The GMAT Quantitative Section - Algebra

 Translating Expressions and Equations Translating sentences and word problems into mathematical expressions and equations is similar to trans- lating two different languages. The key words are the vocabulary that tells what operations should be done and in what order. Use the following chart to help you with some of the key words used on the GMAT® quan- titative section. SUM MORE THAN ADDED TO PLUS INCREASED BY PRODUCT TIMES MULTIPLIED BY QUOTIENT DIVIDED BY EQUAL TO TOTAL DIFFERENCE LESS THAN SUBTRACTED FROM MINUS DECREASED BY FEWER THAN Ϭ ϩ Ϫϫ ϭ CHAPTER Algebra 21 339 The following is an example of a problem where knowing the key words is necessary: Fifteen less than five times a number is equal to the product of ten and the number. What is the number? Translate the sentence piece by piece: Fift ee n less than five times the numb er equals the pro duct of 10 and x. 5x – 15 = 10x The equation is 5x – 15 = 10x Subtract 5x from both sides: 5x – 5x – 15 = 10x – 5x Divide both sides by 5: –3 = x It is important to realize that the key words less than tell you to subtract from the number and the key word product reminds you to multiply.  Combining Like Terms and Polynomials In algebra, you use a letter to represent an unknown quantity. This letter is called the variable. The number preceding the variable is called the coefficient. If a number is not written in front of the variable, the coeffi- cient is understood to be one. If any coefficient or variable is raised to a power, this number is the exponent. 3x Three is the coefficient and x is the variable. xy One is the coefficient, and both x and y are the variables. –2x 3 y Negative two is the coefficient, x and y are the variables, and three is the exponent of x. Another important concept to recognize is like terms. In algebra, like terms are expressions that have exactly the same variable(s) to the same power and can be combined easily by adding or subtracting the coef- ficients. Examples 3x + 5x These terms are like terms, and the sum is 8x. 4x 2 y + –10x 2 y These terms are also like terms, and the sum is –6x 2 y. 2xy 2 + 9x 2 y These terms are not like terms because the variables, taken with their powers, are not exactly the same. They cannot be combined. –15 5 ϭ 5x 5 – ALGEBRA – 340 A polynomial is the sum or difference of many terms and some have specific names: 8x 2 This is a monomial because there is one term. 3x + 2y This is a binomial because there are two terms. 4x 2 + 2x – 6 This is a trinomial because there are three terms.  Laws of Exponents ■ When multiplying like bases, add the exponents: x 2 × x 3 = x 2 + 3 = x 5 ■ When dividing like bases, subtract the exponents: ■ When raising a power to another power, multiply the exponents: ■ Remember that a fractional exponent means the root: ͙ ෆ x = x ᎏ 1 2 ᎏ and ͙ 3 x ෆ = x ᎏ 1 3 ᎏ The following is an example of a question involving exponents: Solve for x:2 x + 2 = 8 3 . a. 1 b. 3 c. 5 d. 7 e. 9 The correct answer is d. To solve this type of equation, each side must have the same base. Since 8 can be expressed as 2 3 , then 8 3 = (2 3 ) 3 = 2 9 . Both sides of the equation have a common base of 2, so set the expo- nents equal to each other to solve for x. x + 2 = 9. So, x = 7.  Solving Linear Equations of One Variable When solving this type of equation, it is important to remember two basic properties: ■ If a number is added to or subtracted from one side of an equation, it must be added to or subtracted from the other side. ■ If a number is multiplied or divided on one side of an equation, it must also be multiplied or divided on the other side. 1x 2 2 3 ϭ x 2 ϫ 3 ϭ x 6 x 5 x 2 = x 5 – 2 = x 3 – ALGEBRA – 341 Linear equations can be solved in four basic steps: 1. Remove parentheses by using distributive property. 2. Combine like terms on the same side of the equal sign. 3. Move the variables to one side of the equation. 