Algebra Section 1: Simplifying Rules

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I.  Simple Equations
II.  Complex Expressions with Exponents
III. Manipulating Complex Expressions
IV. Inequalities


I.  Simple Equations

A. Fundamental Definitions
B. Single-Variable Equations
C. Solving Two Variables
D. More than Two Variables
E. Solving for Expressions
F. Equation Trick Questions


A. Fundamental Definitions

	NOTE: Shortcuts in Algebra Keep in mind that the GMAT is NOT a standard achievement test. The test measures your resourcefulness and intellect in solving problems, not the methods achieved to do so. This means that you should use shortcuts whenever you can. This means that you should freely use ALL THE ADVANCED TECHNIQUES, such as Plug in, Backsolving and Experiments (see math introductory chapter). Some GMAT questions are designed to be solved this way.... and doing the conventional algebra will take too much time. Don't be afraid to start Backsolving or Plugging in as soon as you hit a brick wall.

B. Solving Single Variable Equations (Beginner)

You goal in algebra is usually to take complex equations and reduce them to something resembling a single variable equals a constant, such as x = 4.

But getting to that point on the GMAT will rarely be easy! You'll have to add, subtract, divide, multiply, square (or even cube) both sides of an equation. View equations as a balancing act. As long as you do something to both sides, you are all set.

In addition to manipulating both sides of the equation, you'll have to Distribute and Combine Like Terms to help you get to the simple final expression of "x = ?".

C. Multiplying Expressions
To multiply a monomial by a monomial, multiply the numerical coefficients and then follow the laws of exponents with the same base. For example:

To multiply a polynomial by a monomial, multiply each term of the polynomial by the monomial. This is illustrated by:

4x(3x² - 2xy + y²) = 12x³ - 8x²y + 4xy²

To multiply two polynomials, multiply one of them by each term of the other and then combine like terms. The following illustrates the process:

x/3 + x/2 = 5

a) 2x + 4y = -10
b) 5x + 3y = -11

How are you going to get that to look like "x = -1" when you have two variables?

There are two methods: Substitution and Addition.

With Substitution, you solve for one variable, and then you substitute it in wherever that variable appears to help solve the other unknowns.

a) 2x + 4y = -10
b) 5x + 3y = -11

Take the first of the two equations:

Ok, now substitute (-5 - 2y) in for x in the next equation.

Solve for x and y using the substitution method:

    x - y = 2
2x + y = -5

Addition

You may solve equations with more than one variable by adding the statements.

1. Multiply one equation by a properly chosen number so that one of the variables has the same coefficient in both equations.
2. Add or subtract the equations so that one of the unknowns is eliminated.
3. Solve the resulting equation for the remaining unknown.
4. Substitute the value of this known unknown into either original equation and solve for the second unknown.

A) 2x + 4y = -10
B) 5x + 3y = -11

Solve for x and y

   x - y = 2
2x + y = -5

*This time use the addition or subtraction method (not the substitution method).*

Solution We simply add the two equations since this will eliminate y:     x - y = 2     + 2x + y = - 5     3x = -3     x = -1 Substitute this value (x = -1) back into the first equation (x - y = 2) and solve for y:     (-1) - y = 2     -y = 3      y = -3

1: x + y - z = 4
2: x - 2y + 3z = -6
3: 2x + 3y + z = 7

Use the same strategy you used before. We need to eliminate a variable and use substitution. If you look at equations 1 and 3, you'll notice a +z and a -z. If you combine the two you will get rid of z and be able to substitute.

Now we have two equations with which to use for solving for x or y.
A: 4x + y = 6
B: 3x + 4y = 11

To solve, multiply the equation A by 4 and we can cancel out the y's:

Now, plug in x = 1 into equation (a) above:

• If x = (5 - y)/3, what is 3x + y?

Instead of trying to isolate a variable on one side such, you must try to isolate the expression the question wants.

Use any trick at your disposal including the squaring of both sides, taking the root of both sides, multiplying, dividing, etc... You may also factor or distribute one side to help break down the expressions into what the question wants.

If x = (5 - y)/3, what is 3x + y?

By now you should have a feel for Mr. GMAT's personality. The GMAT doesn't ask questions that are as simple as just two equations and two variables... no, the test isn't that easy. Instead the GMAT will throw curve ball after curve ball at you.

