Making Sense of Algebra: The Basic Concepts [OLD VER]

 Before you start to read this, I need you to pretend like you don't know everything about algebra, EVEN IF you do know it.

Imagine you want to buy something which costs $5 at a store. However, you then realize that you only have $2 in your pocket. How much the extra money do you have to bring when you go back to the store later?

When we work in math, we can write it as an equation, something like

$2 + extra money = $5

Now, to ease our calculation, symbolize the extra money with a letter, say \(x\) and remove the dollar sign, since our calculation involves USD and USD and produces a result in USD, so we can just write the dollar sign at the final answer.

$$ 2+x=5 $$

You realize, to get a $5 stuff, the money you currently have, which is $2, added together with the extra money you'll take must be $5. So, you will instinctively guess, "Well, it's $3."

Let's prove your instinct, shall we? Replace \( x \) with the guessed amount of extra money, $3, and see what happens.

$$ \begin{aligned} 2+3&=5 \\ 5&=5 \end{aligned} $$

Then, you conclude that your guess is true because it satisfies the condition, since if you put $2 and $3 together, it will give you $5.

Why do I have to make you imagining those kind of seemingly daily activities? I just want you to realize, that even you, who might hate algebra, or math in general deep in your heart, often unconsciously do an algebra inside your own head... eh, wait a minute! You still don't know what the heck algebra is, do you? Sorry for that.

Algebra comes from an arabic word "al-jabr" which means "calculation". Looking at our calculation earlier, you can say, algebra is "a branch of mathematics dealing with symbols and the rules for manipulating these symbols." Sorry, not really "manipulating", but rather "operating efficiently".

So, algebra is about how we deal with symbols and how we can operate these symbols efficiently, so our calculation will produce the desired result.

What if you currently have $3 rather than $2? How much the extra money do you have to bring now? "Well, it's $2."

Our calculation will change to

$$ 3+x=5 $$

Now, let's prove your instinct again, shall we? Replace \( x \) with 2, and see what happens.

$$ \begin{aligned} 3+2&=5 \\ 5&=5 \end{aligned} $$

Since it satisfies the condition, then you conclude that the extra money you have to bring is indeed $2.

You realize that this \( x \), the amount of extra money you have to bring, can have a various value, in accordance with the money you currently have and the amount of money you really need.

Since the value of \( x \) is able to hold various number, thus you call this \(x\) as a "variable". You can imagine a variable as a mysterious box that you don't know its real content, unless we know the whole condition. Remember, in mathematics, variables are often written in just one or two letter(s). And since the value of your original money in your pocket stays the same, $3, then you call this as a "constant" since it's constant, not changing with time.

Now, what if you say, "Welp, let's just buy it tomorrow." Then, tomorrow, you bring the extra money, but unfortunately, the price of the thing you want to buy now have been increased by $1. Now you say to yourself, "Damn! Now I have to take more $1!" and you go home grudgingly to take it.

Will the amount of (previously) extra money still satisfy the condition if we add it to $1? Let's prove it together, shall we?

$$ \begin{aligned} 3+x+1 &= 5+1 \\ 4+x&=6 \\ 4+2&=6 \\ 6&=6 \end{aligned} $$

Now, maybe you realize, if you added both sides with the same number, which in this case is 1, the \( x \) will still be the valid value which satisfies the condition.

Then, let's move to another case.

Imagine you have 2 box of apple that will give you, say 16 apples. How many apples in each box?

You will instinctively say, "8 apples". But, is that really true? Of course, prove it!

Label the box as \( b \). Then, our calculation should be

$$ 2b=16 $$

Replace \( b \) with the number of apples you've guessed before.

$$ \begin{aligned} 2b &= 16 \\ 2(8)&=16 \\ 16&=16 \end{aligned} $$

Hey! It does satisfy the condition! So, the number of apples in each box is indeed 8. Now, the number of the boxes stays the same throughout time, so it's constant. But wait! It's indeed constant, but there's something special about this "constant". This constant is used to multiply a variable. Thus, you think you need a special name for that.

