Add the two equations work down to a sum for each term, or subtract one equation from the other, to eliminate one variable.. LINEAR EQUATIONS THAT CANNOT BE SOLVED Never assume that one
Trang 16 for p in either equation to find q (we’ll use the second equation):
q 5 10 225
6
q 560
6 2
25 6
q 535
6
The question asks forp
q, so do the division:
p
q5
25
6
35
6
525
35, or
5 7
The Addition-Subtraction Method
Another way to solve for two unknowns in a system of two equations is with the
addition-subtraction method Here are the five steps:
Make the coefficient of either variable the same in both equations (you can
disre-gard the sign) by multiplying every term in one of the equations
Make sure the equations list the same variables in the same order
Place one equation above the other
Add the two equations (work down to a sum for each term), or subtract one equation
from the other, to eliminate one variable
You can repeat steps 123 to solve for the other variable
2, then x 5
5
3
5
Trang 2The correct answer is (B) To solve for x, you want to eliminate y You can multiply each
term in the second equation by 2, then add the equations:
3x 1 4y 5 28 2x 2 4y 5 1 5x 1 0y 5 27 x527
5
Since the question asked only for the value of x, stop here If the question had asked for both
x and y (or for y only), you could have multiplied both sides of the second equation by 3, then
subtracted the second equation from the first:
3x 1 4y 5 28 3x 2 6y 53
2
0x 1 10y 5 291
2
10y 5 219
2
y 5 219
20
Which Method Should You Use?
Which method, substitution or addition-subtraction, you should use depends on what the equations look like to begin with To see what we mean, look again at this system:
2
5p 1 q 5 3q 2 10
q 5 10 2 p
Notice that the second equation is already set up nicely for the substitution method You could use addition-subtraction instead; however, you’d just have to rearrange the terms in both the equations first:
2
5p 2 2q 5 210
p 1 q 5 10
Now, look again at the following system:
3x 1 4y 5 28
x 2 2y 51
2
Notice that the x-term and y-term already line up nicely here Also, notice that it’s easy to match the coefficients of either x or y: multiply both sides of the second equation by either 3 or 2 This
NOTE
If a question
requires you to
find values of
both unknowns,
combine the two
methods For
example, after
using
addition-subtraction to
solve for x,
substitute the
value of x into
either equation
to find y.
Trang 3system is an ideal candidate for addition-subtraction To appreciate this point, try using
substi-tution instead You’ll discover that it takes far more number crunching
LINEAR EQUATIONS THAT CANNOT BE SOLVED
Never assume that one linear equation with one variable is solvable If you can reduce the
equation to 0 5 0, then you can’t solve it In other words, the value of the variable could be
any real number The test makers generally use the Data Sufficiency format to cover
this concept
20 If 3x 2 3 2 4x 5 x 2 7 2 2x 1 4, then what is the value of x ?
(1) x 21
(2) x , 1
The correct answer is (E) All terms on both sides subtract out:
3x 2 3 2 4x 5 x 2 7 2 2x 1 4
2x 2 3 5 2x 2 3
0 5 0
Thus, even considering both statements together, x could equal any real number between 21
and 1 (not just the integer 0)
In some cases, what appears to be a system of two equations with two variables might
actually be the same equation expressed in two different ways In other words, what you’re
really dealing with are two equivalent equations that you cannot solve The test makers
generally use the Data Sufficiency format to cover this concept
21 Does a 5 b ?
(1) a 1 b 5 30
(2) 2b 5 60 2 2a
The correct answer is (E) An unwary test taker might assume that the values of both a
and b can be determined with both equations together, because they appear at first glance to
provide a system of two linear equations with two unknowns Not so! You can easily
manipulate the second equation so that it is identical to the first:
2b 5 60 2 2a
2b 5 2~30 2 a!
b 5 30 2 a
a 1 b 5 30
As you can see, the equation 2b 5 60 2 2a is identical to the equation a 1 b 5 30 Thus,
a and b could each be any real number You can’t solve one equation in two unknowns, so the
correct answer must be (E)
TIP
To solve a system
of two linear equations with two variables, use addition-subtraction
if you can quickly and easily eliminate one of the variables Otherwise, use substitution.