4. Solve the one- or two-step equation that remains, remembering the two previous properties. Examples Solve for x in each of the following equations: a. 3x – 5 = 10 Add 5 to both sides of the equation: 3x – 5 + 5 = 10 + 5 Divide both sides by 3: x = 5 b. 3 (x – 1) + x = 1 Use distributive property to remove parentheses: 3x – 3 + x = 1 Combine like terms: 4x – 3 = 1 Add 3 to both sides of the equation: 4x – 3 + 3 = 1 + 3 Divide both sides by 4: x = 1 c. 8x – 2 = 8 + 3x Subtract 3x from both sides of the equation to move the variables to one side: 8x – 3x – 2 = 8 + 3x – 3x Add 2 to both sides of the equation: 5x – 2 + 2 = 8 + 2 Divide both sides by 5: x = 2  Solving Literal Equations A literal equation is an equation that contains two or more variables. It may be in the form of a formula. You may be asked to solve a literal equation for one variable in terms of the other variables. Use the same steps that you used to solve linear equations. 5x 5 ϭ 10 5 4x 4 ϭ 4 4 3x 3 ϭ 15 3 – ALGEBRA – 342 Example Solve for x in terms of a and b:2x + b = a Subtract b from both sides of the equation: 2x + b – b = a – b Divide both sides of the equation by 2:  Solving Inequalities Solving inequalities is very similar to solving equations. The four symbols used when solving inequalities are as follows: ■ Ͻ is less than ■ Ͼ is greater than ■ Յ is less than or equal to ■ Ն is greater than or equal to When solving inequalities, there is one catch: If you are multiplying or dividing each side by a negative number, you must reverse the direction of the inequality symbol. For example, solve the inequality –3x + 6 Յ 18: 1. First subtract 6 from both sides: 2. Then divide both sides by –3: 3. The inequality symbol now changes: Solving Compound Inequalities A compound inequality is a combination of two inequalities. For example, take the compound inequality –3 Ͻ x + 1 Ͻ 4. To solve this, subtract 1 from all parts of the inequality. –3 – 1 Ͻ x + 1 – 1 Ͻ 4 – 1. Simplify. –4 Ͻ x Ͻ 3. Therefore, the solution set is all numbers between –4 and 3, not including –4 and 3. x ՆϪ4 –3x –3 Յ 12 –3 –3x ϩ 6 Ϫ 6 Յ 18 Ϫ 6 x ϭ a Ϫ b 2 2x 2 ϭ a Ϫ b 2 – ALGEBRA – 343  Multiplying and Factoring Polynomials When multiplying by a monomial, use the distributive property to simplify. Examples Multiply each of the following: 1. (6x 3 )(5xy 2 ) = 30x 4 y 2 (Remember that x = x 1 .) 2. 2x (x 2 – 3) = 2x 3 – 6x 3. x 3 (3x 2 + 4x – 2) = 3x 5 + 4x 4 – 2x 3 When multiplying two binomials, use an acronym called FOIL. F Multiply the first terms in each set of parentheses. O Multiply the outer terms in the parentheses. I Multiply the inner terms in the parentheses. L Multiply the last terms in the parentheses. Examples 1. (x – 1)(x + 2) = x 2 + 2x – 1x – 2 = x 2 + x – 2 F O I L 2. (a – b) 2 = (a – b)(a – b) = a 2 – ab – ab – b 2 F O I L Factoring Polynomials Factoring polynomials is the reverse of multiplying them together. Examples Factor the following: 1. 2x 3 + 2 = 2 (x 3 + 1) Take out the common factor of 2. 2. x 2 – 9 = (x + 3)(x – 3) Factor the difference between two perfect squares. 3. 2x 2 + 5x – 3 = (2x – 1)(x + 3) Factor using FOIL backwards. 4. 2x 2 – 50 = 2(x 2 – 25) = 2(x + 5)(x – 5) First take out the common factor and then factor the difference between two squares.  Solving Quadratic Equations An equation in the form y = ax 2 + bx + c,where a, b, and c are real numbers, is a quadratic equation. In other words, the greatest exponent on x is two. – ALGEBRA – 344 Quadratic equations can be solved in two ways: factoring, if it is possible for that equation, or using the quadratic formula. By Factoring In order to factor the quadratic equation, it first needs to be in standard form. This form is y = ax 2 +bx + c. In most cases, the factors of the equations involve two numbers whose sum is b and product is c. Examples Solve for x in the following equation: 1. x 2 – 25 = 0 This equation is already in standard form. This equation is a special case; it is the difference between two perfect squares. To factor this, find the square root of both terms. The square root of the first term x 2 is x. The square root of the second term 25 is 5. Then two factors are x – 5 and x + 5. The equation x 2 – 25 = 0 then becomes (x – 5)(x + 5) = 0 Set each factor equal to zero and solve x – 5 = 0 or x + 5 = 0 x = 5 or x = –5 The solution is {5, –5}. 