You're accidentally a modern Latin expert, so you take the word "com" which means "together" and "efficient" which means "produce", so you start to call it a "coefficient" since it works together with the variable to produce a certain result.

Now, what if we double what we have. Thus, we now have 4 box of apples, which should give you 32 apples. Is the value of \( b \) still valid, meaning that the number of apples in each box is still 8? Of course it is, if we talk about logic. But still, let's prove it together.

$$ \begin{aligned} 2(2b) &= 2(16) \\ 4b&=32 \\ 4(8)&=32 \\ 32&=32 \end{aligned} $$

Do you see the pattern?

From the 1st case and the 2nd case, you conclude that if you add or multiply the both sides of the equation, it will still give you the same solution. Thus, if we add or multiply the both sides of the equation, it'll produce another equation that has the exact same solution, thus we call the previous and the latter equations as "equivalent".

But then you ponder to yourself, "If I subtract or divide the both sides instead, will I still get an equivalent equation?"

If you're wondering, why don't you just try it to convince yourself?

CASE I: Subtraction

Suppose you have an equation

$$ x+5=8 $$

Your instinct will immediately tell you that \( x=3 \). See if it satisfies the condition.

$$ \begin{aligned} x+5&=8 \\ 3+5&=8 \\ 8&=8\end{aligned} $$

\( x \) indeed is 3. Then, you subtract, say 3, from the both sides of your original equation, and see if \( x=3 \) still satisfies the condition.

$$ \begin{aligned} x+5-3&=8-3 \\ x+2&=5 \\ 3+2&=5 \\ 5&=5\end{aligned}$$

Thus, you conclude that if we subtract the both sides with the same number, you'll get an equivalent equation.

CASE II: Division

Suppose you have an equation

$$ 4x=20 $$

Your instinct will immediately tell you that \( x=5 \). See if it satisfies the condition.

\begin{aligned} 4x&=20 \\ 4(5)&=20 \\ 20&=20\end{aligned}

\( x \) indeed is 5. Then, you divide the both sides of your original equation, say by 3, and see if \( x=5 \) still satisfies the condition.

$$ \begin{aligned} \frac{4x}{2}&=\frac{20}{2} \\ 2x&=10 \\ 2(5)&=10 \\ 10&=10\end{aligned}$$

Thus, you conclude that if we subtract the both sides with the same number, you'll get an equivalent equation.

From those 2 cases, you then further conclude, that if you treat the both sides with the exact same operation, be it addition, subtraction, multiplication, and division, and maybe even root, square root, etc., you'll get an equivalent equation.

Then, from that conclusion, you think to yourself, "Oh, then I can treat the both sides the hell I want, can't I? So, I can use whatever operation to 'eliminate' what I don't need. NEAT!"

Then, you suppose an equation

$$ x+6=10 $$

You think, "I can just 'erase' this 6 so we can immediately get the value of \( x \) without guessing. But how can I do it? Ah, ahah! Of course, I just have to subtract 6 from it!"

You then subtract 6 from the both sides.

$$ \begin{aligned} x+6-6&=10-6 \\ x&=10-6 \\ x&=4 \end{aligned} $$

You then prove to ensure that it really satisfies the condition.

$$ \begin{aligned} x+6&=10 \\ 4+6&=10 \\ 10&=10 \end{aligned} $$

Yes, it does satisfy the condition.

Then, you suppose another equation

$$ 4x=20 $$

You think, "I just have to 'erase' this 4 behind \( x \) so I can immediately get the value of \( x \) without guessing. But how can I do it? Ah, ahah! Of course, I just have to divide it by 4!"

$$ \begin{aligned} \frac{4x}{4}&=\frac{20}{4} \\ x&=\frac{20}{4} \\ x&=5 \end{aligned} $$

Do you see the pattern? When you wish to eliminate a term, you can subtract the same value from it so it'll give you 0, and when you wish to eliminate a coefficient or maybe variable, you can divide it by the same value so it'll give you 1. But, you also realize, that this will make as if the constant, coefficient, variable, or term is "moved" to the another side of equation. 