Trang 4Whenever you encounter a Data Sufficiency question that calls for solving one or more linear equations, stop in your tracks before taking pencil to paper Size up the equation to see whether it’s one of the two unsolvable kinds you learned about here If so, unless you’re given more information, the correct answer will be (E)
FACTORABLE QUADRATIC EXPRESSIONS WITH ONE VARIABLE
A quadratic expression includes a “squared” variable, such as x2 An equation is quadratic if you can express it in this general form:
ax21 bx 1 c 5 0,
where:
x is the variable
a, b, and c are constants (numbers)
a Þ 0
b can equal 0
c can equal 0 Here are four examples (notice that the b-term and c-term are not essential; in other words, either b or c, or both, can equal zero):
Quadratic Equation
Same Equation, but in the form:
ax21 bx 1 c 5 0
2w2
5 16
x2
5 3x 3y 5 4 2 y2
7z 5 2z2
2 15
2w2
2 16 5 0 (no b-term)
x2
2 3x 5 0 (no c-term)
y2
1 3y 2 4 5 0 2z2
2 7z 2 15 5 0 Every quadratic equation has exactly two solutions, called roots (But the two roots might be the same.) On the GMAT, you can often find the two roots by factoring To solve any factorable
quadratic equation, follow these three steps:
Put the equation into the standard form: ax21 bx 1 c 5 0.
Factor the terms on the left side of the equation into two linear expressions (with no exponents)
Set each linear expression (root) equal to zero and solve for the variable in each one
Trang 5Factoring Simple Quadratic Expressions
Some quadratic expressions are easier to factor than others If either of the two constants
b or c is zero, factoring requires no sweat In fact, in some cases, no factoring is needed at all:
2x25 x
2x22 x 5 0
x~2x 2 1! 5 0
x 5 0, 2x 2 1 5 0
x 5 0,1
2
2x22 4 5 0
2~x22 2! 5 0
x22 2 5 0
x25 2
x 5=2, 2=2
Factoring Quadratic Trinomials
A trinomial is simply an algebraic expression that contains three terms If a quadratic
expression contains all three terms of the standard form ax2
1 bx 1 c, then factoring becomes
a bit trickier You need to apply the FOIL method, in which you add together these terms:
(F) the product of the first terms of the two binomials
(O) the product of the outer terms of the two binomials
(I) the product of the inner terms of the two binomials
(L) the product of the last (second) terms of the two binomials
Note the following relationships:
(F) is the first term (ax2) of the quadratic expression
(O 1 I) is the second term (bx) of the quadratic expression
(L) is the third term (c) of the quadratic expression
You’ll find that the two factors will be two binomials The GMAT might ask you to recognize
one or both of these binomial factors
2 x 2 6 ?
(A) (x 1 6)
(B) (x 2 3)
(C) (x 1 1)
(D) (x 2 2)
(E) (x 1 3)
factoring into two binomials easier Set up two binomial shells: (x )(x ) The product of the
two missing second terms (the “L” term under the FOIL method) is 26 The possible integral
pairs that result in this product are (1,26), (21,6), (2,23,), and (22,3) Notice that the second
term in the trinomial is 2x This means that the sum of the two integers whose product is 26
must be 21 The pair (2,23) fits the bill Thus, the trinomial is equivalent to the product of
the two binomials (x 1 2) and (x 2 3).
ALERT!
When dealing with a quadratic equation, your first step is usually
to put it into the
general form ax2
1 bx 1 c 5 0 But
keep in mind: The only essential
term is ax2
Trang 6To check your work, multiply the two binomials using the FOIL method:
~x 1 2!~x 2 3! 5 x22 3x 1 2x 2 6
5 x22 x 1 6
If the preceding question had asked you to determine the roots of the equation x22 x 2 6 5 0, you’d simply set each of the binomial factors equal to 0 (zero), then solve for x in each one The solution set (the two possible values of x) includes the roots 22 and 3.
5 4x 2 1
contain?
