Sequence of steps to solve $5-\dfrac{y}{7}=19$?

Question

The equation $5-\dfrac{y}{7}=19$ is solved by cancelling out 5 on both sides and then $-\dfrac{1}{7}$. However, why can the $-\dfrac{1}7$ not be cancelled out first. If this is done, the answer is -138. The actual answer is -98 which makes sense with the first method. Why is the second method wrong?

First method:

$5-\dfrac{y}{7}=19$

$-\dfrac{y}{7}=19-5$

$y=14\times -7$

$y=-98$

Second method:

$5-\dfrac{y}{7}=19$

$5+y=19\times -7$

$y=19\times -7-5$

$y=-138$

Answer 1

Remember to multiply the $5$ with $-7$ as well, making it $$ 5-\frac y7=19\\ -35+y=-133\\ y=-98 $$ As mentioned above, "cancel" is a bad word to use about an action, because there are many different things which may need to be cancelled differently. Using the same word to refer to all of these different active may make it difficult to keep track of what you're actually doing.

In other words, I think "cancel" is a goal, not an action. For instance, you want to cancel the $5$, and in order to do that, you need to subtract $5$ from both sides. You want to cancel the $-\frac{}{7}$, and in order to do that, you multiply both sides by $-7$.

Answer 2

$$5-\dfrac{y}{7}=19\\-7\times\left(5-\dfrac{y}{7}\right)=-7\times19$$Becuase addition/subtraction is distributive, $$-7\times 5-\left(-7\times\dfrac{y}{7}\right)=-7\times19.$$

Answer 3

One way to see why we get a contradiction is to notice that we must unconsciously change the equation (at least if we are consistent in our application of the laws of addition and multiplication in $\mathrm R$).

By trying to apply the product cancellation law first one has to (in order to apply the law correctly, which you did not do -- the law is that if $ab=ac$ for elements $a,b,c$ of some field $\mathcal F$ and $a$ is not the additive identity of $\mathcal F$, then $b=c$) treat $$5-{y\over7}=19$$ as $$\left(5-{1\over7}\right)y=19,$$ which is completely different from the initial equation, as is easy to see. Indeed, if they were equivalent, then we must have the identity $$5-{y\over7}=\left(5-{1\over7}\right)y,$$ which is false if you take $y=0$, for example, so that it is not an identity. This is a contradiction.

Answer 4

Notice $$5-\frac y7=19\to5(7)-\frac{\not{7}y}{\not{7}}=19(7)$$ Hence $$y=5(7)-19(7)=-14(7)=-98$$