2. x 2 + 6x = –9 This equation needs to be put into standard form by adding 9 to both sides of the equation. x 2 + 6x + 9 = –9 + 9 x 2 + 6x + 9 = 0 The factors of this trinomial will be two numbers whose sum is 6 and whose product is 9. The fac- tors are x + 3 and x + 3 because 3 + 3 = 6 and 3 × 3 = 9. The equation becomes (x + 3)(x + 3) = 0 Set each factor equal to zero and solve x + 3 = 0 or x + 3 = 0 x = –3 or x = –3 Because both factors were the same, this was a perfect square trinomial. The solution is {–3}. 3. x 2 = 12 + x This equation needs to be put into standard form by subtracting 12 and x from both sides of the equation. x 2 – x – 12 = 12 – 12 + x – x x 2 – x – 12 = 0 – ALGEBRA – 345 Since the sum of 3 and –4 is –1, and their product is –12, the equation factors to (x + 3) (x – 4) = 0 Set each factor equal to zero and solve: x + 3 = 0 or x – 4 = 0 x = –3 or x = 4 The solution is {–3, 4}. By Quadratic Formula Solving by using the quadratic formula will work for any quadratic equation, especially those that are not fac- torable. Solve for x: x 2 + 4x = 1 Put the equation in standard form. x 2 + 4x – 1 = 0 Since this equation is not factorable, use the quadratic formula by identifying the value of a, b, and c and then substituting it into the formula. For this particular equation, a = 1, b = 4, and c = –1. The solution is . The following is an example of a word problem incorporating quadratic equations: A rectangular pool has a width of 25 feet and a length of 30 feet. A deck with a uniform width sur- rounds it. If the area of the deck and the pool together is 1,254 square feet, what is the width of the deck? ͭ –2 ϩ 2 5 , –2 – 2 5 ͮ x ϭ –2 ; 2 5 x ϭ –4 2 ; 22 5 2 x ϭ –4 ; 2 20 2 x ϭ –4 ; 2 16 ϩ 4 2 x ϭ –4 ; 2 4 2 Ϫ 41121–12 2112 x ϭ –b ; 2 b 2 –4ac 2a – ALGEBRA – 346 Begin by drawing a picture of the situation. The picture could be similar to the following figure. Since you know the area of the entire figure, write an equation that uses this information. Since we are trying to find the width of the deck, let x = the width of the deck. Therefore, x + x + 25 or 2x + 25 is the width of the entire figure. In the same way, x + x + 30 or 2x + 30 is the length of the entire figure. The area of a rectangle is length × width, so use A = l × w. Substitute into the equation: 1,254 = (2x + 30)(2x + 25) Multiply using FOIL: 1,254 = 4x 2 + 50x + 60x + 750 Combine like terms: 1,254 = 4x 2 + 110x + 750 Subtract 1,254 from both sides: 1,254 – 1,254 = 4x 2 + 110x + 750 – 1,254 0 = 4x 2 + 110x – 504 Divide each term by 2: 0 = 2x 2 +55x – 252 Factor the trinomial: 0 = (2x + 63 )(x – 4) Set each factor equal to 0 and solve 2x + 63 = 0 or x – 4 = 0 2x = –63 x = 4 x = –31.5 Since we are solving for a length, the solution of –31.5 must be rejected. The width of the deck is 4 feet.  Rational Expressions and Equations Rational expressions and equations involve fractions. Since dividing by zero is undefined, it is important to know when an expression is undefined. The fraction is undefined when the denominator x – 1 = 0; therefore, x =1. 5 x Ϫ 1 x x x x 25 30 – ALGEBRA – 347 You may be asked to perform various operations on rational expressions. See the following examples. Examples 1. Simplify . 2. Simplify . 3. Multiply . 4. Divide . 5. Add . 6. Subtract . 7. Solve . 8. Solve . Answers 1. 2. 3. 4. 5. 6. 3x ϩ 18 Ϫ x ϩ 2 3x ϭ 2x ϩ 20 3x 1 ϩ 3x xy a1a ϩ 2 2 1a ϩ 1 21a ϩ 2 2 ϫ 21a ϩ 1 2 a1a Ϫ 3 2 ϭ 2 a Ϫ 3 4x1x ϩ 4 2 2x 2 1x Ϫ 4 21x ϩ 4 2 ϭ 2 x1x Ϫ 4 2 1x ϩ 3 21x Ϫ 3 2 31x Ϫ 3 2 ϭ 1x ϩ 3 2 3 x 2 b x 3 b 2 ϭ 1 xb 1 x ϭ 1 4 ϩ 1 6 2 3 x ϩ 1 6 x ϭ 1 4 x ϩ 6 x – x –2 3x 1 xy ϩ 3 y a 2 ϩ 2a a 2 ϩ 3a ϩ 2 Ϭ a 2 –3a 2a ϩ 2 4x x 2 –16 ϫ x ϩ 4 2x 2 x 2 Ϫ 9 3x Ϫ 9 x 2 b x 3 b 2 – ALGEBRA – 348 [...]... 