These types of trick questions are very common on the Data Sufficiency section. They prey on your false assumptions that as long as you have two equations, you have enough information to solve for two variables.


Trick #1: Repeating Equations

Is it possible to solve for x using these equations?

A: 4x + 12y = 24

B: x/3 + 1y = 2

The quick assumption is that the two statements can solve for x. The reality is that, if you look closely, both statements are the same thing: A is simply 12 times B. In reality, you don't have two equations, you simply have two different versions of the same one. This means that the two statements aren't sufficient while normally they would be.


Trick #2: Solve for Three Variables with Two Equations

You need two equations to solve for two variables, right? Three equations can solve for three variables? - not right. The GMAT will specifically use tricks to throw you off.

Can you solve this equation for z?

A: 2x + 4y = 12

B: z/(x + 2y) = 2

Now surely we can't solve for z? There are three variables (x, y, z), yet only two equations?

The reality is that there is a trick here. The 2x + 4y = 12 in equation A may be divided by 2 to make x + 2y = 6 (which is in equation B). The result is that we have z/6 = 2 in statement B. So you have enough information to determine that z = 12.

Algebra Section 2: Complex Expressions with Exponents

Table of Contents

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A. Exponents and Multiple Answers
B. Simplifying Bases
C. The Products of Monomials and/or Polynomials
D. Quadratic Equations: Multiplying Binomials
E. Factoring Algebraic Equations

A. Exponents and Multiple Answers
Life is never easy for MBA applicants. One of the GMAT's more difficult tricks is the "dual answer". You would think that a question like x² = 25 would be simple, but it is not because there are two answers: -5 and +5. Any number raised to an even numbered exponent will always be positive. The reason for this is that -5 × -5 is 25.

	The "no answer" trick x2 + 25 = 0 This is a "trick" statement because there isn't an answer to it. This breaks down into x2 = -25. x2 can't be a negative number. Therefore, x2 + 25 = 0 can't have a real answer.

Solving for Dual Answers
Absolute value and even exponent equations questions will make any negative value positive, so they require you to construct two possible answers.

1. Try to isolate the expressions within the absolute value or the even exponent.
2. Solve for two scenarios, one where the value within the absolute value/even exponent is negative, one where it is positive.
3. Solve for the negative value by replacing the absolute value with a parenthesis and a negative sign outside of it.

Try to solve this equation for both solutions: 6 - 5|x - 1| = 1

6 - 5(x - 1)² = 1

B. Simplifying Bases
When you are dealing with exponents on both sides of an equation, try to get terms on the same base or exponent so that you may properly compare them. When the bases are the same and the exponents are variables, you may get rid of the bases and simply solve for the exponents as variables in an equation.

Solve for x:
(9^x) = 81^{x - 1}

C. The Products of Monomials and/or Polynomials
A polynomial is an expression in which constants and variables are combined using only addition, subtraction, and multiplication. There are different types of polynomials, such as a binomial, which is a polynomial with two terms such as (x + 3) or (x + a).

(x + 3)(x - 5) =

E. Factoring Polynomials
If (x + 4)(4x + 4) = 4x² + 20x + 16, then aren't (x + 4) and (4x + 4) factors of 4x² + 20x + 16 in the same way that 3 and 2 are factors of 6?

Factor this equation: x² - 2x - 15

Common Factoring Shortcuts
There are several algebraic expressions that occur frequently. You should be able to recognize their factors immediately. Here are five common shortcuts.

Factor 4x² - 9

Factor r² - 8r + 16

Factor y² + 6y + 9

x² + 8x + 16 = 0 is a clear example of a quadratic equation. But, obviously, the Mr. GMAT is rarely so nice as to put it in that friendly form. To solve these equations you will have to take equations and convert them into the standard ax² + bx + c = 0 format.

	NOTE: Very Sneaky Unsolvable Equations Does this equation have a solution? (x2 - 4x + 4) = 0 (x - 2) If you solve the numerator, you get x = 2 (easy enough?). BUT, did you notice the (x - 2) in the denominator? Guess what, if the value is 2, then (x - 2) is ZERO in the denominator! You know what that means.... the answer is U N D E F I N E D. You can't have zero in the denominator. If you are clever you can find the exception here: (x2 - 4x + 4) (x - 2) Unless Mr. GMAT is pulling another fast one on me, using +2 for x means that (x - 2) in the denominator = 0, which means..... U N D E F I N E D. (x - 2)(x - 2) (x - 2) BUT, factor x2 - 4x + 4 into (x - 2)(x - 2) and look what happens.... (x - 2)(x - 2) (x - 2) Because we canceled out the (x - 2) in the denominator, guess what? +2 is actually a valid answer and the solution! +2 equals 0 in the numerator and there is no denominator to worry about because we canceled it out.

