Practice Question What is the equation represented in the graph above? a. y ϭ x 2 ϩ 3 b. y ϭ x 2 Ϫ 3 c. y ϭ (x ϩ 3) 2 d. y ϭ (x Ϫ 3) 2 e. y ϭ (x Ϫ 1) 3 Answer b. This graph is identical to a graph of y ϭ x 2 except it is moved down 3 so that the parabola intersects the y-axis at Ϫ3 instead of 0. Each y value is 3 less than the corresponding y value in y ϭ x 2 , so its equation is therefore y ϭ x 2 Ϫ 3. x y 123456 1 2 3 4 5 –1 –2 –6 –5 –4 –3 –1–2–3–4–5–6 x y 1234567 1 2 3 4 5 –1 –2 –3 –1–2–3–4–5–6–7 –ALGEBRA REVIEW– 84  Rational Equations and Inequalities Rational numbers are numbers that can be written as fractions (and decimals and repeating decimals). Similarly, rational equations are equations in fraction form. Rational inequalities are also in fraction form and use the sym- bols <, >, ≤, and ≥ instead of ϭ. Example Given ϭ 30, find the value of x. Factor the top and bottom: ϭ 30 You can cancel out the (x ϩ 5) and the (x Ϫ 2) terms from the top and bottom to yield: x ϩ 7 ϭ 30 Now solve for x: x ϩ 7 ϭ 30 x ϩ 7 Ϫ 7 ϭ 30 Ϫ 7 x ϭ 23 Practice Question If ϭ 17, what is the value of x? a. Ϫ16 b. Ϫ13 c. Ϫ8 d. 2 e. 4 Answer e. To solve for x, first factor the top and bottom of the fractions: ϭ 17 ϭ 17 You can cancel out the (x ϩ 8) and the (x Ϫ 2) terms from the top and bottom: x ϩ 13 ϭ 17 Solve for x: x ϩ 13 Ϫ 13 ϭ 17 Ϫ 13 x ϭ 4 (x + 8)(x + 13)(x Ϫ 2) ᎏᎏᎏ (x + 8)(x Ϫ 2) (x + 8)(x 2 + 11x Ϫ 26) ᎏᎏᎏ (x 2 + 6x Ϫ 16) (x + 8)(x 2 + 11x Ϫ 26) ᎏᎏᎏ (x 2 + 6x Ϫ 16) (x + 5)(x + 7)(x Ϫ 2) ᎏᎏᎏ (x + 5)(x Ϫ 2) (x + 5)(x 2 + 5x Ϫ 14) ᎏᎏᎏ (x 2 + 3x Ϫ 10) –ALGEBRA REVIEW– 85  Radical Equations Some algebraic equations on the SAT include the square root of the unknown. To solve these equations, first iso- late the radical. Then square both sides of the equation to remove the radical sign. Example 5͙c ෆ ϩ 15 ϭ 35 To isolate the variable, subtract 15 from both sides: 5͙c ෆ ϩ 15 Ϫ 15 ϭ 35 Ϫ 15 5͙c ෆ ϭ 20 Next, divide both sides by 5: ᎏ 5 ͙ 5 c ෆ ᎏ ϭ ᎏ 2 5 0 ᎏ ͙c ෆ ϭ 4 Last, square both sides: (͙c ෆ ) 2 ϭ 4 2 c ϭ 16 Practice Question If 6͙d ෆ Ϫ 10 ϭ 32, what is the value of d? a. 7 b. 14 c. 36 d. 49 e. 64 Answer d. To solve for d, isolate the variable: 6͙d ෆ Ϫ 10 ϭ 32 6͙d ෆ Ϫ 10 ϩ 10 ϭ 32 ϩ 10 6͙d ෆ ϭ 42 ϭ ͙d ෆ ϭ 7 (͙d ෆ ) 2 ϭ 7 2 d ϭ 49 42 ᎏ 6 6͙d ෆ ᎏ 6 –ALGEBRA REVIEW– 86  Sequences Involving Exponential Growth When analyzing a sequence, try to find the mathematical operation that you can perform to get the next number in the sequence. Let’s try an example. Look carefully at the following sequence: 2,4,8,16,32, Notice that each successive term is found by multiplying the prior term by 2. (2 ϫ 2 ϭ 4, 4 ϫ 2 ϭ 8, and so on.) Since each term is multiplied by a constant number (2), there is a constant ratio between the terms. Sequences that have a constant ratio between terms are called geometric sequences. On the SAT, you may be asked to determine a specific term in a sequence. For example, you may be asked to find the thirtieth term of a geometric sequence like the previous one. You could answer such a question by writ- ing out 30 terms of a sequence, but this is an inefficient method. It takes too much time. Instead, there is a for- mula to use. Let’s determine the formula: First, let’s evaluate the terms. 