Sat - MC Grawhill part 32 pot

300 MCGRAW-HILL’S SAT Equations as Balanced Scales Algebra is really common sense once you start think- ing of every equation as a balanced scale. The terms on either side of the equals sign must be “balanced,” just like the weights on a scale. The laws of equations come from the common- sense rules for keeping a scale balanced. Imag- ine that you are the keeper of a scale, and you must keep it balanced. What would you do if someone took weights from one side of the scale? You’d remove an equal weight from the other side, of course. This is a law of equality: anything you do to one side of the equation, you must do to the other to maintain the balance. Example: If 12x − 8 = 28, then what is the value of 3x − 2? Don’t worry about solving for x, because that’s not what the question is asking for. Notice that the expression you are given, 12x − 8, is 4 times the expression you are looking for, 3x − 2. So to turn the given expression into the one you want, divide by 4. Of course, you must do the same to the other side to keep the balance: Solving as Unwrapping Solving simple algebraic equations is basically the same thing as unwrapping a present. (And it’s just as fun, too, right? Okay, maybe not.) Wrapping a present involves a sequence of steps: 1. Put the gift in the box. 2. Close the box. 3. Wrap the paper around the box. 4. Put the bow on. Here’s the important part: unwrapping the present just means inverting those steps and revers- ing their order: 1. Take off the bow. 2. Unwrap the paper. 3. Open the box. 4. Take out the gift. Example: Solve for x:5x 2 − 9 = 31 The problem is that x is not alone on the left side; it is “wrapped up” like a gift. How is it wrapped? Think of the order of operations for turning x into 5x 2 − 9: 1. Square it: x 2 2. Multiply by 5: 5x 2 3. Subtract 9: 5x 2 − 9 12 8 4 28 4 327 x x − =−= So to “unwrap” it, you reverse and invert the steps: 1. Add 9: (5x 2 − 9) + 9 = 5x 2 2. Divide by 5: 5x 2 /5 = x 2 3. Find the square roots (both of them!): =±͉x͉ If you perform these steps to both sides, 5x 2 − 9 = 31 transforms into Watch Your Steps To solve that last equation, we had to perform three operations. Many equations, as you probably know, require more steps to solve. It is very important that you keep track of your steps so that you can check your work if you need to. In other words, the equation we just solved should really look like this on your scratch paper: Check by Plugging Back In Always check your answer by plugging it back into the original equation to see if it works. Remember that solving an equation means simply finding the value of each unknown that makes the equation true. Example: Are solutions to 5x 2 − 9 = 31? Plug them in: − 9 = 5(8) − 9 = 40 − 9 = 31 (Yes!) 58 2 ± () ± 8 (): 5931 1 2 x −= Step Add 9 Step 2 (Divide by 5) : ++ = 99 540 2 x 5 Step 3 (Simplify): x 2 5 40 5 = x 2 8= Step 4 (Square r ooot): 8x =± x =± 8. ± x 2 Lesson 1: Solving Equations There’s a lot of detail to learn and understand to do well on the SAT. For more tools and resources that will help, visit our Online Practice Plus at www.MHPracticePlus.com/SATmath. CHAPTER 8 / ESSENTIAL ALGEBRA I SKILLS 301 Concept Review 1: Solving Equations 1. Explain the laws of equality. 2. Are there any operations you can perform to both sides of an equation that will not yield a true equation? Explain. Show your steps and check your work in solving the following equations. 3. 9x − 12 + 5x = 3x 4. (x − 4) 2 = 5 5. 6. Solve the following equations for the given expression by performing one operation on both sides. 