Let’s look at the most basic example. I purchased three horses, the total cost of which is $4,200 dollars. Assuming the cost of each horse is equal, how much did each horse cost? Yes – this is a simple example as all you need to do is divide $4,200 by three, but how would we write this from an algebraic perspective?

This example clearly demonstrates what we mean – in the sense that x represents what we are looking for, and we have constructed a valid equation to work the cost out. In everyday discourse, of course, we do not need to construct algebraic formulae to work out these simple problems, but the purpose here is to flag how and why we use algebra. In the examples below, the equations are just as valid as this horse example, only they are in a slightly more complex form. You should attempt to understand the equation in the same way you understand this simple horse example.

With this in mind, we will proceed to look at several key areas of algebra that will appear on your ASVAB Mathematics Knowledge exam, beginning with the most fundamental equations that appear year on year in the test.

In this type of question, you are given the value of x. Knowing this, your first step should always be to insert 6 into wherever you find the value x. When we complete this step, all we need to do is tidy up both sides of the equation, isolate y, and find the final value.

If you are new to algebra, you should probably only try one step per line. This is the method chosen in the example above. With time, though, you will develop the ability to do multiple steps at the same time, meaning the equation becomes simpler, quicker, and easier to solve. We would advise against rushing to this stage, as it’s simply not worth the extra potential mistakes. We do advise, however, that you learn the basic structural steps in solving this type of problem in the first place, steps such as:

We solved the brackets first before turning to any other part of the equation. This is due to the BOMDAS Rule which states you order how you answer questions in the following way: Brackets, of Multiplication, Division, Addition, Subtraction. Always follow the BOMDAS Rule if you’re unsure how to start and progress through an algebra problem.

Notice how we solve one side of the = side at a time. This eliminates confusion about what’s going on, while isolating what we require (the value of y) on its own on the left-hand side.

This allows us to eliminate the 3y into y as we divide the other side of the equation by 3. In essence this last step involves dividing both sides of the equation by 3 in order to eliminate it from the equation. After all, 3 divides into 3 to leave 1, meaning we’re left with 1y which is the same as y.

By following the logical steps of algebra, with particular emphasis on using the BOMBAS Rule, we can easily arrive at the correct answer.

Example 2 – Dealing with Common Factors

What is the highest common factor of the expression 6xy + 2x^2?

A common factor is something that divides into two terms. So, take the numbers 4 and 6 – the highest common factor is 2 because this is the highest number that divides equally between 4 and 6. When it comes to algebra, the same thing is true. In this example, you need to extract the highest common factor. When you have found this factor, you isolate it from the rest of the expression. In this example 2x is the highest common factor, as it’s the highest factor to divide into 6xy, while is also divides into 2x^2.

If we multiply 2x by everything in the brackets, we will arrive back at the form stipulated in the question.

Factor the following equation x^2 – 12x + 20

This is slightly more complex, but nothing you cannot handle. You simply need to follow three simple steps:

Now we can fill in our brackets: (x – 2)(x – 10)

This means, that if we multiply these brackets out, we will produce the equation referenced back in the original question.

There are numerable ways to approach this question, but we’ll provide two of the leading and simplest techniques here. The first technique involves finding a common denominator for all numbers. So, in this case, we find that 40 is a common denominator. This means that all the bottom numbers (denominators such as 5, 8, 10, and 20) all divide equally into the number 40. If we divide by the bottom and multiply by the top, these numbers become much more manageable. Take a look:

Based on this analysis, we easily determine (d) as the largest number and, hence, the largest fraction.

Example 2

Add, subtract, multiply, and divide the following fractions: 1/4 , 2/3

You won’t be asked this type of question with multiple components, but you may be asked one of these. To make things a little easier for comparative purposes, we will look at all three here. If you want to add these two fractions, all you need to do is this:

Convert one or both fractions such that they share a common denominator.

Let’s not forget the top of the fraction is known as the numerator, while the bottom of the fraction is known as the denominator. We have two different denominators in this case – 4 and 3. The best way of converting to a shared denominator is to multiply these two numbers, giving 12.

In this case, we divided 4 into 12 to arrive at 3, and then let 12 be the consequent denominator. If we do the same with 2/3, we will arrive at 8/12. This means the fraction has changed to the following form:

Given we now have a common denominator, all we need to do is add the numerators, to arrive at 11/12. Simple! Let’s take a look at subtraction…

We can do exactly the same thing. Let’s replace these numbers with the new denominator form we calculated above:

You can see that the same steps were required, the only difference here is that we randomly decided to take ¼ from 2/3 purely because the latter was the larger number, and it makes it easier to demonstrate as an example. We can now move onto the penultimate part of the question, which asks us to multiply these two fractions. Let’s take a little look about which we can go about this:

This is probably the easiest example, as all we need to do is multiply the numerators (2 and 1) together and multiplies the denominators (3 and 4) together.

