Algebra for the DAT
/Learn key DAT concepts about algebra, plus practice questions and answers
Table of Contents
Part 1: Introduction to algebra
Part 2: Expressions and equations
a) Basics
b) Factoring
Part 3: Inequalities and absolute values
a) Inequalities
b) Absolute values
Part 4: Square roots, exponents, and logarithms
a) Square roots
b) Exponents
c) Logarithms
Part 5: Ratios and proportions
Part 6: Helpful terms
Part 7: Questions and answers
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Part 1: Introduction to algebra
Algebra skills are crucial on the DAT. You will likely get some algebra-specific questions on the QR section, and understanding algebra will help you set up QR word problems. You may even find algebra useful in the general chemistry section.
Whether you took advanced math classes in college or haven’t done any math since high school, you can do well on the QR section. This guide will help you review the basics of algebra as well as important DAT-relevant algebra rules. Once you’ve reviewed the concepts, practice makes progress. Be sure to work through each of the practice questions at the end of this guide.
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Part 2: Expressions and equations
a) Basics
Algebra will show up in various forms throughout the quantitative reasoning section of the DAT, so it’s important to have the basics down. Simply put, algebra is math expressions and equations that use symbols or letters to represent numbers. This commonly includes solving equations for x or y, but it can get more complicated. Let’s start with some definitions and then go over some basic functions.
Algebraic expressions are made up of terms. A term is a number, letter, or other variable. Terms sometimes include coefficients, which are numbers that multiply variables. Terms are strung together to form expressions, and like terms can be combined to simplify expressions and solve equations. Like terms have the same variable raised to the same exponent, but may have different coefficients. For example, 3x and 17x are like terms because they both contain an x to the power of 1. 3x and 3x2 are unlike terms because x is raised to a different power.
FIGURE 1: LABELED EXPRESSION
Multiple terms can be strung together using operators like addition or multiplication signs to form an expression. A common algebraic expression is a polynomial. A polynomial consists of multiple terms connected by an operator, but they usually include variables like x or y, especially when raised to a higher power.
Let’s look at how to combine like terms for polynomial expressions, starting with addition. Here is an example:
To simplify this expression, remove the parentheses and add the like terms—it’s that simple.
Now for an example using subtraction:
Notice the placement of the parentheses. Because there are parentheses after the subtraction sign, all of the terms inside the parentheses are being subtracted. In other words, we need to distribute the negative sign to all terms in the parentheses and then combine like terms. Don’t forget that distributing a negative sign to a negative number makes the number positive.
\[2x^2-3x-4x^2-2x+4\]
\[-2x^2-5x+4\]
Now, we have a much simpler expression. Adding and subtracting polynomial expressions is fairly straightforward, so let’s look at a few examples of algebraic multiplication.
\[3(3x+6)\]
In order to expand this expression, each term inside the parentheses must be multiplied by the term outside. This is known as distributive property. Just like when subtracting polynomials and distributing the negative sign to each term within the parentheses, we need to multiply each term inside the parentheses by 3.
\[9x+18\]
Sometimes, you will need to multiply two polynomial expressions together. Here is an example.
\[(x+3)(2x-2)\]
In this scenario, like we did previously, each term in the parentheses needs to be multiplied. This example, however, has multiple terms in both parentheses. Therefore, we need to multiply each term in the first set of parentheses by each term in the second, and then combine like terms.
FIGURE 2: FOIL METHOD FOR MULTIPLYING POLYNOMIAL EXPRESSIONS
Dividing polynomials can become complicated and in some cases is beyond the scope of the DAT, but you should be familiar with the following concept. If the same polynomial is found in the numerator and denominator, they cancel each other out. For example:
\[\frac{(x + 2)(x - 4)(x + 3)}{(x + 2)(x + 3)}\]
Notice that both \((x+2)\) and \((x+3)\) are found in the numerator and denominator, so they can both be removed from the expression and we are left with \((x-4)\).
Now, let’s apply what we’ve learned about expressions to an equation. An equation is a statement that one expression is equal to another expression. To solve equations, move like terms to the same side of the equals sign and combine. The principle to remember is that whatever you do to one side, you must do to the other. Let’s look at an example.
\[4x-6=2\]
Start by adding 6 to both sides of the equation. Doing so will cancel the negative 6 from the left side of the equation.
\[4x=8\]
Now, divide both sides of the equation by 4 to solve for x.
\[x=2\]
Another type of problem that you need to know for the DAT relates multiple equations in a system. A system of equations is two or more equations with variables that have a common value. For example, \(x+3y=5\) & \(4x-3y=5\) is a system of equations. Notice that each equation has two variables, x and y, and that we need both equations to solve for each variable. The value of x will be the same for the first equation as it is for the second equation; the same goes for the y value.
There are two methods to solve systems of equations-elimination and substitution. Each method aims to create a solvable equation with one variable instead of two.
To use the elimination method, start by adding both equations and adding like terms. You may need to multiply one or both of the equations in order to eliminate a variable. Let’s look at the previous example:
\[x+3y=5\]
\[4x-3y=5\]
If we wanted to eliminate the x variable first, we could multiply each term of the top equation by -4. This would give us -4x for the first equation and 4x for the second, eliminating x entirely. We don’t need to do that though because the current y terms combine to 0 and can be eliminated without manipulating either equation. After adding both equations, we have one equation that we can easily solve.
\[5x=10\]
\[x=2\]
After solving for one variable, we need to solve for the other. To determine the y value, plug the x value into either equation. Let’s plug x=2 into the first equation:
\[2+3y=5\]
Now we can combine like terms and solve for y.
\[3y=3\]
\[y=1\]
Our answer is x=2, y=1. This can also be expressed as (2,1). You can always check your answer by plugging these values back into one of the equations.
