Equality Principle

Example 1

Solve this simple equation:

$$x+2=10$$

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Now, what most people would describe on how to resolve this is to do the following:

$$x=10-2$$ $$x=8$$

As a teacher, you would've asked why this was made, and probably the awnser would be something in the lines of:

"- Since 2 is positive, we pass it to the other side subtracting."

But by being more inquisitive, you could've asked: Why?

"- I don't know."

This is where the problem resides, we've been so used to simply following this mantra that we don't really know WHY we are doing it in the first place. We could be here all day long talking about the flaws in teaching math the basic education system or whatnot, but for simplicity sake let's see the proper way to solve (and understand) equations:

So let's go back to the equation above:

$$x=10-2$$

Firstly we need to understand that when we operate a equation, we operate it as a whole, making the same operation in both sides. (that's why we call it a equality!).

$$x+2+(-2)=10+(-2)$$

In this sense we can assume that for operating the number 2, we need to operate it on the equation as a whole. So instead of saying "sums pass subtracting", we SUM -2 on both sides, so that we can eliminate 2 by using the opposite operations:

$$x=8$$

and now we have the same result

A worry in lots of teachers is that when presented to this principle, the student would say " - Its the same thing [as doing the other way].". In fact the result does remains the same, however the problem resides in the understandings of how equations really work.

Example 2

Solve the equation:

$$2x-3=7$$

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Based on what we've seen now, we can already figure out how to properly do this equation:

$$2x-3(+3)=7+3$$ $$2x=10$$

So what we do now? Since we have 2 multiplying, we can DIVIDE it by 2 on both sides, it's the same concept seen on the first example:

$$\frac{2x}{2}=\frac{10}{2}$$

We do the opposite operation with the number in question on both sides of the equation in order to eliminate it.

$$x=5$$

Example 3 - Inequation

This example gets a little more tricky, but shows well the principle:

$$-3 \lt \dfrac{x-5}{2} \lt 3$$

In order to remove number 2 in the middle, since it's dividing we multiply by 2 in all the instances:

$$-3(*2) \lt (2*)\dfrac{x-5}{2} \lt 3(*2)$$

If we were to make the operation in the middle, we can see that both the number 2 that is multiplying and the one dividing null themselves:

$$(2*)\dfrac{x-5}{2}=\dfrac{2x-10}{2}=x-5$$

so with that:

$$-6 \lt x-5 \lt 6$$

On the same idea, we need to eliminate number 5:

$$-6+5 \lt x-5+5 \lt 6+5$$ $$-1 \lt x \lt 11$$

Result: x is set between -1 and 11.