Dividing both sides by 5, when solving $x = 5(\sin y) + 2$ for $y$?

Question

How do I know which operations to do when solving an equation? For example, $$x = 5(\sin y) + 2 \tag1$$ If I wanted to make $y$ a function of $x$, $$\begin{align} x &= 5(\sin y) + 2 \tag2 \\[4pt] x - 2 &= 5(\sin y) \tag3 \\[4pt] \frac{x - 2}{5} &= \sin y \tag4 \\[4pt] y &= \sin^{-1}\left(\frac{x - 2}{5}\right) \tag5 \end{align}$$ I believe this is the correct answer, but I was wondering why it isn't this way:

$$\begin{align} x &= 5(\sin y) + 2 \tag6 \\ \frac{x}{5} &= (\sin y) + 2 \tag7 \\ \frac{x}{5} - 2 &= \sin y \tag8 \\ y &= \sin^{-1}\left(\frac{x}{5} - 2\right) \tag9 \end{align}$$

Why is equation $(9)$ wrong?

Answer 1

@player3236's comment has already correctly pointed out the issue in your calculation: you simply make a mistake from (6) to (7) because you forgot to divide $2$ by $5$ as well.

In general, when you divide a nonzero constant into both sides of a equation, you need to do it for each term. To see a simpler example, if you have $$ x=5y+2 $$ then, after dividing by $5$ on both sides, you get $$ \frac15x=y+\frac25 $$ instead of $$ \frac15x=y+2 $$

To fix your problem, check the following correct lines:

$$\begin{align} x &= 5(\sin y) + 2 \tag6 \\ \frac{x}{5} &= (\sin y) + \color{red}{\frac{2}{5}} \tag{7'} \\ \frac{x}{5} - \color{red}{\frac{2}{5}} &= \sin y \tag{8'} \\ y &= \sin^{-1}\left(\frac{x}{5} - \color{red}{\frac{2}{5}}\right) = \sin^{-1}\left(\color{red}{\frac{x-2}{5}}\right) \tag{9'} \end{align}$$

Answer 2

A lot of algebra boils down to doing the same thing to both sides of an equation. For instance, if I started with the equation: $$x=y+1$$ I could subtract $1$ from each side to get $$x-1=y+1-1$$ which doesn't look that useful - except that using what I know about addition and subtraction, I can figure out that $y+1-1$ must equal $y$ to deduce a more useful expression: $$x-1=y.$$ It's easy to learn this sort of manipulation by rote, but what we're doing is just cleverly using a fundamental property of equality: if two things are equal, I can do the same operation to both, and they will remain equal.

I could do other correct manipulations, if I wished - I could double each side to get $$2\cdot x = 2\cdot (y+1)$$ which would simplify to $$2\cdot x = 2\cdot y + 2.$$ Would this be useful? Probably not - but it's certainly legal. The trick with solving algebraic equations is that you choose your manipulations cleverly - you try to sort of "unwrap" the variable you wish to solve for, usually by undoing whatever operations were applied to it. I mention this because it's important to remember that we're not just shuffling terms about from side to side - to get from $x=y+1$ to $x-1=y$, I didn't magically move a $1$ from one side to the other - I subtracted from both sides in a way that conveniently made $y+1-1$ and then cancelled.

In the given example, removing the $\sin$ since it isn't at fault here, if I start with: $$x=5y+2$$ It's absolutely true that there are multiple ways to start: One way is to start by subtracting $2$ from both sides giving $$x-2=5y+2-2$$ $$x-2=5y$$ and then dividing both sides by $5$ to get $$\frac{x-2}5=\frac{5y}5$$ $$\frac{x-2}5=y.$$ This is a nice sequence to go about it, because $5y+2$ means "take $y$, multiply it by $5$, then add $2$" and if I want to undo that sequence of operations, I just undo each operation in reverse order - that is, "subtract $2$, then divide by $5$".

But, you're right that you can start in other ways: if I divide both sides by $5$, I get $$\frac{x}5=\frac{5y+2}5$$ which would then simplify as $$\frac{x}5=y+\frac{2}5.$$ This step - from (6) to (7) in your reasoning - appears to be where the issue arises. From there, I could subtract $\frac{2}5$ from both sides as $$\frac{x}5-\frac{2}5=y+\frac{2}5-\frac{2}5$$ $$\frac{x}5-\frac{2}5=y$$ which is equivalent.

As long as you take both sides and do the exact same thing to each of them, you'll arrive at a correct statement at the end*. Succinctly put, the property we're using is that if $L=R$ then $f(L)=f(R)$ for any function $f$ - where that function could be "add $2$" or "divide by $5$" or anything else we might dream of. Algebra is just knowing the correct thing to do so that at the end your result is useful.

(*Caveat: Sometimes students will assume this goes both ways, but if you do something like "square both sides" you can end up with extraneous solutions. It's still true that if $L=R$ then $L^2=R^2$, but it's not true that if $L^2=R^2$ then $L=R$. This doesn't come up with adding and subtracting stuff from both sides or multiplying by non-zero coefficients since you can undo those operations)