7. If , then 10x + 12 = Operation: ______________________ Solution: ______________________ 8. If 18y + 12 = 7, then 6y + 4 = Operation: ______________________ Solution: ______________________ 5 2 37 x += xx+ = −2 3 5 2 2 3 2 2 x x= 302 MCGRAW-HILL’S SAT SAT Practice 1: Solving Equations 1. If 5d + 12 = 24, then 5d − 12 = (A) −24 (B) −12 (C) 0 (D) 12 (E) 24 2. What number decreased by 7 equals 5 times the number? (A) (B) (C) (D) (E) 3. If , then y + 5 = 4. If 2x 2 − 5x = 9, then 12x 2 − 30x = (A) −54 (B) −6 (C) 18 (D) 36 (E) 54 ( p + 2) 2 = ( p − 5) 2 5. The equation above is true for which of the fol- lowing values of p? (A) −2 and 5 (B) 2 and −5 (C) 0 and 1 (D) 1.5 only (E) 3.5 only 2 5 2 2 y y= 4 7 − 4 7 − 5 7 − 7 5 − 7 4 6. If what is the value of x? (A) (B) −7 (C) (D) (E) 7 7. The product of x and y is 36. If both x and y are integers, then what is the least possible value of x − y? (A) −37 (B) −36 (C) −35 (D) −9 (E) −6 8. The graph of y = f(x) contains the points (−1, 7) and (1, 3). Which of the following could be f(x)? I. f(x) = ͉5x − 2͉ II. f(x) = x 2 − 2x + 4 III. f(x) =−2x + 5 (A) I only (B) I and II only (C) I and III only (D) II and III only (E) I, II, and III 9. If , which of the following gives all possible values of x? (A) 9 only (B) −9 and 9 (C) 81 only (D) 81 and −81 (E) 961 10. For all positive values of m and n, if then x = (A) (B) (C) (D) (E) 3 22mn− 2 32 m n+ 32 2 + n m 23 2 m n − 22 3 mn− 3 2 x mnx− = , 20 11−=x − 7 5 − 24 7 − 25 2 57 5 1 x +=, 1 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 CHAPTER 8 / ESSENTIAL ALGEBRA I SKILLS 303 5. Multiply by 3: 2x 2 = 3x 2 Subtract 2x 2 :0 = x 2 Take square root: 0 = x 6. Cross-multiply: 2x + 4 = 3x − 15 Subtract 2x: 4 = x − 15 Add 15: 19 = x 7. Operation: Multiply both sides by 4 Solution: 10x + 12 = 28 8. Operation: Divide both sides by 3 Solution: 6y + 4 = 7/3 xx+ = −2 3 5 2 2 3 2 2 x x= Concept Review 1 1. The laws of equality say that whatever you do to one side of an equation you must do to the other side, to maintain the balance. 2. Yes. Dividing by 0 and taking the square root of a negative number are “undefined” operations in the real numbers. Be careful, then, when dividing both sides of an equation by an unknown, to check that the unknown could not possibly be 0. 3. 9x − 12 + 5x = 3x Commutative law: 14x − 12 = 3x Add 12, subtract 3x: 11x = 12 Divide by 11: x = 12/11 4. (x − 4) 2 = 5 Take square root: Add 4: x =±45 x −=±45 Answer Key 1: Solving Equations SAT Practice 1 1. C 5d + 12 = 24 Subtract 24: 5d − 12 = 0 2. A Translate into an equation: x − 7 = 5x Subtract x: −7 = 4x Divide by 4: −7/4 = x You can also “test” the choices and see which one works, but that’s probably more time-consuming. 3. 5 It’s easiest to solve the equation for y, then add 5 Multiply by 5: 2y 2 = 5y 2 Subtract 2y 2 :0 = 3y 2 Divide by 3: 0 = y 2 Take the square root: 0 = y Add 5: 5 = y + 5 4. E 2x 2 − 5x = 9 Multiply by 6: 12x 2 − 30x = 54 5. D Plugging in and checking is perhaps easiest here, but you could do the algebra too: (p + 2) 2 = (p − 5) 2 FOIL: p 2 + 4p + 4 = p 2 − 10p + 25 Subtract p 2 :4p + 4 =−10p + 25 Subtract 4: 4p =−10p + 21 Add 10p: 14p = 21 Divide by 14: p = 1.5 2 5 2 2 y y= 6. A Multiply by 5x: 25 + 7x = 5x Subtract 7x: 25 =−2x Divide by −2: −25/2 = x 7. C Guess and check here. If x =−36 and y =−1, or x = 1 and y = 36, then x − y =−35. 8. E Just plug in the points (−1, 7) and (1, 3) to the equations, and confirm that the points “satisfy” all three equations. 