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 22 1a ϩ 121a ϩ 2 2 ϫ 21a ϩ 12 a1a Ϫ 32 ϭ 2 a Ϫ 3 4x1x ϩ 42 2x 2 1x Ϫ 421x ϩ 4 2 ϭ 2 x1x Ϫ 42 1x ϩ 321x Ϫ 3 2 31x Ϫ 32 ϭ 1x ϩ 32 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 7. Multiply each term by the LCD = 12. 8x + 2x = 3 10x = 3 x = 8. Multiply each term by the LCD = 12x. 3x + 2x = 12 5x = 12 x = Coordinate Graphing The coordinate plane is divided into four quadrants that are created by the intersection of two perpendicu- lar signed number lines: the x- and y-axes. The quadrants are numbered I, II, III, and IV as shown in the diagram. 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 down. When graphing linear equations (slope and y-intercept), use the y = mx + b form, where m 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– 349 Slope The slope between two points (x 1 , y 1 ) and (x 2 , y 2 ) can be found by using the following formula: Here are a few helpful facts about slope and graphing linear equations: ■ Lines that slant up to the right have a positive slope. ■ Lines that slant up to the left have a negative slope. ■ Horizontal lines have a slope of zero. ■ Vertical lines have an undefined slope or no slope. ■ Two lines with the same slope are parallel and will never intersect. ■ Two lines that have slopes that are negative reciprocals of each other are perpendicular. To find the midpoint between any two points (x 1 , y 1 ) and (x 2 , y 2 ), use the following formula: To find the distance between any two points (x 1 , y 1 ) and (x 2 , y 2 ), use the following formula: Systems of Equations with Two Variables When solving a system of equations, you are finding the value or values where two or more equations equal each other. This can be done in two ways algebraically: by elimination and by substitution. Elimination Method Solve the system x – y = 6 and 2x + 3y = 7. Put the equations one above the other, lining up the xs, ys, and the equal sign. x – y = 6 2x + 3y = 7 21x 1 – x 2 2 2 ϩ 1y 1 – y 2 2 2 1 x 1 ϩ x 2 2 , y 1 ϩ y 2 2 2 change in y change in x ϭ y 1 Ϫ y 2 x 1 Ϫ x 2 – ALGEBRA– 350 Multiply the first equation by –2 so that the coefficients of x are opposites. This will allow the xs to can- cel 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 equations vertically. –2x + 2y = –12 2x + 3y = 7 5y = –5 Divide both sides by 5. To complete the problem, solve for x by substituting –1 for y into one of the original equations. x – y =6 x – (–1) = 6 x + 1 = 6 x + 1 – 1 = 6 – 1 x =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 y ϭ –1 5y 5 ϭ –5 5 – ALGEBRA– 351 Combine like terms. Remember x = 1x. –3x + 14 = 5 Subtract 14 from both sides and then divide by –3. –3x + 14 –14 = 5 – 14 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 with this type of problem, first begin by figuring out what you need to solve for and defining your variable as that unknown. Then write and solve an equation that matches the question asked. Mixture Problems How many pounds of coffee that costs $4.00 per pound need to be mixed with 10 pounds of coffee that costs $6.40 per pound to create a mixture of coffee that costs $5.50 per pound? a. 4 b. 6 c. 8 d. 10 e. 16 For this type of question, remember that the total amount spent in each case will be the price per pound 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, –3x –3 ϭ –9 –3 – ALGEBRA– 352 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. 4.00(x) + 6.40(10) = 5.50(x + 10) Multiply through the equation: 4x + 64 = 5.5x + 55 Subtract 4x from both sides: 4x – 4x + 64 = 5.5x – 4x + 55 Subtract 55 from both sides: 64 – 55 = 1.5x + 55 – 55 Divide both sides by 1.5: 6 = x You need 6 pounds of the $4.00 per pound coffee. The correct answer is b. Distance Problems Most problems that involve motion or traveling will probably use the formula distance = rate×time. Wendy drove 4 hours in a car to reach a conference she was attending. On her return trip, she followed the same route but the trip took her 1 ᎏ 1 2 ᎏ hours longer. If she drove 220 miles to conference, how much slower was her average speed on the return trip? a. 10 b. 15 c. 25 d. 40 e. 55 Use the formula distance = rate × time and convert it to = rate. Remember that the distance was 220 miles for each part of the trip. Since it took her 4 hours to reach the conference, then 4 + 1 ᎏ 1 2 ᎏ = 5 ᎏ 1 2 ᎏ hours for the return trip. = 40 miles per hour. However, the question did not ask for the speed on the way back; it asked for the difference between the speed on the way there and the speed on the way home. The speed on the way there would be = 55 miles per hour and 55 – 40 = 15 miles per hour slower on the return trip. The correct answer is b. 220 4 220 5.5 distance time 9 1.5 ϭ 1.5x 1.5 – ALGEBRA– 353 . 32 ϭ 2 a Ϫ 3 4x1x ϩ 42 2x 2 1x Ϫ 421x ϩ 4 2 ϭ 2 x1x Ϫ 42 1x ϩ 32 1x Ϫ 3 2 31 x Ϫ 32 ϭ 1x ϩ 32 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. 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 22 1a ϩ 121a ϩ 2 2 ϫ 21a ϩ 12 a1a Ϫ 32 ϭ 2 a. 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– 34 8 7. Multiply each term by the LCD = 12. 8x + 2x = 3 10x = 3 x = 8. Multiply each term by the LCD = 12x. 3x +