Algebra Problems Go-Through: Part 1, Super-Easy

As a follow-up to the popular Welcome to Algebra series, here comes a whole new series of lessons, going through algebra sample problems step-by-step for you to follow. We start off with the really easy ones.

Oh no, equations!

Solving equations is easier than it seems. Our sample one-variable degree 1 equation to start off with will be: . Looks scary? Have no fear, I’ll help you out:

1. When dealing with fractions, get rid of them first. In this case, multiply both sides by 2:
.

2. Great. Now, let’s isolate all the x’s on one sides and all the free-variables (no x’s) on the other by subtracting 5 from both sides:

.

3. Great. Now, finally, get rid of that 3 to find x:

.

You’ve just solved an equation! cool. Let’s move on to a word problem:

Word problems? How do we deal with those?

Word problems are just the same as equations if you know to deal with them. Here’s a sample problem: “Mary had some sheep. If Mary had 5 more sheep than 3 times the number of sheep she has now, and then lost half of her sheep, she’d have 16 sheep. How many sheep does Mary have?”

The first thing you need to do is declare your variables. Let’s call our variable x the number of sheep Mary has. Now, if Mary had 5 sheep more means +5. But it’s 5 more than 3 times her current number of sheep, so it’s . Now Mary loses half, so we remain with half the sheep. That’s . And this equals 16 from the question, so to find x all we need to do is solve .

Wait a minute, we’ve done this a moment ago! So . This means Mary has 9 sheep, so the answer is 9. Not so bad, is it?

Ok, now let’s do 2 variables!

Sure thing. Say we have 2 equations: and . How do we do this? There are 2 ways.

Way 1: Substitution

If you have anywhere variables with no coefficient, like the x in the second equation, substitution is nice and fun. First of all, throw everything else to the other side. In this case, add 3y to both sides:

.
Now, plug in the value of x into the first equation, and solve:




Finally, plug y back into the first equation to solve for x:

. And you’re done.

Way 2: Elimination

If didn’t like that, here’s the second way to go- work your equations so that one of the variables has the same coefficient in both equations, then subtract them to eliminate that variable, and find the other one. It works like this:

First, let’s say we want to eliminate x. We’ll multiply the second equation by 2 to make both coefficients the same:

. Now we subtract this equation from the first:



Now we just plug y into one of the equations and solve for x:


.

Good, this is how you solve a set of 2 equations with 2 unknowns. Again, not so bad.

Now a word problem again

Yes, another word problem: “Mike has 2 types of sheep, white and black. If he had twice the amount of white sheep he’s got now with the same number of black sheep, he’s have 4 sheep. But if a wolf ate 3 times the black sheep he has without touching the white ones, Mike would owe the sheep’s owner 5 sheep.

Ok, let’s break it down. Remember step 1? declare variables. I feel like calling the number of white sheep Mike has now x and the number of black ones y. Now step 2 is to read through the problem and construct an equation. If Mike had twice the white sheep (2x) and the same number of black ones (y), he’d have 4 sheep. That means . If a wolf ate 3 black sheep (- 3y) without touching the white ones (which means x, not 0- x isn’t changed here), Mike you owe 5 sheep (- 5). So .

We’ve just done this- and . We now know that Mike has 1 white sheep and 2 black ones.

Wasn’t that a piece of cake?

Next time we’ll try some quadratic equations and harder word problems (without sheep this time).

Thanks for reading this Algebra Problems Go-Through lesson