9. C Subtract 20: Multiply by −1: Square both sides: x = 81 10. D = 2 Multiply by m − nx: 3x = 2(m − nx) Distribute on right: 3x = 2m − 2nx Add 2nx: 3x + 2nx = 2m Factor left side: x(3 + 2n) = 2m Divide by (3 + 2n): x m n = + 2 32 3x mnx− x = 9 −=−x 9 20 11−=x 57 5 1 x += 304 MCGRAW-HILL’S SAT Systems A system is simply a set of equations that are true at the same time, such as these: Although many values for x and y “satisfy” the first equation, like (4, 0) and (2, −3) (plug them in and check!), there is only one solution that works in both equations: (6, 3). (Plug this into both equations and check!) The Law of Substitution The law of substitution simply says that if two things are equal, you can always substitute one for the other. Example: The easiest way to solve this is to substitute the sec- ond equation (which is already “solved” for y) into the first, so that you eliminate one of the unknowns. Since y = x + 1, you can replace the y in the first equation with x + 1 and get 3x + (x + 1) 2 = 7 FOIL the squared binomial: 3x + x 2 + 2x + 1 = 7 Combine like terms: x 2 + 5x + 1 = 7 Subtract 7: x 2 + 5x − 6 = 0 Factor the left side: (x + 6)(x − 1) = 0 Apply the Zero Product Property: x =−6 or x = 1 Plug values back into 2nd equation: y = (−6) + 1 =−5 or y = (1) + 1 = 2 Solutions: (−6, −5) and (1, 2) (Check!) Combining Equations If the two equations in the system are alike enough, you can sometimes solve them more easily by combining equations. The idea is sim- ple: if you add or subtract the corresponding sides of two true equations together, the result should also be a true equation, because you are adding equal things to both sides. This strategy can be simpler than substitution. 37 1 2 xy yx += =+ ⎧ ⎨ ⎩ 3212 315 xy xy −= += Example: Add equations: 5x = 30 Divide by 5: x = 6 Plug this back in and solve for y: 2(6) − 5y = 7 Simplify: 12 − 5y = 7 Subtract 12: − 5y =−5 Divide by 5: y = 1 Special Kinds of “Solving” Sometimes a question gives you a system, but rather than asking you to solve for each unknown, it simply asks you to evaluate another expression. Look carefully at what the question asks you to evaluate, and see whether there is a simple way of combining the equations (adding, subtracting, multiplying, dividing) to find the expression. Example: If 3x − 6y = 10 and 4x + 2y = 2, what is the value of 7x − 4y? Don’t solve for x and y! Just notice that 7x − 4y equals (3x − 6y) + (4x + 2y) = 10 + 2 = 12. “Letter-Heavy” Systems An equation with more than one unknown, or a system with more unknowns than equations, is “letter-heavy.” Simple equations and systems usually have just one solution, but these “letter- heavy” equations and systems usually have more than one solution, and you can often easily find solutions simply by “plugging in” values. Example: If 2m + 5n = 10 and m ≠ 0, then what is the value of You can “guess and check” a solution to the equation pretty easily. Notice that m =−5, n = 4 works. If you plug these values into the expression you’re evaluating, you’ll see it simplifies to 2. 4 10 5 m n− ? 25 7 3523 xy xy −= += ⎧ ⎨ ⎩ Lesson 2: Systems Adding the correspond- ing sides will eliminate the y’s from the system CHAPTER 8 / ESSENTIAL ALGEBRA I SKILLS 305 Concept Review 2: Systems 1. What is a system? 2. What are two algebraic methods for solving systems? 3. How do you check the solution of a system? Solve the following systems by substitution, and check your solution. 4. 3x − 4y = 25.x 2 − 2y = 10 6. x = y − 2 y = 3x − 9 Solve the following systems by combination, and check your solution. 7. a − b = 58.−3x − 5y = 20 9. a + b = 12 −3x − 4y = 14 Give three different solutions to each of the following “letter-heavy” systems. 10. 