How can I transfer number with exponent to other side of equation without losing its sign?

Question

I just begun learning some math and I stumbled upon this issue.. I am currently programming a little game, where I need to deal with intersection of circle and a line and this is where I need standard circle equation like this one: $$(x-5)^2+(y-5)^2 = 5^2$$ Now I need to expand it into general form, which is: $$x^2 - 10 x + y^2 - 10 y + 25 = 0$$ This is what WolframAlpha spits and I trust this tool, but I do not understand, how & why EXACTLY it dealt with the right side: $$5^2$$ Because if I try to simplify my issue as: $$y=5^2$$ Why Wolfram spits: $$y - 25 = 0$$ Which was done as: $$y=25$$ But why it does not do this: $$y-5^2=0$$ Which would lead to: $$y+25=0$$ Is it simply, that I should treat 5^2 as variable: (5^2) .. so that it will preserve sign, when transferred? I only saw people doing it "correctly", but I could not find any evidence, why the latter approach is illegal.

I apologize if this is something very basic, that I am lacking, but I am unable to find a cure, so thanks a lot for possible help.

Answer 1

The exponent is evaluated first. You have the following order of evaluation:

1. Parenthetic subexpressions
2. Exponentiation
3. Multiplication and Division
4. Addition and Subtraction

source

Meaning $-5^2$ is evaluated to $-25$. $(-5)^2$ would be evaluated to $25$.

Edit: As I have already mentioned in the comments, order of evaluation is not only important for evaluating equations or an arithmetic operations, but for mathematical expressions in general.

Answer 2

This is not really an order of operations issue. The entire right side, whatever it is, needs to be moved to the left side with parentheses, because you are moving the right side as a single quantity. If the right side were 5+2 instead, when you move it to the other side, it would be $y - (5+2) = 0$, not $y - 5 + 2 = 0$.

Answer 3

I notice that you seem to have some confusion about what the symbol $-$ means, so I'm going to write $\text{minus}$ instead of $-$. Note that the word $\text{minus}$ always means "the thing on the right is subtracted from the thing on the left"; it never means "negative."

The right way to do it

Let's start with the equation

$$y = 5^2.$$

We can subtract $5^2$ from both sides, which gives us

$$y \text{ minus } (5^2) = (5^2) \text{ minus } (5^2).$$

Notice that I've put parentheses around $5^2$ in order to remind us that the $5$ and the $^2$ go together. It's always okay to put parentheses around something that you're treating as a 'block."

Now, of course, $(5^2) \text{ minus } (5^2)$ is $0$, so we can write our equation as

$$y \text{ minus } (5^2) = 0.$$

And, of course, the number $5^2$ is $25$, so we can rewrite this equation one last time as

$$y \text{ minus } 25 = 0,$$

which is exactly the same thing that Wolfram Alpha gave you.

As you noticed, another way to perform the task is to first rewrite $5^2$ as $25$, and then subtract 25 from both sides. Both ways give the same answer.

But why not $y + 25$?

You asked why we don't go from $y - 5^2 = 0$ to $y + 25 = 0$. To be honest, I'm not entirely sure why you would expect that to be legal or correct. It's possible that you saw the $-$ and the $5$ next to each other and mistakenly thought "negative five," and then if you were to square negative five, the result would be positive twenty-five. But what $y - 5^2 = 0$ actually means is $y \text{ minus } (5^2) = 0$; there's no negative five here anywhere.