Algebra Basics for Graduate School Entrance Exams: A Comprehensive Guide, Study Guides, Projects, Research of Mathematics

Download Algebra Basics for Graduate School Entrance Exams: A Comprehensive Guide and more Study Guides, Projects, Research Mathematics in PDF only on Docsity!

Chapter 4

The Basics of Algebra

ALGEBRA

If you’re planning to attend graduate school, you’ve probably had some sort of algebraic training in the dark reaches of your past. Algebra is the fine art of determining how variable quantities relate to each other within complex functions, and it dates back more than 3,000 years to ancient Babylon.

If it seems like it’s been 3,000 years since you last studied algebra, or even if you flunked algebra last year, fear not. This chapter is devoted to reintroducing you to the basic algebraic operations that you will need to execute on test day.

In the next chapter, we will outline for you ways to subvert the rules and solve algebraic problems with strategies designed to reduce the amount of algebra you have to do. But, it is impossible, and not in your best interests, to ignore algebra altogether. And even on problems that algebra can be subverted, many times these problems can be made easier to work with after applying some basic algebra work.

Know the Lingo

Any letter in an algebraic term or equation is called a variable ; you don’t know what its numerical value is. It varies. Until you solve an equation, a variable is an unknown quantity. Any number that’s directly in front of a variable is called a coefficient , and the coefficient is a constant multiplied by that variable. For example, 3 x means “three times x ,” whatever x is.

Combining Like Terms

If two terms have the same variables or series of variables in them, they’re referred to as like terms. You can combine them like this:

6 a + 4 a = 10 a

13 x − 7 x = 6 x

For example, if you have six apples in one hand and four in the other, you have a total of 10 apples (and a pair of humongous hands).

Solving an Equation

Whenever a variable appears in an equation and you have to find the value of the variable, you have to “isolate” it. To isolate a variable, you must use mathematical operations to put all the terms that contain the variable on one side of the equals sign and all terms that don’t contain the variable on the other. Then you’ll need to manipulate the side of the equation with the variable to find the value the question asks for.

In order to solve an equation, you must rely on the following paramount rule of algebraic manipulation:

You can do anything you want to an equation as long as you do exactly the same thing to both sides.

4 m + 7 ( + 2 m ) = 16 − 2 m ( + 2 m )

6 m + 7 = 16

6 m + 7 ( − 7) = 16 ( − 7)

6 m = 9

Now isolate the variable by dividing both sides of the equation by 6.

Check your work just to make sure.

Solving an Equation Quick Quiz

Solve for x in each of the following equations:

− −

− − −

Explanations for Solving an Equation Quick Quiz

1. Add 5 to both sides of the equation and then divide both sides of the equation by 2 to find that x = 8.
2. Subtract 3 from both sides and then divide both sides of the equation by −5 to find that x = −2.

3. Subtract 12 x from both sides of the equation to find that 4 = −8 x. Divide both sides of the equation by −

to find that x = −.

4. Subtract x from both sides of the equation to find that 7 x − 9 = −2. Now add 9 to both sides of the equation to find that 7 x = 7. Divide both sides of the equation by 7 to find that x = 1.
5. Subtract 5 from both sides of the equation and then multiply both sides of the equation by 2 to find that x = −28.

Inequalities

Inequality symbols are used to convey that one number is greater than or less than another.

The symbols used in inequalities are as follows:

means “is greater than” < means “is less than” ≥ means “is greater than or equal to” ≤ means “is less than or equal to”

Even though the two sides of an inequality aren’t equal, you can manipulate them in much the same way as you do the expressions in regular equations when you have to solve for a variable.

− ○ − ○ ○ ○ ○

Here’s How to Crack It Adding and subtracting take place as usual, like this:

5 b − 3 > 2 b + 9

3 b − 3 > 9

3 b > 12

At this point, because the coefficient of b is positive, divide both sides by 3 and get the final range of values for b.

Any value that is greater than − is a possible value of p , so you should check the box next to 0 and every other box

with a number that’s greater than zero. See how important that little rule is? If you didn’t know about it, you might

have picked all the numbers that were less than − , which would be the exact opposite of the correct answers. And

that would have been unfortunate.

Range of Inequalities

Some GRE questions present two inequalities and ask for the range of possible values when the inequalities are combined through addition, subtraction, multiplication, or division. The range of possible values is the greatest distance between two possible values that satisfy the inequalities. So, answering a question that asks for the range of possible values for two separate inequalities is as easy as finding the least and the greatest possible values and placing those values as a part of an inequality.

To find the least and greatest possible values, you’ll need to combine the end ranges of the individual inequalities. Combine the least number for each inequality, the least number of the first inequality with the greatest number of the second inequality, the greatest number of the first inequality with the least number of the second inequality, and the greatest number for each inequality. This will result in four numbers. The range of the combined inequalities will be the difference between the greatest and the least of those four numbers.