2x + 5y = 40 11. −2a + 5b + c = 10 a + b = 7 x = _______ y = _______ a = _______ b = _______ c = _______ x = _______ y = _______ a = _______ b = _______ c = _______ x = _______ y = _______ a = _______ b = _______ c = _______ xy xy 33 9 43 2 −=− += 553 2 5 mn mn += = 306 MCGRAW-HILL’S SAT SAT Practice 2: Systems 6. If and , then which of the following expresses m in terms of y? (A) (B) (C) (D) (E) 7. The sum of two numbers is 5 and their difference is 2. Which of the following could be the differ- ence of their squares? (A) −17 (B) −3 (C) 3 (D) 10 (E) 21 8. If 7x + 2y − 6z = 12, and if x, y, and z are pos- itive, then what is the value of (A) 1/12 (B) 1/6 (C) 1/4 (D) 5/12 (E) 7/12 9. If and , then (A) (B) (C) (D) (E) 9 35 17 70 8 35 9 70 2 35 a c = 4 3 1 7 b c = a b2 3 5 = 2 72 + + z xy ? 18 6 3 − y y 2 y y 2 y 3 18 18 3 y m y 5 2 6 = m y 6 3 = 1. If 3x + 2y = 72 and y = 3x, then x = (A) 6 (B) 7 (C) 8 (D) 9 (E) 10 2. The difference of two numbers is 4 and their sum is −7. What is their product? (A) −33.0 (B) −28.0 (C) −10.25 (D) 8.25 (E) 10.5 3. If 4m − 7n = 10 and 2m + 2n = 4, what is the value of 2m − 9n? 4. If 9p = 3a + 1 and 7p = 2a − 3, then which of the following expresses p in terms of a? (A) (B) (C) (D) (E) 5. The cost of one hamburger and two large sodas is $5.40. The cost of three hamburgers and one large soda is $8.70. What is the cost of one hamburger? (A) $1.50 (B) $1.95 (C) $2.40 (D) $2.50 (E) $2.75 a+ 4 2 7 9 a 2 63 a 23 9 a− 31 7 a+ 1 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 CHAPTER 8 / ESSENTIAL ALGEBRA I SKILLS 307 Answer Key 2: Systems 7. (8.5, 3.5) a − b = 5 + a + b = 12 Add the equations: 2a = 17 Divide by 2: a = 8.5 Plug in to find b: (8.5) − b = 5 Subtract 8.5: −b =−3.5 Multiply by −1: b = 3.5 8. (10/3, −6) −3x − 5y = 20 −(−3x − 4y = 14) Subtract the equations: −y = 6 Multiply by −1: y =−6 Plug in to find x: −3x − 5(−6) = 20 Simplify: −3x + 30 = 20 Subtract 30: −3x =−10 Divide by −3: x = 10/3 9. (−12, 15) Add the equations to get Combine fractions: Multiply by 12: 7x =−84 Divide by 7: x =−12 Plug in and solve for y: y = 15 10. There are many solutions. Here are a few: (0, 8); (20, 0); (10, 4); (5, 6) 11. There are many solutions. Here are a few: (1, 6, 0); (3, 4, −4); (2, 5, 31); (7, 0, 24) 7 12 7 x =− xx 34 7+=− Concept Review 2 1. Any set of equations that are true at the same time. 2. Substitution and combination. 3. Plug the solutions back into the equations and check that both equations are true. 4. (−10, −8) Substitute: 3(y − 2) − 4y = 2 Distribute: 3y − 6 − 4y = 2 Combine: −y − 6 = 2 Add 6: −y = 8 Multiply by −1: y =−8 Plug in and solve for x: x = y − 2 = (−8) −2 =−10 5. (4, 3) and (2, −3) Substitute: x 2 − 2(3x − 9) = 10 Distribute: x 2 − 6x + 18 = 10 Subtract 10: x 2 − 6x + 8 = 0 Factor: (x − 4)(x − 2) = 0 (Look over Lesson 5 if that step was tough!) Zero Product Property: x = 4 or x = 2 Plug in and solve for y: y = 3x − 9 = 3(4) − 9 = 3 or = 3(2) − 9 =−3 So the solutions are x = 4 and y = 3 or x = 2 and y =−3. 6. (2, 5) Substitute: Simplify: 2n + 5 = 3n Subtract 2n: 5 = n Plug in to find m: m = () = 2 5 52 5 2 5 53nn ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ += SAT Practice 2 1. C Substitute: 3x + 2(3x) = 72 Simplify: 9x = 72 Divide by 9: x = 8 2. D Translate into equations: x − y = 4 x + y =−7 Add the equations: 2x =−3 Divide by 2: x =−1.5 Substitute: −1.5 + y =−7 Add 1.5: y =−5.5 (−1.5)(−5.5) = 8.25 3. 6 Subtract them: 2m − 9n = (4m − 7n) − (2m + 2n) = 10 − 4 = 6 4. E Subtracting gives 2p = a + 4 Divide by 2: p = (a + 4)/2 5. C Translate: h + 2s = 5.40 3h + s = 8.70 Multiply 2nd eq. by 2: 6h + 2s = 17.40 −(h + 2s = 5.40) Subtract 1st equation: 5h = 12.00 Divide by 5: h = 2.40 6. A Divide the first equation by the second: Simplify: 7. D Translate: x + y = 5 and x − y = 2 Although you could solve this system by combin- ing, it’s easier to remember the “difference of squares” factoring formula: x 2 − y 2 = (x + y)(x − y) = (5)(2) = 10 m y yy =× = 3618 23 m m y y 6 5 2 3 6 =÷ 308 MCGRAW-HILL’S