Introduction to One-Variable Linear Equations

Hi everyone!

Let’s review how to solve basic one-variable linear equations. In other words, we’re solving for the value of a particular unknown variable \( x \). Typically, we don’t know what the value of \( x \) is (or else we would have the solution!), so we can usually do some manipulations to both sides to solve for it.

Hold up… what is a one-variable linear equation?

Let’s break the term up.

“One-variable” means that there is one unknown quantity that we want to solve for. Remember that variables are typically represented as letters, like \( a \), \( x \) or \( y \). As the name suggests, when we call something “one-variable”, we usually mean that we’re only interested in expressions that use single variable repeatedly.

“Linear” means that all the instances of that one-variable have a power of one. For example, if that variable is \( x \), we won’t be dealing with expressions like \( x^2 \) (which is just \( x \times x \)), \( x^3 \) or even \( x^{\frac{1}{2}} \).

Thus, when we are solving “one variable linear equations”, we are typically solving for the value of one variable. We are only dealing with “x’s” and constants, so we won’t need to worry about funky things like radicals or terms like \( x^2 \) yet.

This is an important skill that will be extremely fundamental to your mathematical toolbox, so it’s essential that you master it!

Let’s start with an easy example: Solve for \( x \) when \( 6 \times x = 36 \).

When we say “solve for x”, we mean to find the numerical value of x. In our original expression, we don’t know what the value of x is yet.

In plain English, we would say: six times x equals thirty-six. To find the value of x, we should ask: what number, when multiplied by six, gives us thirty-six?

We can obviously find the answer using trial and error or memorization. Six times six is thirty-six, so the number is six. Therefore, \( x=6 \), since \( 6 \times x = 6 \times 6 = 36 \). Simple enough.

However, is there a more precise way to find the value of a variable?

What if I gave you \( 15c = 345 \) and I told you to solve for \( c \)?

Unless you’re a whiz and you’ve memorized \( 15 \cdot 23 = 345 \), you wouldn’t be able to find the value of \( c \) easily using trial and error or times tables.

Here is the cool part: we can actually find the value of a variable by applying operations to both sides. That is, we can add, subtract, multiply or divide an equation by a certain number, and still achieve a true statement.

Think of linear equations as a see-saw. Take the equation \( x = 6 \), and imagine that the see-saw’s hinge is the equal sign and quantities on each side are placed on their respective benches:

Of course, we know that the value of \( x \) is 6 in this situation. The two have the same numerical value.

Now, it’s apparent that whatever we do to one side, we must do to the other. That’s kind of obvious if you’ve ever played with an actual scale or see-saw in a playground. One side is too heavy, and so you must balance it with the other side.

Let’s add one to the left hand side. What happens?

Oops… we made the left hand side too “heavy”. Here, the thing \(x+1\) is NOT equal to \( 6 \), so \(x+1=6\) cannot be a true statement. We should add one to the right hand side to balance things out.

The two sides are now equal. Since \( 6+1=7 \), this new equation we’ve formed is \( x+1=7 \). Here, we CAN say that \( x+1 \) is equal to \( 7 \) because both sides of the “see-saw” are balanced.

We can obviously subtract, multiply or divide by the same quantity, as long as we do it to both sides to maintain that “see-saw”.

The principle that we can use to solve many one-variable linear equations is the following:

In an equation, what you do to one side, you must do to the other.

We can add two to both sides, subtract by 5 to both sides, multiply by 10 to both sides, and divide by 1000 to both sides. For now, if you see an equal sign, you should remember that you can do anything to one side as long as you do it to another, just like a see saw.

Armed with this concept, let’s solve our originally proposed equation.

Here it is: Solve for \(x \) if \(6\times x=36\).

Here, we want \( x \) by itself. Remember in the above demonstration that we knew the value of \( x \) from the beginning ( \(x=6 \), so \(x \) is 6 ). \(x \) won’t always be six for every math equation in the world, so here, we want to solve for the value of \( x \) in this current problem. Math gets confusing because most of us reuse symbols like a,b,x,y, but remember that a new equation means a new variable. Just because \( x \) is 6 in one problem doesn’t mean that \( x \) will be 6 in the next problem! (If that happens to be the case, it’s a coincidence. Rely on the math!)

Again, we want to know the number \( x \) such that when multiplied by six, gives us the value of 36. How do we get \( x \) by itself?

This is where another cool technique comes in, and relies on the fact that we can do anything to both sides of the equal sign.

I start with zero. I add one, and I subtract one. What does that give you?

Zero.

I start with one (I won’t start with zero since \( 0 \) times anything is still \( 0 \)). Multiply by five and divide by five. What do you get?

One.

You’ve probably heard it in school, but this is where it comes in handy! Consider doing inverse operations to eliminate coefficients and constants, in order to isolate the variable.

Multiplication is is the inverse of division (vice versa), and addition is paired with subtraction (and vice versa). How does this apply to \( 6 \times x =36 \)?

Remember that \( 6 \times x = x \times 6 \) since the order we do our multiplication doesn’t matter! (go ahead and review Week 1 Notes if you’re not sure why)

So now think about it. If you start with x and multiply by six, by the same analogy as our second example, won’t you get x itself if you divided that result by six?

To get “x” by itself (and not 6x, or 5x, or 7x, or (sid)x, as none of these directly give us the value of “x”), we should divide both sides by six, as follows:

\[ 6x=36 \]

\[ 6x\div 6=36 \div 6 \]

\[ x=36\div 6=6.\]

The step \( 6x \div 6 = x \) on the left hand side comes from the fact that multiplication and division are inverse operations. Remember our analogy? Start with x, multiply and divide by six to still keep x. Alternatively, you can also think about it as dividing six x’s among six people; each person will get one x.

Sounds cool? Try a few practice problems.

Practice Problems

1. Solve for \( c \): \(c+1=5\).
2. Solve for \( x \): \( 5x=30 \).
3. Solve for \( y \): \( y-32=24 \).
4. Solve for \( t \): \( \frac{t}{5}=10 \).
5. Solve for \( t \): \( 8t=196 \).
6. Solve for \( b \): \( 6b+7=55 \).

Let us know if you have any questions!