Algebra Practice Set Answers and Explanations

1. Basic

1. 1
   None of the operations in parentheses can be completed, so the first step is to distribute the factors across the terms in parentheses: 3x – 18 + 8 – 2 + 4x = 7 – 4x – 8. Combine like terms: 7x – 12 = –1 – 4x. Add 4x and 12 to each side of the equation: 7x + 4x – 12 + 12 = –1 + 12 – 4x + 4x. So, 11x = 11 and x = 1.
2. (a – 2b)(c + d)
   Notice that the first 2 terms each contain the variable c, so convert these to (c)(a – 2b) by factoring out c. The last 2 terms each contain the variable d, so convert these to (d)(a – 2b). The expression can be restated as (c)(a – 2b) + (d)(a – 2b). Now (a – 2b) is a common term, so factor it out to get (a – 2b)(c + d).                                                         
3. 4
   Substitute 3 for x in the expression: 3³ – 3² – 7(3 – 1) = 27 – 9 – 7(2) = 4.
4. 2
   Add 7 to both sides of the inequality to get 3x ≥ 6. Divide both sides by 3 to determine that x ≥ 2. Since “≥” means “greater than or equal to,” the minimum value of x is 2.
5. –16
   Substitute 2x for y in the equation: 3(2x + 6x) – 7(2(2x) + 3) = 11. Simplify: 3(8x) – 7(4x + 3) = 11. Complete the multiplications: 24x – 28x – 21 = 11. Finally, add 21 to both sides of the equation and combine like terms to get –4x = 32. Divide both sides by –4 to get x = –8. The question asks for the value of y, which is 2x, so multiply –8 by 2 to get y = –16.

6. Intermediate

6. 
   First get rid of the fraction by multiplying everything by 4: . The equation becomes 4xy + 7 + 3x – 8y = 4. Add 8y and subtract 7 from both sides to get 4xy + 3x = 4 + 8y – 7. Factor out x and combine like terms: x(4y + 3) = 8y – 3. Finally, divide both sides by (4y + 3), so  .
7. 2, –2
   This is the difference of two squares, a pattern equation that is likely to appear on the GRE. Remember that a² – b² = (a + b)(a – b). Since 9x² = (3x)² and 36 = (6)², the equation becomes (3x + 6)(3x – 6) = 0. If 3x + 6 = 0, then 3x = –6 and x = –2; if 3x – 6 = 0, then 3x = 6 and x = 2. A shortcut would be to add 36 to both sides to get 9x² = 36 then dividing both sides by 9 to get x² = 4. Remember that x can be either +2 or –2! 
8. 3

   This question presents 2 linear equations with 2 variables. Systems of equations such as this can be solved by combination or substitution. To combine the 2 equations, adapt one of the equations to make the absolute value of one of its variable’s coefficients the same as the absolute value of that variable’s coefficient in the other equation. Here, the terms of the first equation can be multiplied by 2 to get 2x + 8y = 38. Subtract the second equation from this so that the 2x cancels out:

   To solve using substitution, rearrange the first equation to get x = 19 – 4y. Substitute that for x in the second equation: 2(19 – 4y) – y = 11, so 38 – 8y – y = 11, 27 = 9y, and y  = 3.

9. 6
   Symbolism questions such as this are algebra questions that can be solved by substitution. In order to find the value of ♫(r), first determine the value of r. Since 3r – 4 = 5, 3r = 9 and r = 3. Substitute 3 for x to get  . Simplify to 
   
10. 14
    A series of numbers beginning with the third number in the series is composed of the previous number in the series plus the second prior number plus 3. So, in order to calculate the value of any number in the series, the value of the 2 preceding numbers is needed. To find the value of s₄, the values of s₃ and s₂ must first be determined. The question provides values for s₁ and s₂, so s₃ = s₂ + s₁ + 3, which is  4 + 0 + 3 = 7. So s₄ = 7 + 4 + 3 = 14.

11. Advanced

11. 5

    The question has three variables and three linear equations, so you can solve for any of the variables as long as the equations are distinct. Start by simplifying the first equation: x – 2z = 2y – 2z, so x =  2y. Now substitute 2y wherever x appears in the other equations to end up with two equations with two variables: 2(2y) – 6y + z = 1 simplifies to z – 2y = 1. Similarly, 3(2y) + y – 2z = 4 becomes 7y – 2z = 4. Multiply the first of these two equations by 2 to get 2z – 4y = 2. Now add the equations: 

    

    Plug that value into 2z – 4y = 2: 2z – 8 = 2, so 2z = 10 and z = 5.

12. 3
    When confronted with a function of a function, as in this question, start from the inside and work outward. Substitute 2 for y to get ##(2) = 2² – 2(2) + 3 = 4 – 4 + 3 = 3. Now evaluate  .
13. 
    First rearrange the equation into standard quadratic form so that it can be factored: 6x² – 11x – 10 = 0. Identify the factors of the last term and those of the coefficient of the first term. The factors of –10 are (–10 and 1), (10 and –1), (–5 and 2), and (5 and –2). The factors of 6 are (1 and 6), (–1 and –6), (2 and 3), and (–2 and –3). Find the pairs that will result in an algebraic sum of –11, which is the coefficient of the middle term. Note that 3 × (–5) = –15 and 2 × 2 = 4; –15 + 4 = –11. So (2x – 5)(3x + 2) are the correct factors. Set each factor equal to zero to find the possible values of x. If 2x – 5 = 0, then  ; if 3x + 2 = 0, then .
14. 85
    To predict a forward value for the sequence, determine the equation that quantifies the relationship between any value and “the two prior values.” Looking at the overall trend, the values always increase, and the rate of increase accelerates. Start with s_3, which could be the simple sum of s₁ and s₂. However, moving along to s₄, that value is greater than the sum of s₂ and s₃, so there must be a multiplier or exponent in the formula. Continuing forward, try to identify a pattern relating to how much greater each value is than the sum of the two prior values. One way to do this is to set up a table showing the “missing” quantities.                                                                                    

    
    

    Seq	Value	sx–1 + sx–2	Difference
    s5	5	3 + 1 = 4	5 –4 = 1
    s6	11	5 + 3 = 8	11 – 8 = 3
    s7	21	11 + 5 = 16	21 – 16 = 5
    s8	43	21 + 11 = 32	43 – 32 = 11

    Look at the values on the far right: 1, 3, 5, and 11. Those are the values of s_{x – 2}. So the equation for this series is s_x = s_{x – 1} + s_{x – 2} + s_{x – 2} = s_{x – 1} + 2(s_{x – 2}). Apply this formula to calculate the value of s₉: s₉ = 43 + (2)21 = 43 + 42 = 85.

    
    
15. 37
    Represent the three people’s number of tokens by the first letters of their names and translate the information given into algebraic equations: R = T + 3, C = 2R, and C = 3T – 1. Since there are three distinct linear equations and three variables, you can solve this system of equations. Substitute 2R for C in the third equation: 2R = 3T – 1. Double the first equation to get 2R = 2T + 6. This gives two different equations for 2R, so set them equal to each other: 2T + 6 = 3T – 1. Subtract 2T from both sides and add 1 to both sides to get T = 7. This means that R has 10 tokens, 3 more than T, and C  has twice as many as R, which is 20. The total number of tokens among the three people is 7 + 10 + 20 = 37.