Solving equations with x on one side

"And then the devil said, put the alphabet in math!"

"Dear Algebra, stop asking us to find your $x$. She's not coming back, and I don't know $y$."

These type of jokes are commonplace, because for many Americans, Algebra and "solving for x" is where math started to be especially confusing. But it doesn't have to be that way!

In Algebra, we use x, y, and other letters in two ways: (1) to represent specific but unknown values, and (2) to represent relationships between multiple quantities. When we talk about "finding x," we're talking about the first type - $x$ is used as a placeholder for a number we don't know, and we're trying to find what number it is. Giving that unknown quantity a name, like x, allows us to do mathematical operations with it to help us figure out what it is.

For example, suppose you're planning a party, and you want to serve chips. You know that you have some chips already in your pantry, but not enough for all the party guests. You could just go to the store and buy a few more bags, but you don't know how many you need - what if you don't buy enough? What if you buy too many and you have a ton of leftovers? Instead of just guessing and hoping for the best, you can use $x$ to represent the number of containers you need to buy, and set up an equation and solve it to figure out exactly how many you need.

So how do you actually solve an equation?

When we're solving an equation, the goal is to end up with "x = something" — x is alone on one side of the equation, and the other side is whatever x is equal to. To get x alone, we need to get rid of everything else around x. I like to think of it as unwrapping a present - x is the gift itself, but in order to see what it is, we have to remove the layers of wrapping paper surrounding it. We do that by undoing each layer, or doing the opposite of what we did on that layer. If we subtracted, we undo it by adding; if we multiplied, we undo it by dividing; if we added, we undo by subtracting; and so on.

Consider, for example, the equation $7(x-8)^3+4=60.$

Spoiler alert: the answer to this problem will turn out to be x=10. So, let's think about how we would work with the left side of this equation if the 10 was actually there instead of the x. We'd start out with $7(10-8)^3+4$; then, following the order of operations, we'd do the parentheses first (10-8, giving us 2), then the exponent (2³, which becomes 8), then multiplication (7 times 8 gives us 56), and finally the addition (56 plus 4 is 60).

Now, go back to the original equation, where we didn't know what x was. We've "wrapped the present" in PEMDAS order, but to "unwrap the present" and reveal x, we have to take off the outer layer first. The outer layer was "add 4," so we have to undo that by doing the opposite: subtract 4. The next layer to unwrap is the multiplication: we multiplied by 7, so undoing it is dividing by 7. The next layer was the exponent - to undo cubing something, we take a cube root. Finally, the innermost layer was the subtraction in the parentheses: to undo subtracting 8, we have to add 8. 

There's one other thing we need to take into account, though. We can't just randomly tack on the "subtract 4," "divide by 7" etc. on the left side of the equation. If we did that, the left side would end up being x alone like we want, but the right side wouldn't change - so we'd have x=60 at the end instead of the x = 10 that we're supposed to have!

In order for this process to work, we have to do each of the "unwrapping" steps to both sides of the equation at the same time - whatever we do to the left side of the equals sign, we have to also do to the right side of the equals sign as well, so that the equation continues to match the original situation and gets the correct result.

Notice how the numbers that we get on the right side of the equation when we do each of the "undoing" steps match the numbers we got on each step when we did PEMDAS on $7(10-8)^3+4.$ By undoing each step on both sides, we're revealing exactly what the steps were that got to 60 in the first place!

Of course, when you solve an equation, you usually won't be told the answer from the beginning and work both forward and backward the way we did here. Instead, you're just going to do the backward version - doing all the "undo" steps on both sides to discover what x is for that equation. But just like we did here, we'll undo the operations in reverse order of the PEMDAS order in which they're applied, to "unwrap the present" and get x alone.

So here's another example: $\frac{(2x+6)^5}{8}-11=-7.$ Since we need to do reverse PEMDAS order, addition and subtraction come first. The equation has it subtracting 11, so we undo that by adding 11 to both sides, giving us $\frac{(2x+6)^5}{8}=4.$ The next step is multiplication and division. The equation includes division by 8, so we multiply both sides by 8 to undo it, resulting in $(2x+6)^5=32.$ Exponents are next, so we undo the 5th power by taking a 5th root of both sides, giving us $2x+6=2.$ Now we’re on the parentheses step - so we work on undoing the steps that were inside the parentheses at the beginning. To handle those inner steps, we again work in reverse PEMDAS order. Addition and subtraction first: to undo adding 6, we subtract 6 from both sides, giving us $2x= -4.$ And finally, multiplication and division: to undo multiplication by 2, we divide both sides by 2, resulting in $x=-2$ as the final answer.

Now it's your turn! Try solving this equation: $81(7x-25)^3+5=2192.$ Tell me your answer in the comments, or on Facebook or Instagram, and I'll tell you if you got it right!

Present photo by Jess Bailey on Unsplash