


Intermediate Algebra
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 onesixth of your
math score.
In this section, we’ll present the intermediate
algebra topics to you in the following order:
 1. Solving and Factoring Quadratic Equations
 Solving Systems of Equations
 Relationships between the Sides of an Equation
 Functions
 Matrices
 Logarithms
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.
Solving and Factoring Quadratic Equations
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.
Definition of a Quadratic Equation
A quadratic equation is a seconddegree 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
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.
The Quadratic Formula
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 .
Solving Systems of Equations
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:
 When you first solve for one variable, make sure you solve for it in its lowest form (solve for x rather than 2x).
 When you substitute, make sure you correctly apply the distributive law of multiplication: 3(2y – 6) = 6y – 18.
 Always answer the question the ACT asks. For example, in the sample above, the question asked for the value of x – y. But it’s certainly possible that after doing all the work and figuring out that x = 4 you might forget to carry out the final simple operation of 4 – 5 = –1, and instead incorrectly answer 4.
Systems of Equations with Infinite Solutions
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.
Relationships between the Sides of an Equation
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.
Functions
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.
Compound Functions
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:
Matrices
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.
Adding and Subtracting Matrices
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.
Logarithms
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.