Thus, this is the origin of a rule, that basically says, "If you want to 'move' something from a side of equation to another, the sign will be 'flipped'."

Okay, what I mean by "flipped" is the "opposite operation". So, if you want to move, say a constant of positive 4, to another side, then it'll be a negative 4 and vice versa, and if you want to move a coefficient or a separate a variable from its coefficient, then you move it to divide the another side.

$$ \begin{aligned} x+y&=z \\ x&=z-y \end{aligned} $$

$$ \begin{aligned} x-y&=z \\ x&=z+y \end{aligned} $$

$$ \begin{aligned} xy&=z \\ x&=\frac{z}{y} \end{aligned} $$

$$ \begin{aligned} \frac{x}{y}&=z \\ x&=zy \end{aligned} $$

PITFALL PREVENTION

First of all, I still haven't told you about what is a term, so I'll tell you. A term is a single mathematical expression. HOWEVER, a term can be said as a term on its own if they add or subtract (add as a negative number) another term. For example, if we have an equation

$$2(4x)=8$$

On the left side, you can't say that 2 is a term and 4 is a term on their own.

However, if the equation is like this

$$ 2+4x=8 $$

On the left side, you now can say that 2 is a term, and \( 4x \) is a term on their own.

Now, what if the equation is like this

$$ 4x-2=8 $$

What are the terms on the left side? It's \(4x\) and -2. But why negative? Because, a subtraction is just the same as an addition with a negative number.

CASE: Multiple Terms

Suppose you have an equation like this

$$ 3x+2=x+10 $$

What should you do? First of all, of course move each term according to what they contain. So, the terms that contain variable \( x \) will be on a side of the equation, and terms that contain only constant, then calculate them.

$$ \begin{aligned} 3x+2&=x+10 \\ 3x&=x+10-2 \\ 3x-x&=10-2 \\ 2x&=8 \end{aligned} $$

Then, when you've got the simplest equivalent equation, just do "whatever you want" to get the value of \( x \)

$$ \begin{aligned} x&=\frac{8}{2} \\ x&=4 \end{aligned} $$

Let's prove it once again to ensure its validity.

$$ \begin{aligned} 3x+2&=x+10 \\ 3(4)+2&=4+10 \\ 12+2&=4+10 \\ 14&=14 \end{aligned} $$

ANOTHER PITFALL PREVENTION

Suppose you have an equation like this

$$ 4x+4=16 $$

You can't just simply move the coefficient 4 to the right side like this

$$ x+4=\frac{16}{4} (wrong!) $$

Don't believe it? Sure, I'll prove it.

$$ \begin{aligned} 4x+4&=16 \\ 4(\frac{16}{4})+4&=16 \\ 4(4)+4&=16 \\ 16+4&=16 \\ 20&=16 (contradiction) \end{aligned} $$

If you want to move a coefficient, the side it's on must only contain one term, or you have to make it technically one term, so you can do something like 

$$ 4(x+1)=16 $$

since 4 multiplied by \( x \) equals \( 4x \) and 4 multiplied by 1 equals 4. Now, you can move the "4" to the right side.

$$ \begin{aligned} x+1&=\frac{16}{4} \\ x+1&=4 \\ x&=4-1 \\ x&=3 \end{aligned} $$

Let's prove it to ensure its validity.

$$ \begin{aligned} 4x+4&=16 \\ 4(3)+4&=16 \\ 12+4&=16 \\ 16&=16 \end{aligned} $$

CONCEPT STRENGTHENING

If there is an equation like this

$$ x+2=8 $$

Then you do the same exact operation to the both sides, say add them with 2, it becomes

$$ x+2+2=8+2 $$

But, you basically say

$$ 8+2=8+2 $$

since \( x+2=8 \). It makes sense that adding the both sides will result an equivalent equation, right?

Of course, all of these are just the basics, so brace yourself for more complex algebra.