problem. Even if I used a different algorithm, I would still work it out with algebra first. Algebra gets to the core of arithmetic reasoning: there are operators (+, -, X, /, exponents, roots) and operands (numerals, letters to represent numerals).
You want to know the relationship between operators and operands. It is simply a game of logic with numbers. If you follow a few simple rules, arithmetic will be your servant, not your master.
Remember to follow the order of operations: PEMDAS
Parenthesis: Things inside parentheses are done first.
Exponents are done next (there are none in your problem).
Multiplication and Division are done next, left to right.
Addition and Subtraction are done last, left to right.
Be careful if you are using a calculator: most if not all scientific calculators will use PEMDAS, but "four function" arithmetic calculators will not; they use "chain calculation" which means just working left to right, whichever operator comes first is done first. You will get different answers from each calculator. Use the scientific calculator.
The down side is that your teacher, or school, or state department of education may not know the difference or even care. Hopefully you will care, because you will not be able to communicate mathematically if you do not follow PEMDAS. (This issue has actually come up where I live, so I think you should be aware of it as a potential problem anywhere.)
If you are not using algebra now, you ought to start now: algebra is generalized arithmetic and will make arithmetic much more powerful (and easier) for you.
As an exercise you can plug in your numbers where a, b, c, d occur below.
a / b + c / d = (ad + bc) / bd
a / b - c / d = (ad - bc) / bd
Multiply both fractions by one made of products of denominators:
(bd) / (bd) = 1
Multiplying a number by one does not change its value:
(bd / bd)(a / b) + (bd / bd)(c / d)
(bda / bdb) + (bdc / bdd)
In each fraction, cancel selected factors common to numerator and denominator. In this case, cancel (b / b) in first fraction, cancel (d / d) in the second fraction. I do not assume you know why that is, so I will tell you: n / n = 1 for all n not equal to zero. The extra factor of one is not essential and does not change the value of the fraction, so we need not include it any more:
ad / bd + bc / bd
We have two fractions with same denominator.
For reasons too elementary to explain, we can add the numerators and combine them into one fraction with a same denominator:
(ab + bc) / bd
The same holds true for subtracting fractions.
You would probably get this a lot quicker if you wrote it out.
I'm probably confusing you more than helping.
Wouldn't the LCD method result in 2/3+2/3 = 4/3 ?
Yes it would.
You may also want to write the answer as a mixed number because the fraction is improper.
In other words, the numerator (top number) is bigger than the denominator (bottom number).
How many thirds are there in 1 (whole)?
So we have .