(D) Four (E) Infinitely many
Notice that the c-term is 1 The only two integral pairs that result in this product are (1,1) and (21, 21) Since the b-term (24x) is negative, try using (21, 21) Set up a binomial shell:
(? 2 1)(? 2 1)
Notice that the a-term contains the coefficient 4 The possible integral pairs that result in this
product are (1,4), (2,2), (21,24), and (22,22) A bit of trial-and-error reveals that only the
pair (2,2) works Thus, in factored form, the equation becomes (2x 2 1)(2x 2 1) 5 0 To check
your work, multiply the two binomials using the FOIL method:
~2x 2 1!~2x 2 1! 5 4x22 2x 2 2x 1 1
5 4x22 4x 1 1
Since the two binomial factors are the same, the two roots of the equation are the same In
other words, x has only one possible value (Although you don’t need to find the value of x in order to answer the question, solve for x in the equation 2x21 5 0; x 5 1
2 Note that the
negative form of the binomial 2x 2 1, that is, 22x + 1, can be used and will yield the same
result.)
Stealth Quadratic Equations
Some equations that appear linear (variables include no exponents) may actually be quadratic Following, you will see the two GMAT situations you need to be on the lookout for The same variable inside a radical also appears outside:
=x 5 5x
~=x!25 ~5x!2
x 5 25x2 25x22 x 5 0
Trang 7The same variable that appears in the denominator of a fraction also appears elsewhere
in the equation:
2
x 5 3 2 x
2 5 x~3 2 x!
2 5 3x 2 x2
x22 3x 1 2 5 0
In both scenarios, you’re dealing with a quadratic (nonlinear) equation with one variable So,
in either equation, there are two roots (Both equations are factorable, so go ahead and find
their roots.) The test makers often use the Data Sufficiency format to cover this concept
24 What is the one, unique value of x?
(1) 6x 5=3x
(2) x 0
The correct answer is (C) An unwary test taker might assume that the equation in
statement (1) is linear—because x is not squared Not so! Clear the radical by squaring both
sides of the equation, then isolate the x-terms on one side of the equation and you’ll see that
the equation is quite quadratic indeed:
36x25 3x
36x22 3x 5 0
To ferret out the two roots, factor out 3x, then solve for each root:
3x~12x 2 1! 5 0
3x 5 0; 12x 2 1 50
x 5 0, 1
12
Since there is more than one possible value for x, statement (1) alone is insufficient to answer
the question Statement (2) alone is obviously insufficient But the two together eliminate the
root value 0, leaving 1
12as the only possible value of x.
THE QUADRATIC FORMULA
For some quadratic equations, although rational roots exist, they’re difficult to find For
example, 12x2
1 x 2 6 5 0 can be solved by factoring, but the factors are not easy to see:
12x2
1 x 2 6 5 (3x 2 2)(4x 1 3)
Trang 8Faced with a quadratic equation that’s difficult to factor, you can always use the quadratic
formula, which states that for any equation of the form ax21 bx 1 c 5 0:
x 5 2b 6=b22 4ac
2a
In the equation 12x2
1 x 2 6 5 0, for example, a 5 12, b 5 1, and c 5 2 6 Plugging these
values into the quadratic formula, you’ll find that the two roots are2
3 and 2
3
4. Some quadratic equations have no rational roots (solutions) Referring to the quadratic formula, if =b22 4ac turns out to be a negative number, then its square root will be imaginary, and hence so will the roots of the quadratic equation at hand But the GMAT
doesn’t test you on imaginary numbers In other words, you’ll find only real-number roots in any GMAT quadratic equation
NONLINEAR EQUATIONS WITH TWO VARIABLES
In the world of math, solving nonlinear equations with two or more variables can be very
complicated, even for bona-fide mathematicians But on the GMAT, all you need to remember are these three general forms:
Sum of two variables, squared: (x 1 y)2
5 x21 2xy 1 y2 Difference of two variables, squared: (x 2 y)2
5 x2 2 2xy 1 y2 Difference of two squares: x2
2 y25 (x 1 y)(x 2 y)
You can verify these equations using the FOIL method:
(x 1 y)2
5 (x 1 y)(x 1 y)
5 x21 xy 1 xy 1 y2
5 x21 2xy 1 y2
(x 2 y)2
5 (x 2 y)(x 2 y)
5 x22 xy 2 xy 1 y2
5 x22 2xy 1 y2
(x 1 y)(x 2 y)
5 x21 xy 2 xy 2 y2
5 x22 y2
For the GMAT, memorize the three equations listed here When you see one form on the exam, it’s a sure bet that your task is to rewrite it as the other form
(D) 50
Trang 9The correct answer is (D) If you’re on the lookout for the difference of two squares, you can
handle this question with no sweat Use the third equation you just learned, substituting 2 for
(x 1 y), then solving for (x 2 y):
x22 y25 ~x 1 y!~x 2 y!