Note in this example how we multiplied the numerators and denominators – but subsequently reduced the fraction down even further. The factor 2 divides into both 2 and 12 evenly, so we arrived at the final finished form of 1/6. You always need to consider whether or not your fraction is in its final form, reduced to the point where no common factor can break it down any further. With this in mind, we can now advance onto our last fraction form – division.

To answer this question, we need to take a slightly different approach. You simply need to inverse the fraction you are dividing by and change to multiply the fractions. Take a look below to see how this works:

All we need to do now is multiply the fractions as per normal.

We cannot leave 8/3 in this form. There is a much simpler form. To determine this, we need to divide three into eight, and the maximum even division is twice. The remainder is, of course, 2. So, let’s take a look at what happens next:

In other words, we divided the denominator into the numerator, which left is with the integer 2 you can see on the other side of the equation. But, by dividing we could only evenly divide twice, leaving 2 as a remainder (as 2 x 3 = 6). Thus, we had to represent this remainder as 2/3 after the integer 2.

In these types of question, solve the question as if there were a standard = sign in place of the ≥ (greater than or equal to sign). We can henceforth resolve the problem as follows:

The useful thing about questions like this is that you can verify whether you have the correct answer. In this case, simply substitute 2 back into x in the original question, where you’ll find 4 ≥ 4, which is true (remember the sign refers to both greater than and equal to!).

Exponents are a natural branch of algebra but they often appear regularly as individual questions on the ASVAB Mathematics Knowledge exam. They are actually quite simple problems if you know the basic rules of addition, multiplication, subtraction, and division. As they’re quite short, we’re going to take you through a couple of examples to highlight the similarities and differences that lie within.

If x = -3 and y = 7, what is the value of x^2y^3

While these questions may appear daunting at first, you’d be surprised how easy and short they actually are. Let’s take each example in turn and discover what the underlying principle behind each of them is.

Example 1: (2b^3)^2

In this example, let’s first note that 2b3 is inside brackets. This observation is very important because it’s the entire contents of the brackets which are squared. This means we have to multiply everything inside the brackets by itself.

(2b^3)^2 = 4b^6

Notice how we square the 2 (to make 4) and we square the exponent b^3 to make b^6. One of the fundamental principles of exponents is that when squaring exponents we must multiply them.

Example 2: (-2b)^3</p.

This example is slightly different as we need to cube a negative number -2. So, (-2)^3 is exactly the same as (-2) x (-2) x (-2). When we multiply the first two we arrive at 4 (as multiplying 2 negative numbers produces a positive number), but when we multiply the 4 by the remaining (-2), we arrive at -8. Thus, the answer to this question is -8b^3.

Example 3: (5y^4)^2

You should understand how to solve this problem by now. By following the rules described above, we’ll arrive at the answer of 25y^8. Don’t forget – when squaring exponents we’re actually multiplying them, so it isn’t 4 x 4, but rather 4 x 2. On the other hand, when we square the 5, it’s 5 x 5 and not 5 x 2.

Example 4: 2x^4 . 3x^3

Always remember that a ( . ) in mathematics stands for multiplication. When multiplying exponents, we actually add them. So, in this example, all we need to do is multiply the integers (2 and 3), but add the exponents (4 and 3), allowing us to arrive at the answer of 6x^7.</p.

Example 5: 3x^-2 . 2x^3

This is a similar example but the only difference is that you have a negative 2 as an exponent. However, the rules are precisely the same as the previous example, allowing us to arrive at an answer of 6x^-6.</p.

Example 6: 2(3x^2)^3

This is a slightly different example as we have now got a 2 placed before the brackets. Ignore the 2 for now and, instead, solve the brackets as per normal to arrive at 2(27x^6). Only now should you consider the 2 before the brackets, and simply multiply the 27 accordingly, meaning our final answer is 54x^6.

Example 7: If x = -3 and y = 7, what is the value of x^2y^3

You will face no exponent question that is not in some way related to the 7 examples described above. With a little practice, you should gain easy marks in your ASVAB Mathematics Knowledge exam.

This is a simple slot in the numbers example. When we slot in the numbers, we arrive at an equation looking like (-3)2(7)3. (-3) x (-3) = + 9, while 7 x 7 x 7 = 343. This means the equation will look like (9).(343), so we simply multiply the brackets to arrive at the final answer of 3,087.