− ≤ ≤ − ≤ ≤ ○ − ≤ ≤ ○ − ≤ ≤ ○ − ≤ ≤ ○ ≤ ≤ ○ ≤ ≤

Here’s How to Crack It For the range of a , there are two numbers: −5 and 12. Make sure that each of those numbers combines with each of the two numbers from the range of b. a – b (−5) – (−10) = 5 (−5) – (25) = − (12) – (−10) = 22

Notice how each number from a was paired up with each number from b. The least result for a – b is −30, and the greatest is 22. The full range is −30 ≤ a – b ≤ 22, and the correct answer is (A).

Inequalities Quick Quiz

For questions 1−3, find the range of values for x.

−

− ≥ −

≥ − − ○ ○ ○ ○ ○

− ≤ ≤ − ≤ ≤

Explanations for Inequalities Quick Quiz

1. Subtract 7 from both sides of the inequality and divide both sides by 3 to find that x > 5.
2. Subtract 8 from both sides to find that −3 x < 5. Now divide both sides of the inequality by −3 and flip the

sign to find that x > −.

3. Subtract 2 from both sides of the inequality and then multiply both sides by −2, remembering to flip the sign, to find that x ≤ −10 x + 22. Now add 10 x to both sides to find that 11 x ≤ 22. Divide both sides of the inequality by 11 to find that x ≤ 2.

Solving Quadratic Equations

When quadratics and equals signs come together, the result is a quadratic equation. On the GRE, quadratics are usually set equal to zero, and you’ll have find an equation’s solutions, or roots, by factoring.

− −

Here’s How to Crack It To answer this question, its necessary to find the value of x. When dealing with a quadratic, this is called finding the roots of the equation. There is more than one possible root for a quadratic equation, so find them both by factoring the original equation. Factoring is basically the opposite of FOILing, and it usually requires a little trial and error. In this case, the challenge lies in finding two numbers whose sum is −5 (the middle coefficient) and whose product is −6 (the last term). When you find those numbers, place them inside parentheses with the variable x.

x^2 − 5 x − 6 = 0

( x )( x ) = 0

( x − 6)( x + 1) = 0

In order for the product of two numbers to be 0, one of them must be 0. So set both factors equal to 0 and solve for x.

x − 6 = 0 x + 1 = 0

x = 6 x + 1 = 0 x = −

Check your answers to make sure you didn’t make any unfortunate slip-ups.

62 − 5(6) − 6 = 0 (−1)^2 − 5(−1) − 6 = 0

Now, 6 and −1 are the roots of the equation, but the question asks for only the positive root. So the answer is 6.

Common Quadratics

As basic as FOILing is, there are a few very common multiplications of binomials that come up so frequently that you’re better off memorizing them. The writers of standardized tests like them a lot, because they’re great for making easier problems seem a lot more difficult.

( x + y )^2 = x^2 + 2 xy + y^2

( x – y )^2 = x^2 − 2 xy + y^2

( x + y )( x – y ) = x^2 – y^2

Knowledge of the last formula—which is commonly referred to as a “difference of squares”—is especially useful if you come across a question that looks like this:

Here’s How to Crack It

Rather than resort to cross-multiplication here, recognize that the fraction on the left is in the form of ; in this

case, y is the constant. Use the difference of squares to simplify the numerator. Because. Cancel

out x − 3 from both the numerator and denominator leaving only x + 3. Now, the original question asks for the value

of x if x + 3 = 7, so x = 4.

Here’s another one on which a little quadratic knowledge can save you some time.

○ ○ ○ ○

Here’s How to Crack It The numerator of Quantity A is a difference of squares, so it can be rewritten as (836 + 835)(836 − 835). Cancel out 836 + 835 from both the numerator and denominator of Quantity A. Therefore, the value of Quantity A is 836 − 835 =

1. The quantities are equal and the correct answer is (C).

From here, you can determine that x = 5.

When There’s Less Work Than You Think

ETS likes to build its simultaneous equation questions to look a lot more daunting and time-consuming than they actually are. Take this problem for example.

At first glance, you might think you have to solve for a and b individually and then add them together to get your final answer. But that isn’t the case.

Here’s How to Crack It Add the two equations together to find a new equation.

Factor out the 2 from the left side of the equation and divide each side of the equation by 2.

2( a + b ) = 18

a + b = 9

The question asks for the value of a + b , so the correct answer is 9.

Simultaneous Equations Quick Quiz

− − −

− −

Explanations for Simultaneous Equations Quick Quiz

1. Solve for y by stacking the equations and finding a way to eliminate the x ’s. After stacking, multiply the second equation by −2 so that it becomes −2 x + 10 y = 12. Now, add the equations to eliminate the x ’s and solve for y.
2. Solve for a first by stacking the equation and multiplying the first equation by 2 to yield 6 a − 2 b = 400. Now add the equations for eliminate b and find the value of a.

Find the value of b by replacing the variable a with 40 in either of the original equations to find that 3(40)

• b = 200, so b = −80. Therefore, a – b = 40 – (−80) = 120.

3. Stack and add the equations to find the value of m – n.
4. This is a simultaneous equation problem masquerading as a word problem. If three pens and five notebooks cost a total of $19.02, then 3 p + 5 n = 19.02. Similarly, the second equation can be rewritten as 2 p + 3 n = 11.63. To find the value of n , stack the equations and find a way to eliminate the p ’s.

Each notebook costs $3.15.