act math algebra: everything you need to know

of the 60 questions on the act math section, about 21 of them will consist of algebra problems. that’s about 1/3 of the total math test, so you’ll need to have a firm grasp on the concepts here in order to do well.

first, we’ll take a quick look at the elementary algebra topics you need to know for the act, conveniently listed so that you can go through and pick out the topics you need to work on. a list of intermediate algebra topics follows. finally, we’ll go into greater depth on intermediate algebra and the top three content areas you need to know for act math.

photo by felixleong

act math: elementary algebra topics

solving expressions with substitution

here you need to be very comfortable to manipulating equations, solving a problem with given expressions and numbers, and plugging in numbers. this is your basic bread and butter of algebra,.

simplifying algebraic expressions

know how to combine like terms and how to factor. in order to do this, you’ll
need to know basic math properties such as the associative and distributive
property.

writing/solving expressions and equations

on some problems, you’ll need to take a word problem and convert it into
simple linear equation or expression in order to solve it. to solve equations,
the most reliable way to get to the answer is by isolating the variable on one
side of the equation and the numbers on the other side.

multiplying binomials

when you see a binomial, the first thing you should think about is foil. this
stands for first, outer, inner, and last.

inequalities

you can treat these the same as equations, but remember that the sign flips and points the other way whenever you multiply or divide both sides by a negative number.

act math: intermediate algebra topics

quadratic equations

most of the time, you can factor a quadratic equation problem on the act. however, you should know the quadratic formula just in case you need it. for more on quadratic equations, see below.

systems of equations

there are multiple ways to solve a system of equations – so make sure you
know how to do it several ways. depending on the problem, one way will be
quicker than another. more info on systems of equations follows below!

relationships among variables in an equation

you’ll sometimes come across some tougher questions that ask you to think
about what happens to one variable as another variable increases or
decreases. if you aren’t sure, you can try plugging in a few values and
check the answer. beware of variables to a power higher (or lower) than 1.

functions

the act likes to throw in multistep problems that involve figuring out the answer to one function and then plugging that answer into a second function. make sure you don’t confuse one function with the other!

logarithms

you won’t see too many of these on the act, but make sure you know how
they work and the basics of logarithms. if you don’t, scroll down and take a look at intermediate algebra in-depth.

matrices

again, it’s a topic that is not tested very often, but make sure you familiarize
yourself with them so that you don’t waste too much time trying to remember
how they work.

act math: intermediate algebra in-depth

three concepts—quadratics, systems of equations, and logarithms—are probably ones with which you’re less a bit less familiar (especially logarithms), but they come up fairly regularly on the act math test. in these types of problems, you’ll use some of the exact same skills you did for elementary algebra, but in more complex ways. here are the basics you’ll need to know to master these types of questions.

quadratic equations have three terms and are in the form ax² + bx + c. an example of a quadratic is x² – 5x + 6. to find the factors of this equation, we must set up our set of two parentheses: ( )( ). the first term in both parentheses must be x, since x multiplied by x is the only way to get x². then we look at the coefficient of the second term, -5. it’s important to include the sign in front of the integer as part of the coefficient. one of the rules of quadratic equations is that the second terms in the two factors must add together to equal the middle term’s coefficient. so we need to think of two numbers that add together to give us -5.

already, we can think of many combinations: -6 and 1, -2 and -3, -200 and 105. so which pair is it? now we have to look at the integer that’s the third term of the quadratic. here it’s + 6. another rule of quadratic equations is that the third term of the quadratic equation will equal the product of the second terms in the two factors. so not only do we need the two numbers to add together to equal -5, but we need them to multiply together to equal + 6. therefore the factors must be: (x – 2) (x – 3). the “roots” or the “solutions” for this quadratic would be 2 and 3.

the act math test will often present you with two or more equations with multiple variables. remember the “n equations with n variables rule.” if there are 2 variables in an equation (for example, x and y), then there must be 2 equations that each contain those variables in order to solve. the two common ways to solve are substitution and combination.

logarithms are a unique way of writing exponents. we’re used to seeing exponents in a format like y = x^a. in “logs” that equation is equal to log_x(y) = a. this is the most essential piece of information you’ll need to solve logarithms. you can get more practice with logarithms on purple math!

let’s try a practice logarithm problem, just like the ones you might see on test day:

given that log_xa = 2 and log_xb = 3, what is the value of log_x(ab)³?

1. 6
2. 15
3. 36
4. 54
5. 216

here, the term we are interested in, log_x(ab)³, is equivalent to 3log_x(ab).

this can also be expressed as 3log_xa + 3log_xb, and since we know the values of log_xa and log_xb, we can substitute to find the answer. log_x(ab)³ = 3log_xa + 3log_xb = 3(2) + 3(3) = 15 (choice b).

if you don’t know these logarithmic identities, you can still solve the problem by finding values for x, a, and b that satisfy the conditions. then, simply calculate the value of log(x)(ab)³. the easiest way to do this is to work with a base of 10, which would mean that x = 10, a = 100, and b = 1,000. we can then calculate the answer:

log_x(ab)³ = log(10)(100 * 1,000)³ = log(10)(1,000,000,000,000,000) = log(10)(10¹⁵) = 15. the answer is (b).

act algebra review: what to do next

now it’s time for you to put your head down and study hard. if you have a mountain of studying to do, don’t despair! i strongly recommend that you break it down into manageable, small sections so that you can learn effectively while avoiding burnout.

good luck!

ready for more advanced stuff? check out our act trigonometry guide!