Elementary Algebra
Elementary Algebra
Unlike the section on pre-algebra, we’ve organized this elementary algebra section in terms of increasing difficulty:
  1. Substitution
  2. Simplifying Algebraic Expressions
  3. Writing Expressions and Equations
  4. Solving Linear Equations
  5. Multiplying Binomials
  6. Inequalities
This is also the order of the topics in terms of decreasing frequency on the test, with one exception: problems involving inequalities pop up more often than problems involving binomial multiplication.
Before covering these topics, however, we will address a question brought up by the teachings of some other test prep companies.
To Algebra or Not to Algebra
There are many ways to answer most algebra problems. You can use algebra—setting up and working out equations—or you can plug numbers into equations to try and avoid using algebra. In some cases, you might even be able to solve a question by being a particularly intuitive genius and finding a magnificent shortcut.
We want to stress that none of these methods is necessarily better than another. Which method is best for you depends on your math ability and your target score. Trying to solve problems with algebra is more conceptually demanding, but can take less time. Plugging in numbers makes questions easier to understand, but will likely take more time. In general, if you are uncomfortable with algebra, you should try to use the plugging-in method. If you are comfortable with algebra, using it is probably the best way to go. Still, these suggestions are not carved in stone. If you are generally comfortable with algebra but come upon a question that is stumping you, try plugging in answers. If you usually prefer plugging in answers but come upon a question you can answer using algebra, use algebra. When you study your practice tests and look at the algebra questions you got wrong, you should think about the method you employed. Did you plug in when you should have used algebra? Did you use algebra when you should have plugged in? As for being an intuitive math genius, it just can’t be taught—though we will show you how one might think.
Here’s a sample algebra question:
A man flipped a coin 162 times. The coin landed with the heads side up 62 more times than it landed with tails up. How many times did the coin land heads?
A. 100
B. 104
C. 108
D. 112
E. 116
Solving by Plugging In
If you were to answer this problem by plugging in numbers, you would pick the middle number or C, 108, as the first number to try, since if it does not happen to be the answer, you can discard the numbers smaller than it or larger than it. If the coin came up heads 108 times, then how many times did it land tails? It landed tails 162 – 108 = 54 times. Is 108 heads landings 62 more than 54 tails landings? No, 108 – 54 = 54. In order for the problem to work out, you need more heads landings. You can eliminate A and B as possibilities. Let’s say we choose D, 112, as our next plug-in number: 162 – 112 = 50. Does 112 – 50 = 62? Yes. D is the answer.
Solving with Algebra
If you answer this question with algebra, you realize that if heads are represented by the variable x, then tails are represented by (x – 62). Therefore,
As you can see, there’s simply less math to do for this problem when you use algebra. Using algebra will only take you longer than plugging in if you have trouble coming up with the equation x + (x – 62) = 162.
Therefore, if you can quickly come up with the necessary equation, then use algebra to solve algebra problems. If you have the feeling that it will take you a while to figure out the correct equation, then plug in.
Solving by Being an Amazing Genius
It is quite possible that you just looked at this problem and said to yourself, “Other than the 62 more heads, all the other flips were equally heads and tails. So if you take the 62 out of the total of 162, then you know that the other 100 flips were 50 heads and 50 tails. Now I can just add 62 + 50 = 112. Man, I am an amazing genius!”
The Bottom Line on Using Algebra
Hopefully, our example has convinced you that there isn’t any “right way” to answer a question dealing with algebra. There are faster ways and slower ways, and it always benefits you to use the faster way if you can, but the most important thing is getting the question right. Therefore, when you come to a question, don’t insist on using only one method to try to answer it. Just do what you have to do in order to answer the question correctly in as little time as possible.
Now we’ll begin to cover the topics of elementary algebra tested on the ACT.
Substitution questions are the simplest algebra problems on the ACT. These questions provide you with an algebraic expression and the value of a variable within the equation, and ask you to calculate the value of the equation. For example,
If 2y + 8x = 11, what is the value of 3(2y + 8x)?
You might see this question with all its variables and panic. But, in truth, this is a simple problem. Since 2y + 8x = 11, all you have to do is substitute 11 for 2y + 8x in the expression 3(2y + 8x), and you get 3(11) = 33.
For some substitution questions, you will have to do some simple math either before or after the substitution.