Let’s solve a different problem using the substitution method.
\[3x-y=8\]
\[x+3y=16\]In order to use the substitution method, we need to determine what one of the variables is equal to. Then plug that value into the other equation, substituting one of the variables for the other. This enables us to combine like terms and solve for one of the variables.
The bottom equation can be rearranged by subtracting 3y from both sides.
Plug -3y+16 into the first equation where the x variable is.
Multiply and combine like terms to solve for y.
\[-9y+48-y=8\]
\[-10y=-40\]
\[y=4\]
Just like we did when using the elimination method, plug the y value into one of the equations, and solve for x.
\[3x-4=8\]
\[3x=12\]
\[x=4\]
You can use either method to solve systems of equations, but some problems are solved easier using one method over the other. Try to use the method that solves the system the quickest so you have more time for other QR problems.
b) Factoring
Factoring is a useful skill that you need to know for the DAT. Factoring is a way to simplify an expression or show it a different way, and it involves finding the greatest common factor (GCF). The GCF is the largest whole number that can divide each number. Here is a simple example of factoring:
Each term is divisible by 12, so we can factor this expression to 12(x + 3). Notice that we can expand this back to the original expression if needed. You can also factor out variables, such as in this example:
Each term has at least one x, and each coefficient is divisible by 4. Therefore, we can factor out 4x and our expression becomes 4x(x+2).
Factoring isn’t just used to simplify expressions; it can also be used to solve equations more easily. Let’s set the previous example equal to 0 and solve for x.
At first glance, this may seem like the type of problem where you have to guess and check. However, after factoring out 4x, it becomes much simpler to solve.
Because there are two simplified terms containing variables, we set each equal to 0 and solve for x. This particular example has 2 answers because it is a parabola (for a review on graphical analysis, see the guide titled graphing and geometry)
\[4x=0\]
\[x+2=0\]
Here is another factoring example:
To factor this expression, we first need to realize that each term is a perfect square. The square root of x2 is x, and the square root of 36 is 6. This expression factored is to (x+6)(x-6). Note that this method of factoring is known as the difference of perfect squares, and will not work if the perfect squares are added together.
Factoring is also helpful in solving quadratic polynomials. A quadratic polynomial is one that follows the form of ax2+bx+c. In the graphing and geometry guide, we will review the quadratic formula—one method to solve quadratic equations—but for now we will focus on using factoring.
\[x^2+2x-8\]
To factor the expression above, start by looking at the last term, -8. We need two numbers that, when multiplied together, will equal -8. Our options are -1 and 8, 1 and -8, -2 and 4, and 2 and -4. The next step is to determine which of these sets, when added together, will equal the coefficient of the middle term, +2. 4 + (-2) is the only set that adds to 2. x2 can be factored as x and x because, when multiplied together, x and x will equal x2. Therefore, the factor for this expression is (x-2)(x+4). We can check our answer by multiplying these expressions together.
FIGURE 3: EXPANDING A FACTORED QUADRATIC POLYNOMIAL
Let’s look at another expression.
\[2x^2+2x-12\]
In this example, we can see that this polynomial has a gcf. Each coefficient or constant is divisible by 2, so start by factoring out the 2.
\[2(x^2+x-6)\]
Now we can factor this polynomial similar to the previous problem, by finding which two numbers multiply to -6 and add to 1.
\[2(x+3)(x-2)\]
Here is one more example:
\[2x^2+x-6\]
This expression may look similar to the previous one, but notice that this expression does not have a gcf. To factor, start with the 2x2 term. The only way we can get 2x2 is by multiplying 2x and 1x, so let’s put them in separate parentheses.
\[(2x\_\_)(x\_\_)\]
We still need two numbers that, when multiplied, equal -6. Our options are 1 and -6, -1 and 6, -2 and 3, 2 and -3. In a previous example, we chose the pair that, when combined, equaled the coefficient of the middle term. Let’s try that for this problem and then check our answer by expanding the polynomial.
\[(2x-2)(x+3)\]
\[2x^2+6x-2x-6\]
\[2x^2+4x-6\]
Obviously, this expression isn’t the same as what we started with, so let’s try switching the places of the numbers.
\[(2x+3)(x-2)\]
\[2x^2-4x+3x-6\]
\[2x^2-x-6\]
This is almost correct, but the sign of the middle term is wrong. To fix this, let’s change both signs of the factored expression.
\[(2x-3)(x+2)\]
\[2x^2+4x-3x-6\]
\[2x^2+x-6\]
This was the expression we started with, so we know that this is the correct factor. This method of guessing and checking is tedious, but you can use it if you prefer. There is, however, another method to factor this type of polynomial.
This method involves multiplying the a (2) and c (-6) terms. In this case, that would equal -12. Next, determine 2 numbers that when multiplied will equal -12 and when added will equal the coefficient of the b term (1). -3 and 4, in this case, would satisfy both conditions, so we rewrite our expression as follows:
\[2x^2-3x+4x-6\]
Now, separate the first 2 terms from the last 2 and see if each set has a GCF.
\[2x^2-3x\]
\[x(2x-3)\]
\[4x-6\]
\[2(2x-3)\]
After factoring out the GCF for each set, we see that both have 2x-3. That will be the first part of our factored expression. The variable and number that we factored out, when combined, will be the other part of the factored expression, so our answer is (2x-3)(x+2).
FIGURE 4: FACTORED QUADRATIC POLYNOMIAL
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