100 5 ~x 1 y!~x 2 y!
100 5 ~2!~x 2 y!
50 5 ~x 2 y!
SOLVING ALGEBRAIC INEQUALITIES
You can solve algebraic inequalities in the same manner as equations Isolate the variable on
one side of the inequality symbol, factoring and eliminating terms wherever possible
However, one important rule distinguishes inequalities from equations: Whenever you
multiply or divide both sides of an inequality by a negative number, you must reverse the
inequality symbol Simply put: If a b, then 2a , 2b.
12 2 4x , 8 original inequality
24x , 24 12 subtracted from each side; inequality unchanged
x 1 both sides divided by 24; inequality reversed
Here are five general rules for dealing with algebraic inequalities Study them until they’re
second nature to you because you’ll put them to good use on the GMAT
Adding or subtracting unequal quantities to (or from) equal quantities:
If a b, then c 1 a c 1 b
If a b, then c 2 a , c 2 b
Adding unequal quantities to unequal quantities:
If a b, and if c d, then a 1 c b 1 d
Comparing three unequal quantities:
If a b, and if b c, then a c
Combining the same positive quantity with unequal quantities by multiplication or
division:
If a b, and if x 0, then xa xb
Combining the same negative quantity with unequal quantities by multiplication or
division:
If a b, and if x , 0, then xa , xb
TIP
You usually can’t solve quadratics using a shortcut Always look for one of the three common quadratic forms.
If you see it, rewrite it as its equivalent form
to answer the question as quickly and easily
as possible.
Trang 1026 If a b, and if c d, then which of the following must be true?
(A) a 2 b c 2 d
(B) a 2 c b 2 d
(C) c 1 d , a 2 b
(D) b 1 d , a 1 c
(E) a 2 c , b 1 d
The correct answer is (D) Inequality questions can be a bit confusing, can’t they? In this
problem, you need to remember that if unequal quantities (c and d) are added to unequal quantities of the same order (a and b), the result is an inequality in the same order This rule
is essentially what answer choice (D) says
WEIGHTED AVERAGE PROBLEMS
You solve weighted average problems using the arithmetic mean (simple average) formula,
except you give the set’s terms different weights For example, if a final exam score of 90
receives twice the weight of each of two midterm exam scores 75 and 85, think of the final exam score as two scores of 90—and the total number of scores as 4 rather than 3:
WA 575 1 85 1 ~2!~90!
340
4 5 85 Similarly, when some numbers among terms might appear more often than others, you must give them the appropriate “weight” before computing an average
27 During an 8-hour trip, Brigitte drove 3 hours at 55 miles per hour and 5 hours at 65
miles per hour What was her average rate, in miles per hour, for the entire trip?
(C) 61.25
(D) 62.5 (E) 66.25
The correct answer is (C) Determine the total miles driven: (3)(55) 1 (5)(65) 5 490 To
determine the average over the entire trip, divide this total by 8, which is the number of total hours: 490 4 8 5 61.25
A tougher weighted-average problem might provide the weighted average and ask for one of the terms, or require conversions from one unit of measurement to another—or both
28 A certain olive orchard produces 315 gallons of oil annually, on average, during four
consecutive years How many gallons of oil must the orchard produce annually, on average, during the next six years, if oil production for the entire 10-year period is
to meet a goal of 378 gallons per year?
(D) 420
ALERT!
Be careful when
handling
inequality
problems: The
wrong answers
might look right,
depending on
the values you
use for the
different
variables.