Solve x² + 4x = 0

Solve x² - 5x + 4 = 0

Solve 2x - 3 = (4/x) + x.

Solve a(a + t) = b² - bt for t

Algebra Section 3: Manipulating Complex Expressions

A. Division of Algebraic Expressions / Fractions
B. Addition of Algebraic Fractions
C. Inequalities
D. Algebra Word Problems


A. Division of Algebraic Expressions

There are three common divisions that you may be asked to perform.

1. Division by Monomials
2. Division by Binomials
3. Division by Algebraic Fractions.
   

1. Division by Monomials
When you divide by monomials, try to factor out the monomial in the denominator.

3x + 6ax2 + 15xa2

3x

3x + 6ax2 + 15xa2 3x	Factor 3 out of the numerator and denominator
x + 2ax2 + 5xa2 x	Factor x out of the numerator and denominator
1 + 2ax + 5a2

2. Division by Binomials

5x2 - 15x + 10 5x - 10
5(x2 - 3x + 2) 5(x - 2)	Factor out 5
5(x2 - 3x + 2) 5(x - 2)	Then cancel out the 5's
(x - 2)(x - 1) (x - 2)	You may factor out (x2 - 3x + 2) using FOIL. What two numbers added together give - 3 (the middle value) and when multiplied together give 2? They are -2 and -1. Hence, we factor it into (x - 2)(x - 1)
(x - 2)(x - 1) (x - 2)	Now you may cancel out the (x - 2)'s.
x - 1	You are left with x - 1.

Example

2m2n - 6mn 2m
2m2n 2m - 6mn 2m	Split up into two parts.
mn - 3n	Cancel out the 2m denominators into the numerators.

3. Dividing Algebraic Fractions

When dividing algebraic fractions:

1. When dividing, invert the denominator (flip the fraction upside down).
2. Search for common factors common to both numerator and the inverted denominator.
3. Cancel factors.
4. Multiply the numerator by the inverted denominator.
5. Simplify if possible.

Perform the division:

3a - 4 2 - 3a 3ab - 4b 4 - 9a2	Invert the divisor to make: 4 - 9a2 3ab - 4b
3a - 4 2 - 3a × 4 - 9a2 3ab - 4b	Factor any factorable expressions. 4 - 9a2 can be factored and so can 3ab - 4b.
4 - 9a2 = (2 - 3a)(2 + 3a) 3ab - 4b = b(3a - 4) 3a - 4 2 - 3a × (2 - 3a)(2 + 3a) b(3a - 4)	Factor 4 - 9a2 and 3ab - 4b. .
3a - 4 2 - 3a × (2 - 3a)(2 + 3a) b(3a - 4)	Cancel out factors: (3a - 4) and (2 + 3a) All that's left is....
2 + 3a b

B. Addition of Algebraic Fractions

When adding (or subtracting) algebraic fractions, we proceed as in basic arithmetic:

1. Find the lowest common denominator (LCD) of the fractions.
2. Write each fraction using the LCD of the fractions.
3. Add (or subtract) the numerators.
4. Simplify and reduce the resulting fraction.

Express as a single fraction:

Algebra Section 4: Inequalities (Beginner)

4. Inequalities

An inequality is simply a comparison of two quantities or expressions.

Inequalities and Algebra
The inequality 3x + 2 > x - 6 is solved just as an algebraic equation is solved.

There are several rules that we must follow when manipulating inequalities:

• The same number or algebraic expression may be added or subtracted from each side of an inequality.
• The same positive number (or positive algebraic expression) may multiply or divide each side of an inequality.
• Both sides of the same type of inequality may be added and the inequality remains.
  (If x < y and w < z, then x + w < y + z).
• If a negative number (or negative algebraic expression) multiplies or divides each side of an inequality, the inequality sign must be reversed. (Be sure to remember this; it often leads to errors!)
  

Solve the inequality 2x - 2 > x - 5.

Solve the inequality 3r + 5 > 6r - 7.