4 I 3 2 1 x-axis -5 -4 -3 -2 -1 -1 1 2 3 4 5 -2 III -3 -4 IV -5 Each location in the plane is named by a point (x, y) These numbers are called the coordinates of the point Each point can be found by starting at the intersection of the axes, the origin, and moving x units to the right or left and y units up or down Positive directions are to the right and up and negative directions are to the left and... difference between the speed on the way there and the speed on the way home The speed on the way there would be 220 4 = 55 miles per hour and 55 – 40 = 15 miles per hour slower on the return trip The correct answer is b 353 – ALGEBRA – Ratio Word Problems You can often use the ratio to help Three-fifths of the employees at Company A work overtime each week and the other employees do not What is the ratio of... 7 Put the equations one above the other, lining up the xs, ys, and the equal sign x–y =6 2x + 3y = 7 350 – ALGEBRA – Multiply the first equation by –2 so that the coefficients of x are opposites This will allow the xs to cancel out in the next step Make sure that ALL terms are multiplied by –2 The second equation remains the same –2 (x – y = 6) ⇒ –2x + 2y = –12 2x + 3y = 7 ⇒ 2x + 3y = 7 Combine the new... hours Think about how much of the lawn each person completes individually Since Jason can finish in 2 1 hours, in 1 hour he completes 2 of the lawn Since Ciera can finish in 4 hours, then in 1 hour she completes 1 4 of the lawn If we let x = the time it takes both Jason and Ciera working together, then 1 x is the amount of 1 the lawn they finish in 1 hour working together Then use the equation 1 ϭ 1 ϭ x and...– ALGEBRA – 7 Multiply each term by the LCD = 12 8x + 2x = 3 10x = 3 x= 10 3 8 Multiply each term by the LCD = 12x 3x + 2x = 12 5x = 12 x= 12 5 ϭ 2.4 Coordinate Graphing The coordinate plane is divided into four quadrants that are created by the intersection of two perpendicular signed number lines: the x- and y-axes The quadrants are numbered I, II, III, and IV as shown in the diagram y-axis II... times how many pounds are in the mixture Therefore, if you let x = the number of pounds of $4.00 coffee, then $4.00(x) is the amount of money spent on $4.00 coffee, $6.40(10) is the amount spent on $6.40 coffee, 352 – ALGEBRA – and $5.50(x + 10) is the total amount spent Write an equation that adds the first two amounts and sets it equal to the total amount Multiply through the equation: Subtract 4x from... on the return trip? a 10 b 15 c 25 d 40 e 55 Use the formula distance = rate × time and convert it to distance time = rate Remember that the distance was 1 1 220 miles for each part of the trip Since it took her 4 hours to reach the conference, then 4 + 1ᎏ2ᎏ = 5 ᎏ2ᎏ hours for the return trip 220 5.5 = 40 miles per hour However, the question did not ask for the speed on the way back; it asked for the. .. (slope and y-intercept), use the y = mx + b form, where m represents the slope of the line and b represents the y-intercept 349 – ALGEBRA – Slope The slope between two points (x1, y1) and (x2, y2) can be found by using the following formula: change in y change in x y Ϫ y ϭ x 11 Ϫ x22 Here are a few helpful facts about slope and graphing linear equations: ■ ■ ■ ■ ■ ■ Lines that slant up to the right have... complete the problem, solve for x by substituting –1 for y into one of the original equations x–y= x – (–1) = x+1= x+1–1= x= 6 6 6 6–1 5 The solution to the system is x = 5 and y = –1, or (5, –1) Substitution Method Solve the system x + 2y = 5 and y = –2x + 7 Substitute the second equation into the first for y x + 2(–2x + 7) = 5 Use distributive property to remove the parentheses x + –4x + 14 = 5 351 – ALGEBRA. .. both sides and then divide by –3 –3x + 14 –14 = 5 – 14 –3x –3 ϭ –9 –3 x=3 To complete the problem, solve for y by substituting 3 for x in one of the original equations y = –2x + 7 y = –2 (3) + 7 y = –6 + 7 y=1 The solution to the system is x = 3 and y = 1, or (3, 1) Problem Solving with Word Problems You will encounter a variety of different types of word problems on the GMAT quantitative section To help . represents the slope of the line and b represents the y-intercept. 5 4 3 2 1 -1 -2 -3 -4 -5 1 2 3 45 -1 -2 -3 -4 -5 y-axis x-axis III III IV 12 5 ϭ 2.4 10 3 – ALGEBRA. Multiply the first terms in each set of parentheses. O Multiply the outer terms in the parentheses. I Multiply the inner terms in the parentheses. L Multiply the