I'm new to maths and I'm working on simplifying the fraction below:
x^2+xy
y^2+xy
The answer is x over y but i want to know why i cant simply cancel everything out as it is from the start. I have been taught to cancel out like terms, so can't i say, "xy over xy = 1, x^2 over y^2 is simply X x X and Y x Y, so i can divide those by top and bottom and get answer = 1" and ultimately end up with nothing as the answer?
Also, would the answer simply be 1 or would it be 0 if i did it this way? my teacher cancels out like terms a lot but doesn't always write the 1, so i'm a bit confused.
Or at the very minimum, why is that i cant cancel xy over xy out now, but if i write them out as x(x+y) over x(x+y) i can then cancel out the terms in the brackets?
There's a lot going on here; let's take it step by step. I'm assuming that the task is to simplify $$\frac{x^2+xy}{y^2+xy}$$.
It sounds like you are trying to do this: $$ \frac{x^2 \color{red}{+xy}}{y^2 \color{red}{+ xy}} \color{red}{=} \frac{x^2}{y^2}$$ This is wrong. You can cancel common factors in the numerator and denominator but not common terms. If you try it with $x=1$ and $y=2$, then the left-hand side of the above is $\frac{1^2 + 1 \cdot 2}{2^2 + 1 \cdot 2} = \frac{3}{6} = \frac{1}{2}$, while the right-hand side is $\frac{1^2}{2^2} = \frac{1}{4}$.
It sounds like you are trying to do this: $$ \frac{x^2}{y^2} \color{red}{=} \frac{x^2 /x^2}{y^2/y^2} = \frac{1}{1} $$ This is also wrong. You can only divide the top and bottom by the same expression, not different ones.
It sounds like you are equating the $1$ that arrives from canceling everything in sight with $0$ (“nothing"). But a multiplicative nothing is $1$, while $0$ is the additive nothing.
OK, then, turning to your question about the solution which is correct:
You have $$ \frac{x^2+xy}{y^2+xy} = \frac{x\color{blue}{(x+y)}}{y\color{blue}{(y+x)}} = \frac{x}{y} $$ Here we see common factors in the numerator in denominator, so they can be canceled. But there isn't anywhere to go from here, since $x$ and $y$ are independent variables.