SAT 8. B This is “letter-heavy,” so you can guess and check a solution, like (2, 11, 4), and evaluate the expression: . Or you can also just add 6z to both sides of the equation to get 7x + 2y = 12 + 6z, then substitute: 2 72 2 12 6 2 62 1 6 + + = + + () = + + () = z xy z z z z 24 72 211 6 36 1 6 + () () + () == 9. B Multiply the equations: Multiply by 3 2 3 35 3 2 9 70 : a c = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = a b b c a c2 4 3 2 3 3 5 1 7 3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ == ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 335 CHAPTER 8 / ESSENTIAL ALGEBRA I SKILLS 309 Lesson 3: Working with Exponentials What Are Exponentials? An exponential is simply any term with these three parts: If a term seems not to have a coefficient or ex- ponent, the coefficient or exponent is always assumed to be 1! Examples: 2x means 2x 1 y 3 means 1y 3 Expand to Understand Good students memorize the rules for working with exponentials, but great students understand where the rules come from. They come from simply expanding the exponentials and then collecting or cancelling the factors. Example: What is (x 5 ) 2 in simplest terms? Do you add ex- ponents to get x 7 ? Multiply to get x 10 ? Power to get x 25 ? The answer is clear when you expand the exponential. Just remember that raising to the nth power simply means multiplying by itself n times. So (x 5 ) 2 = (x 5 )(x 5 ) = (x⅐x⅐x⅐x⅐x)(x⅐x⅐x⅐x⅐x) = x 10 . Doing this helps you to see and understand the rule of “multiplying the powers.” Adding and Subtracting Exponentials When adding or subtracting exponentials, you can combine only like terms, that is, terms with the same base and the same exponent. When adding or subtracting like exponentials, re- member to leave the bases and exponents alone. Example: 5x 3 + 6x 3 + 4x 2 = (5x 3 + 6x 3 ) + 4x 2 = x 3 (5 + 6) + 4x 2 = 11x 3 + 4x 2 Notice that combining like terms always involves the Law of Distribution (Chapter 7, Lesson 2). Multiplying and Dividing Exponentials You can simplify a product or quotient of exponen- tials when the bases are the same or the exponents are the same. If the bases are the same, add the exponents (when multiplying) or subtract the exponents (when dividing) and leave the bases alone. (5m 5 )(12m 2 ) =(5)(m)(m)(m)(m)(m)(12) (m)(m) = (5)(12)(m)(m)(m)(m)(m)(m)(m) = 60m 7 If the exponents are the same, multiply (or divide) the bases and leave the exponents alone. Example: (3m 4 )(7n 4 ) = (3)(m)(m)(m)(m)(7)(n)(n)(n)(n) = (3)(7)(mn)(mn)(mn)(mn) = 21(mn) 4 Raising Exponentials to Powers When raising an exponential to a power, multi- ply the exponents, but don’t forget to raise the coefficient to the power and leave the base alone. Example: (3y 4 ) 3 = (3y 4 )(3y 4 )(3y 4 ) = (3y⅐y⅐y⅐y)(3y⅐y⅐y⅐y)(3y⅐y⅐y⅐y) = (3)(3)(3)(y⅐y⅐y⅐y)(y⅐y⅐y⅐y)(y⅐y⅐y⅐y) = 27y 12 512 3 51212121212 3333 5 5 () () = ()()()()() ()()()( ))( ) = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 3 5 12 3 12 3 12 3 12 3 112 3 54 5 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = () 6 3 6 3 7 4 p p ppppppp pppp = ()()()()()()() ()()()() = 22 3 p Coefficient Base Exponent 4 x 3 . 300 MCGRAW-HILL’S SAT Equations as Balanced Scales Algebra is really common sense once you start think- ing of every equation as a balanced scale. The. ______________________ Solution: ______________________ 5 2 37 x += xx+ = −2 3 5 2 2 3 2 2 x x= 302 MCGRAW-HILL’S SAT SAT Practice 1: Solving Equations 1. If 5d + 12 = 24, then 5d − 12 = (A) −24 (B) −12 (C). 2m Factor left side: x(3 + 2n) = 2m Divide by (3 + 2n): x m n = + 2 32 3x mnx− x = 9 −=−x 9 20 11−=x 57 5 1 x += 304 MCGRAW-HILL’S SAT Systems A system is simply a set of equations that are true