2,4,8,16,32, Term 1 ϭ 2 Ter m 2 ϭ 4, which is 2 ϫ 2 Ter m 3 ϭ 8, which is 2 ϫ 2 ϫ 2 Ter m 4 ϭ 16, which is 2 ϫ 2 ϫ 2 ϫ 2 You can also write out each term using exponents: Ter m 1 ϭ 2 Ter m 2 ϭ 2 ϫ 2 1 Ter m 3 ϭ 2 ϫ 2 2 Ter m 4 ϭ 2 ϫ 2 3 We can now write a formula: Te r m n ϭ 2 ϫ 2 n Ϫ 1 So, if the SAT asks you for the thirtieth term, you know that: Term 30 ϭ 2 ϫ 2 30 Ϫ 1 ϭ 2 ϫ 2 29 The generic formula for a geometric sequence is Term n ϭ a 1 ϫ r n Ϫ 1 ,where n is the term you are looking for, a 1 is the first term in the series, and r is the ratio that the sequence increases by. In the above example, n ϭ 30 (the thirtieth term), a 1 ϭ 2 (because 2 is the first term in the sequence), and r ϭ 2 (because the sequence increases by a ratio of 2; each term is two times the previous term). You can use the formula Term n ϭ a 1 ϫ r n Ϫ 1 when determining a term in any geometric sequence. –ALGEBRA REVIEW– 87 Practice Question 1,3,9,27,81, What is the thirty-eighth term of the sequence above? a. 3 38 b. 3 ϫ 1 37 c. 3 ϫ 1 38 d. 1 ϫ 3 37 e. 1 ϫ 3 38 Answer d. 1,3,9,27,81, is a geometric sequence.There is a constant ratio between terms. Each term is three times the previous term. You can use the formula Term n ϭ a 1 ϫ r n Ϫ 1 to determine the nth term of this geometric sequence. First determine the values of n, a 1 , and r: n ϭ 38 (because you are looking for the thirty-eighth term) a 1 ϭ 1 (because the first number in the sequence is 1) r ϭ 3 (because the sequence increases by a ratio of 3; each term is three times the previous term.) Now solve: Te r m n ϭ a 1 ϫ r n Ϫ 1 Term 38 ϭ 1 ϫ 3 38 Ϫ 1 Term 38 ϭ 1 ϫ 3 37  Systems of Equations A system of equations is a set of two or more equations with the same solution. If 2c ϩ d ϭ 11 and c ϩ 2d ϭ 13 are presented as a system of equations, we know that we are looking for values of c and d, which will be the same in both equations and will make both equations true. Two methods for solving a system of equations are substitution and linear combination. Substitution Substitution involves solving for one variable in terms of another and then substituting that expression into the second equation. Example Here are the two equations with the same solution mentioned above: 2c ϩ d ϭ 11 and c ϩ 2d ϭ 13 To solve, first choose one of the equations and rewrite it, isolating one variable in terms of the other. It does not matter which variable you choose. 2c ϩ d ϭ 11 becomes d ϭ 11 Ϫ 2c Next substitute 11 Ϫ 2c for d in the other equation and solve: –ALGEBRA REVIEW– 88 c ϩ 2d ϭ 13 c ϩ 2(11 Ϫ 2c) ϭ 13 c ϩ 22 Ϫ 4c ϭ 13 22 Ϫ 3c ϭ 13 22 ϭ 13 ϩ 3c 9 ϭ 3c c ϭ 3 Now substitute this answer into either original equation for c to find d. 2c ϩ d ϭ 11 2(3) ϩ d ϭ 11 6 ϩ d ϭ 11 d ϭ 5 Thus, c ϭ 3 and d ϭ 5. Linear Combination Linear combination involves writing one equation over another and then adding or subtracting the like terms so that one letter is eliminated. Example x Ϫ 7 ϭ 3y and x ϩ 5 ϭ 6y First rewrite each equation in the same form. x Ϫ 7 ϭ 3y becomes x Ϫ 3y ϭ 7 x ϩ 5 ϭ 6y becomes x Ϫ 6y ϭϪ5. Now subtract the two equations so that the x terms are eliminated, leaving only one variable: x Ϫ 3y ϭ 7 Ϫ (x Ϫ 6y ϭϪ5) (x Ϫ x) ϩ (Ϫ 3y ϩ 6y) ϭ 7 Ϫ (Ϫ5) 3y ϭ 12 y ϭ 4 is the answer. Now substitute 4 for y in one of the original equations and solve for x. x Ϫ 7 ϭ 3y x Ϫ 7 ϭ 3(4) x Ϫ 7 ϭ 12 x Ϫ 7 ϩ 7 ϭ 12 ϩ 7 x ϭ 19 Therefore, the solution to the system of equations is y ϭ 4 and x ϭ 19. –ALGEBRA REVIEW– 89 . 3y x Ϫ 7 ϭ 3(4) x Ϫ 7 ϭ 12 x Ϫ 7 ϩ 7 ϭ 12 ϩ 7 x ϭ 19 Therefore, the solution to the system of equations is y ϭ 4 and x ϭ 19. –ALGEBRA REVIEW 89 . 14 c. 36 d. 49 e. 64 Answer d. To solve for d, isolate the variable: 6͙d ෆ Ϫ 10 ϭ 32 6͙d ෆ Ϫ 10 ϩ 10 ϭ 32 ϩ 10 6͙d ෆ ϭ 42 ϭ ͙d ෆ ϭ 7 (͙d ෆ ) 2 ϭ 7 2 d ϭ 49 42 ᎏ 6 6͙d ෆ ᎏ 6 –ALGEBRA REVIEW 86  Sequences. sequence. –ALGEBRA REVIEW 87 Practice Question 1,3 ,9, 27,81, What is the thirty-eighth term of the sequence above? a. 3 38 b. 3 ϫ 1 37 c. 3 ϫ 1 38 d. 1 ϫ 3 37 e. 1 ϫ 3 38 Answer d. 1,3 ,9, 27,81, is