Math before Substitution
If 3x –7 = 8, then 23 – 3x =
In this problem you have to find what 3x equals before you can substitute that value into the expression 23 – 3x. To find 3x, take:
and add 7 to both sides, getting:
Now we can substitute that 15 into
Math after Substitution
If a + b = 7 and b = 3, then 4a =
Here we first have to solve for a by substituting the 3 in for b:
Once you know that a = 4, just substitute it into 4a:
Simplifying Algebraic Expressions
Some ACT Math questions test your ability to simplify or manipulate algebraic expressions. To master either of these skills, you must be able to see how an equation might be expressed differently without changing the value of the expression in any way. There are two primary ways to simplify an equation: factoring and combining like terms.
Factoring and Unfactoring
Factoring an algebraic expression means finding factors common to all terms in an expression and dividing them out. For example, to factor you simply divide out the 3 to get Below are some more examples of factoring:
Unfactoring involves taking a factored expression, such as and distributing one term to the other(s):
Combining Similar Terms
If an expression contains “like terms,” you can combine those terms and simplify the equation. “Like terms” refers to identical variables that have the same exponent value. For example:
  • You can combine:
As long as two terms have the same variable and the same exponent value, you can combine them. Note that when you combine like terms, the variable doesn’t change. If two terms have different variables, or the exponent value is different, the terms are not “like terms” and you cannot combine them.
  • You can’t combine: or
Writing Expressions and Equations
Occasionally, the ACT will throw you a word problem that describes an algebraic expression. You will have to write out the expression in numerical form and perhaps simplify it. For example:
Mary poured g cups of water into a bucket, leaving the bucket with a total of f cups in it. Mary then removed (g – 3) cups of water from the bucket. How many cups of water remain in the bucket?
To answer this question, you have to interpret the word problem. In other words, you have to figure out what is important in the word problem and how it fits into the expression you need to build. This question asks you to generate an expression that describes how many cups of water there are in the bucket after Mary removes (g – 3) cups. It doesn’t matter what g actually equals, because we don’t care how much water was in the bucket before Mary added g cups.
To work out the equation, we take the original number of cups in the bucket and subtract from it what was removed:
Solving Linear Equations
The most common and foolproof way to solve linear equations is to isolate the variable whose value you are trying to determine on one side of the equation.
If you stay alert, you may also be able to find shortcuts that will greatly reduce your time spent per question without affecting your accuracy. Let’s look at an easy example:
If 6p + 2 = 20, then 6p – 3 =
This is an easy problem to solve through the normal algebraic method. First we solve for p:
Next, we plug 3 into the second equation:
But there’s a faster way to answer this question. The secret is that you don’t have to solve for p at all. Instead, notice that both equations contain 6p and that the value of 6p will not change. Therefore, all you have to do in the first equation is solve for 6p. And as you can see above, that simply means subtracting 2 from 20 to get 18. Once you know 6p is 18, you can plug 18 in for 6p in the second equation and get your answer.
When you come upon an algebra question asking you to solve an equation, you should always take a second to look for shortcuts. Look for equations that not only have the same variables, but also the same coefficients attached to that variable (such as 6p and 6p). If you are able to find a good shortcut, your knowledge of algebra will save you time.
Multiplying Binomials
A binomial is an algebraic expression consisting of two terms combined by a plus or minus sign. For instance, x + 4 and y – 11 are both binomials. Multiplying binomials is not a difficult task if you remember the acronym FOIL, which stands for FIRST OUTER INNER LAST. For example, say you are asked to multiply the binomials:
You start by multiplying the first number in each polynomial (x)(x), then the outer numbers (x)(3), then the inner numbers (2)(x), and finally the last numbers (2)(3):
and you get:
The only tricky part to following FOIL is remembering to pay attention to signs. For instance, if you have the polynomials (x + 2)(x – 3), then the –3 comes to play an important part. You always add up the products of FOIL, but look what happens when there’s a negative number involved:
There are three equations involving binomial multiplication that you should know backward and forward before you take the ACT, the most important being the first:
An equation states that the quantities on either side of the equal sign are of the same value. An inequality states that one side of the equation is greater than the other: states that a is less than b, and states that a is greater than b. In other cases, means that a is less than or equal to b, while means that a is greater than or equal to b.
Solving an inequality is basically the same as solving a normal equation: all the rules of simplification and having to do the same thing to both sides still apply. The one rule that does differ when working with inequalities comes when you multiply or divide both sides by a negative. If you do so, you must flip the sign: if x > y, then –x < –y. For example, if you have and multiply the inequality by –2, the result is .
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