Substitute 25 6 for p in either equation to find q (we’ll use the second equation): q 5 10 2 25 6 q 5 60 6 2 25 6 q 5 35 6 The question asks for p q , so do the division: p q 5 25 6 35 6 5 25 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. 19. If 3x 1 4y 528, and if x 2 2y 5 1 2 , then x 5 (A) 212 (B) 2 7 5 (C) 1 3 (D) 14 5 (E) 9 Chapter 10: Math Review: Number Theory and Algebra 253 www.petersons.com The 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 528 2x 2 4y 5 1 5x 1 0y 527 x5 2 7 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 528 3x 2 6y 5 3 2 0x 1 10y 5 29 1 2 10y 5 2 19 2 y 5 2 19 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 5 p 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 5 p 2 2q 5210 p 1 q 5 10 Now, look again at the following system: 3x 1 4y 528 x 2 2y 5 1 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 254 PART IV: GMAT Quantitative Section 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. www.petersons.com system 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 52x 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. Chapter 10: Math Review: Number Theory and Algebra 255 www.petersons.com Whenever 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 x 2 . An equation is quadratic if you can express it in this general form: ax 2 1 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: ax 2 1 bx 1 c 5 0 2w 2 5 16 x 2 5 3x 3y 5 4 2 y 2 7z 5 2z 2 2 15 2w 2 2 16 5 0(nob-term) x 2 2 3x 5 0(noc-term) y 2 1 3y 2 4 5 0 2z 2 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: ax 2 1 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. 256 PART IV: GMAT Quantitative Section www.petersons.com Factoring 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: A quadratic with no c term A quadratic with no b term 2x 2 5 x 2x 2 2 x 5 0 x~2x 2 1!50 x 5 0, 2x 2 1 5 0 x 5 0, 1 2 2x 2 2 4 5 0 2~x 2 2 2!50 x 2 2 2 5 0 x 2 5 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 ax 2 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 (ax 2 ) 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. 22. Which of the following is a factor of x 2 2 x 2 6? (A) (x 1 6) (B) (x 2 3) (C) (x 1 1) (D) (x 2 2) (E) (x 1 3) The correct answer is (B). Notice that x 2 has no coefficient. This makes the process of 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 ax 2 1 bx 1 c 5 0. But keep in mind: The only essential term is ax 2 . Chapter 10: Math Review: Number Theory and Algebra 257 www.petersons.com To check your work, multiply the two binomials using the FOIL method: ~x 1 2!~x 2 3!5x 2 2 3x 1 2x 2 6 5 x 2 2 x 1 6 If the preceding question had asked you to determine the roots of the equation x 2 2 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. 23. How many different values of x does the solution set for the equation 4x 2 5 4x 2 1 contain? (A) None (B) One (C) Two (D) Four (E) Infinitely many The correct answer is (B). First, express the equation in standard form: 4x 2 2 4x 1 1 5 0. 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!54x 2 2 2x 2 2x 1 1 5 4x 2 2 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! 2 5~5x! 2 x 5 25x 2 25x 2 2 x 5 0 258 PART IV: GMAT Quantitative Section www.petersons.com The 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 x 2 x 2 2 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: 36x 2 5 3x 36x 2 2 3x 5 0 To ferret out the two roots, factor out 3x, then solve for each root: 3x~12x 2 1!50 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 12 as the only possible value of x. THE QUADRATIC FORMULA For some quadratic equations, although rational roots exist, they’re difficult to find. For example, 12x 2 1 x 2 6 5 0 can be solved by factoring, but the factors are not easy to see: 12x 2 1 x 2 6 5 (3x 2 2)(4x 1 3) Chapter 10: Math Review: Number Theory and Algebra 259 www.petersons.com Faced 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 ax 2 1 bx 1 c 5 0: x 5 2b 6 = b 2 2 4ac 2a In the equation 12x 2 1 x 2 6 5 0, for example, a 5 12, b 5 1, and c 526. Plugging these values into the quadratic formula, you’ll find that the two roots are 2 3 and 2 3 4 . Some quadratic equations have no rational roots (solutions). Referring to the quadratic formula, if = b 2 2 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 x 2 1 2xy 1 y 2 Difference of two variables, squared: (x 2 y) 2 5 x 2 2 2xy 1 y 2 Difference of two squares: x 2 2 y 2 5 (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 x 2 1 xy 1 xy 1 y 2 5 x 2 1 2xy 1 y 2 (x 2 y) 2 5 (x 2 y)(x 2 y) 5 x 2 2 xy 2 xy 1 y 2 5 x 2 2 2xy 1 y 2 (x 1 y)(x 2 y) 5 x 2 1 xy 2 xy 2 y 2 5 x 2 2 y 2 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. 25. If x 2 2 y 2 5 100, and if x 1 y 5 2, then x 2 y 5 (A) 22 (B) 10 (C) 20 (D) 50 (E) 200 260 PART IV: GMAT Quantitative Section www.petersons.com The 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): x 2 2 y 2 5~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, rewriteitasits equivalent form to answer the question as quickly and easily as possible. Chapter 10: Math Review: Number Theory and Algebra 261 www.petersons.com 26. 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 5 75 1 85 1~2!~90! 4 5 340 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? (A) 58.5 (B) 60 (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? (A) 240 (B) 285 (C) 396 (D) 420 (E) 468 262 PART IV: GMAT Quantitative Section ALERT! Be careful when handling inequality problems: The wrong answers might look right, depending on the values you use for the different variables. www.petersons.com . 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! 2 5~5x! 2 x 5 25x 2 25x 2 2 x 5 0 258 PART IV: GMAT Quantitative. entire 10-year period is to meet a goal of 378 gallons per year? (A) 240 (B) 285 (C) 396 (D) 420 (E) 468 262 PART IV: GMAT Quantitative Section ALERT! Be careful when handling inequality problems:. coefficients of either x or y: multiply both sides of the second equation by either 3 or 2. This 254 PART IV: GMAT Quantitative Section NOTE If a question requires you to find values of both unknowns, combine