Intermediate algebra questions are some of the toughest questions on the ACT Math Test. To compensate for the difficulty of the topic, almost all of the intermediate algebra problems will be in basic form, meaning that you don’t need to sort through a mess of words to find the question. Also, you should be glad to hear that there will be only nine intermediate algebra problems on the Math Test, making them worth less than one-sixth of your math score.

In this section, we’ll present the intermediate algebra topics to you in the following order:

The first two topics in this list appear most frequently on the ACT. You may not encounter a single example of the last four topics, particularly the last two, on a given test. Those topics do appear from time to time, though, so it pays to be prepared for them.

This topic constitutes a major portion of the intermediate algebra questions. You will probably see about three quadratic equations questions per test. Those three questions make up a third of the intermediate algebra questions. If you can master these questions, then you’re well on your way to overcoming intermediate algebra.

A quadratic equation is a second-degree equation with one variable and usually two solutions. If you don’t understand what that means, hold on a second. A quadratic equation on the ACT will almost always appear in the following form:

where a and b are coefficients and That is the standard form of a quadratic equation, and it is the form that the ACT almost always uses. In some cases, you may come across an equation that looks like this:

In this case you can subtract bx from both sides of the equation to get the equation into standard form:

Every quadratic equation contains a variable raised to the second power. In most of the quadratic equations you’ll see on the test, there’ll be two solutions for this variable.

Also, the ACT almost always makes a equal to 1 to simplify solving these equations. (Note that when a = 1, is simply written

Solving a quadratic equation means solving for the variable used in the equation. Almost all quadratic equations appearing on the ACT can be solved by factoring. Solving a quadratic equation by factoring is essentially the reverse of what you do when multiplying binomials. Take the following example:

Solving for x here requires a good degree of intuition, but with time and practice your intuition will become increasingly keen. Try to imagine which binomials would create the equation above. You can do this by considering the factors of 18 (1 and 18; 2 and 9; 3 and 6) and asking yourself which pair of factors adds up to 9. Done that? If so, you see that 3 and 6 add up to 9. So you can factor the equation as:

Whenever you see something in the above form, you can solve it like this:

Either x = –3 or x = –6 satisfies the equation.

Very rarely on the ACT, you may encounter a quadratic equation that cannot be solved by factoring. In that case, you can use the quadratic formula to solve the equation. The quadratic formula is:

where a, b, and c are the same coefficients as in the quadratic equation. You simply plug in the coefficients to determine solutions for x. Just to prove to you that the equation works, we’ll work out the quadratic equation whose roots we already know: Remember that it’s in the form of .

A few times per test, the ACT will give you two equations and ask you to determine the value of a particular variable or some other equation or expression. For example,

The best way to answer this type of question is to use a substitution method: solve for one variable and then substitute that value into the other equation. In looking at the two equations above, it seems obvious that it would be easier to solve for x using the second equation than it would be to solve for y in either of the two equations. All it takes is a little reorganizing:

Next, all we have to do is plug 2y – 6 into the value for x in the first equation:

Now we have only one variable to deal with in the equation, and we can easily solve for it:

Once we know the value of y, we can plug that value into either equation to solve for x:

Now we can answer the question, which asked for x – y. We get 4 – 5 = –1. When you solve systems of equations questions, always be careful of a few things:

Occasionally, the ACT will test your understanding of systems of linear equations by asking you to determine when two equations in a system yield an infinite number of solutions. To answer this sort of question, you only need to know one thing: a system of equations will yield an infinite number of solutions when the two equations describe the same line. In other words, the system of equations will have an infinite number of solutions when the two equations are equal and in y = mx + b form. Here’s an example:

To answer this question, you have to pick a value for b such that the two equations have the same formula fitting the y = mx + b form. The first step in this process is to transfer 3x – 2y = 4 into the y = mx + b form:

Then put 12x – 4by = 16 into the same form:

Since you know that the two equations have to be equal, you know that (3/2)x must equal 3x/b. This means that b = 2, so B is the right answer to the question.

You should understand the relationship between the two sides of an equation. If you have an equation that says where k is a constant, the equation tells you that w varies directly with the square of t; in other words, as t increases, so does w.

If, on the other hand, you have an equation that says where k is a constant, the equation tells you that w varies inversely with the square of t; in other words, as t increases, w decreases.

If you restated y = ax + b as f(x) = ax + b, you would have a function, f(x), which is pronounced “f of x.” On the ACT, you can almost always treat f(x) as you would treat y, but we want you to be aware of the different format.

On rare occasions, the ACT has asked questions about compound functions, in which one function is worked out in terms of another. The notation for a compound function is f(g(x)), or To evaluate a compound function like f(g(x)), first evaluate g at x. Then evaluate f at the results of g(x). Basically, work with the inner parentheses first, and then the outer ones, just like in any other algebraic expression. Try the following example:

Here’s a slightly more complicated example:

This question doesn’t ask you to evaluate the compound function for a given value—it asks you to express the compound function as a single function. To do so, simply plug the formula for f into the formula for g:

You will seldom see a matrix problem on the ACT, and many high school math courses may not have covered matrices by the time you take the test. Still, any matrix problems on the ACT will be very straightforward and fundamental, so you really only need to know the basics of matrices in order to get the right answers. We will cover those basics here.

You will not be asked to do anything more advanced than adding or subtracting matrices. For example,

What is A + B? To answer this question, you simply add the corresponding entries in A and B. The entries in the first row are 2 + (–4) = –2 and 0 + 1 = 1. The entries in the second row are 3 + 6 = 9 and (–5) + 3 = –2. So the resulting matrix is:

If the question had asked you what A – B is, then you would simply subtract the entries in B from the corresponding entries in A.

Like matrices, logarithms rarely appear in the ACT Math Test. But they do pop up occasionally, and you should know how to handle them. Logarithmic functions are inverses of exponential functions. The exponential equation is equivalent to the logarithmic equation

This inverse relationship between logs and exponents is all you need to know in order to answer a logarithm question on the ACT. If you see then you know that You will be able to use this second, more manageable mathematical expression to answer the question.