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GMAT: What is a Linear Equation
Solving Linear Equations With Literal Coefficients

In this lesson, we'll literally learn about literal coefficients. We'll look at how to solve linear equations that contain literal coefficients and practice solving several problems.

Solving Linear Equations With Literal Coefficients

Literally

This lesson is literally the greatest thing you could be watching right now. Literally. Does it bug you when people use 'literally' when they shouldn't? I mean, it's literally the worst thing ever. Well, except for maybe a few things - I don't know, war, climate change, the lack of In-N-Out Burgers where I live. As much as it's literally misused, there are literally good uses of the word 'literal.' And, we're about to literally learn about one in algebra. Literally.

Literal Coefficients

Let's say you're walking down Algebra Street, and you bump into this: ax + 2 = b - 5. Whoa. Hold on. What are you supposed to do with all those letters? And, what kind of street is this where linear equations come to life and wander the streets? That's literally weird. Your first thought is, 'I gotta move to a new neighborhood. Geometry Street has some cool-looking houses.' But wait, before you pack up, let's look again at this algebraic pedestrian. This equation has literal coefficients. A literal coefficient is a symbol that represents a constant, or a fixed number.

Wait - aren't variables just symbols used to represent numbers? Yes! And, literal coefficients are in many ways similar to variables. But in a linear equation, we treat literal coefficients more like numbers, and we're still trying to solve for the variable.

Solving Literal Equations

Let's look at how this literally works. Remember that stranger from Algebra Street? ax + 2 = b - 5. We just want to solve for x. And, how do we do that? We get x alone on one side of the equation.

First, subtract 2 from both sides. ax = b - 7. If that a were a number, like 7, we'd just divide by that number. We do the same thing with the literal coefficient. If we divide by a, we get x = (b - 7)/a. And, that's our answer. We can't go any further. We're basically defining x in terms of b and a.

When you think about that, since we can't do anything with those literal coefficients, there's actually less math to do. If a and b were numbers in that equation, we'd have to keep solving until we got a final number. This makes literal coefficients literally pretty cool. Note that our literal coefficients here were a and b; we usually use the letters from the beginning of the alphabet for our literal coefficients, like a, b, c, and d.

You may have seen those same letters used as ordinary variables, as in 3a = 15. So, how do we know when we have a variable and when we have a literal coefficient? Literally the easiest way is just to look at what the problem says. Problems with literal coefficients will usually say something like, 'if ax = 15, then x = what?' In this case, by the way, we'd divide both sides by a and get x = 15/a.

Practice Problems

Let's get a little practice. Here's one: 6y - c = b. We want to solve for y. First, we move that c over by adding cto both sides. Then, we just divide both sides by 6 to get y = (b + c)/6. That was simple. Maybe Algebra Street isn't going to be so bad after all.

What about this one? 12x + 5 = a + 4x. We need to isolate the x. But, there are xs on both sides. That's okay. We just subtract 4x from both sides to get 8x + 5 = a. Now, let's subtract 5 from both sides. Finally, divide by 8. So, x = (a - 5)/8.

Did you ever go on a quick run to the grocery store and know you need three things, but when you get there, you can only remember two? That's like what we're doing here. It's like x is the list you should've written down but didn't. You know it's something like milk, eggs and something else. That something else is our literal coefficient. If we knew what that was, we'd know everything. Well, not literally everything. But, we wouldn't get home from the store and say, 'I got the milk, eggs and this letter a!'

Okay, let's try another: 3az + 7z = 4. This looks trickier, but we're still following the same protocol. How do we get z alone? Well, 3az + 7z can be factored. If we pull out the z, we have z(3a + 7). And, guess what? Now we just need to divide both sides by 3a + 7. So, our final answer is z = 4/(3a + 7).

How about one more? bx + 5 = ax - 2. Let's first subtract ax from both sides. Then, let's subtract 5 from both sides. Okay, bx - ax = -7. Let's factor out that x. Now, we just divide by b - a. So, x = -7/(b - a). And, we're literally done!