This chapter is divided into four parts:
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
Variable |
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Algebra is all about variables. These are unknown quantities that are usually expressed as a letter such as x or y. |
Coefficient |
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This is a number next to a variable that describes how many of the variable there are. For example, 4x, where 4 is the coefficient. |
Constant |
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These are fixed numbers. 4x + 7, where 7 is the constant. |
Equation |
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This is where at least two expressions separated by an equal sign.
eg. 4x = 4y + 2. |
Solution/Root |
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What values for the variables will solve the equation? |
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NOTE: Shortcuts in Algebra
Keep in mind that the GRE 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 GRE 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.
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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 GRE 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.
2x + 7 = 19 |
Subtract 7 from both sides. |
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2x = 12 |
Divide both sides by 2. |
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x = 6 |
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√x + 1 = 10 |
Subtract 1 from both sides. |
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√x = 9 |
Square both sides. |
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x = 81 |
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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 = ?".
| Distribute |
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a(b + c) = |
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(a × b) + (a × c) = |
Distribute the a value by multiplying b and c. |
Combining Like Terms |
2x + 3x + 7 = 15 |
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5x + 7 = 15 |
Combine the x's.
Note: When you combine terms they must have the same exponent and variable. |
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:
(2rs3)(-3r3s2) = ? |
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| (2 × -3) = -6
(r × r3) = r4
(s3 × s2) = s5 |
Multiply the like terms.
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Muliply together the results |
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:
(2x - y)(2 + x - 3y) |
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2x(2 + x - 3y) - y(2 + x - 3y) |
Distribute 2x and -y and (2 + x - 3y) |
4x + 2x2 - 6xy - 2y - xy + 3y2 |
Multiply out all the content |
4x - 2y + 2x2 - 7xy + 3y2 |
Combine like terms |
Then move all terms containing the unknown to one side and all terms that do not contain an unknown to the other side. Factor out the unknown from all terms that contain the unknown; finally, divide each side by the coefficient of the unknown.
x/3 + x/2 = 5
Solution
Multiply both sides by the Least Common Denominator, which is 6(3 × 2).
x/3 × (6) + x/2 × (6) = 5 × (6) |
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2x + 3x = 30 |
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5x = 30 |
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x = 6 | |
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D. Solving with Two Variables
Take a look at this: 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:
a) 2x + 4y = -10 |
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2x + 4y = -10 |
Divide by 2 |
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x + 2y = -5 |
Subtract -2y from both sides |
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x = -5 - 2y |
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Ok, now substitute (-5 - 2y) in for x in the next equation.
b) 5x + 3y = -11 |
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5x + 3y = -11 |
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Substitute x = (-5 - 2y) for x. |
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5(-5 - 2y) + 3y = -11 |
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Distribute the 5(-5 - 2y) to get -25 - 10y. |
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-25 - 10y + 3y = -11 |
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Combine y's - 10y + 3y. |
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-25 - 7y = -11 |
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+25 to both sides. |
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-7y = 14 |
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Divide by -7. |
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y = -2 |
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Not done yet! You can solve for x. We did get the simple equation for x above (x = -5 - 2y) and plug in y = -2. |
x = -5 - 2y |
Now substitute y = (- 2) for y. |
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x = -5 - 2(-2) |
Combine |
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x = -5 + 4 |
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x = -1 |
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Solve for x and y using the substitution method:
x - y = 2
2x + y = -5
Solution
Using the substitution method, the first equation gives:
x = 2 + y
Substitute this into the second equation and solve for y: |
2(2 + y) + y = -5 |
Distribute 2(2 + y) |
4 + 2y + y = -5 |
Combine y's, subtract 4 from both sides |
3y = -9 |
Divide by 3 |
y = -3 |
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Substitute this value (y = -3) back into the first equation (x - y = 2) and solve for x:
x - (-3) = 2
x = -1
The solution is x = -1, y = -3. |
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Addition
You may solve equations with more than one variable by adding the statements.
- Multiply one equation by a properly chosen number so that one of the variables has the same coefficient in both equations.
- Add or subtract the equations so that one of the unknowns is eliminated.
- Solve the resulting equation for the remaining unknown.
- Substitute the value of this known unknown into either original equation and solve for the second unknown.
Try to solve these two equations:
A) 2x + 4y = -10
B) 5x + 3y = -11
Look for the Least Common Multiples and Divisors for the variables in each equation. Your goal is to combine the equations and eliminate a variable in the process. Here we have 4y and 3y. Their LCM is 12. So set one to +12y and another to -12y to eliminate each other.
| Equation A |
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2x + 4y = -10 |
Multiply both sides by 3 |
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6x + 12y = -30 |
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Equation B |
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5x + 3y = -11 |
Multiply both sides by -4 |
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-20x - 12y = 44 |
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Now add the equations A and B |
6x + 12y = -30 |
Equation A |
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| -20x -12y = 44 |
Equation B |
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-14x = 14 |
Add both equations, canceling out y |
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x = -1 |
Divide by -14 |
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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:
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x - y |
= 2 |
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+ 2x + y |
= - 5 |
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3x |
= -3 |
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x |
= -1 |
Substitute this value (x = -1) back into the first equation (x - y = 2) and solve for y:
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(-1) - y |
= 2 |
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= 3 |
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y |
= -3 |
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D. Solving with More than Two Variables
Now we are going to have more fun..... three equations and three variables!
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.
3) 2x + 3y + z = 7 |
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Equation 3 |
1) x + y - z = 4 |
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Equation 1 |
3x + 4y = 11 |
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Add equation 1 and 3 |
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1) x + y - z = 4 |
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3x + 3y - 3z = 12 |
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Multiply equation 1 by 3 |
2) x - 2y + 3z = -6 |
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Then combine with equation 2 to get rid of z's |
4x + y = 6 |
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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:
4x + y = 6 |
Equation A |
-16x - 4y = -24 |
Multiply by -4 |
3x + 4y = 11 |
Equation B |
-13x = -13 |
Add equation A and equation B |
x = 1 |
Divide by -13 |
Now, plug in x = 1 into equation (a) above:
4x + y = 6 |
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4 + y = 6 |
Plug in 1 for x |
y = 2 |
Subtract 4 from both sides |
E. Solving for Expressions
Sometimes you don't have to solve for x, but for an expression like x + 3y.
- 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?
x = (5 - y)/3 |
multiply by 3 to get rid of fraction |
3x = 5 - y |
+y to get rid of -y |
3x + y = 5 |
you can stop here. |
F. Equation Trick Question Types By now you should have a feel for Mr. GRE's personality. The GRE doesn't ask questions that are as simple as just two equations and two variables... no, the test isn't that easy. Instead the GRE 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 GRE 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.
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