If \(\frac{t+5}{t5}\)= 10, what is the value of t ?

Solution
Multiplying each side of \(\frac{t+5}{t5}\) = 10 by t − 5 gives t + 5 = 10(t − 5). Distributing the 10 over the values in the parentheses yields t + 5 = 10t − 50. Subtracting t from each side of the equation gives 5 = 9t − 50, and then adding 50 to each side gives 55 = 9t. Finally, dividing each side by 9 yields t = \(\frac{55}{9}\).
Choices A, B, and C are incorrect and may be the result of calculation errors or using the distribution property improperly.
If a = 2, what is the solution set of the equation above?

Solution
The question states that \(\sqrt{xa}\) = x − 4 and that a = 2, so substituting 2 for a in the equation yields \(\sqrt{x2}\) = x − 4. To solve for x, square each side of the equation, which gives (\(\sqrt{x2}\))^{2} = (x − 4)^{2}, or x − 2 = (x − 4)^{2}. Then, expanding (x − 4)^{2} yields x − 2 = x^{2} −8x + 16, or 0 = x^{2} − 9x + 18.Factoring the righthand side gives 0 = (x − 3)(x − 6), and so x = 3 or x = 6. However, for x = 3, the original equation becomes \(\sqrt{32}\) = 3 − 4, which yields 1 = −1, which is not true. Hence, x = 3 is an extraneous solution that arose from squaring each side of the equation. For x = 6, the original equation becomes \(\sqrt{62}\) = 6 − 4, which yields √4 = 2, or 2 = 2. Since this is true, the solution set of \(\sqrt{x2}\) = x − 4 is {6}.
Choice A is incorrect because it includes the extraneous solution in the solution set. Choice B is incorrect and may be the result of a calculation or factoring error. Choice C is incorrect because it includes only the extraneous solution, and not the correct solution, in the solution set.
Which of the following equations represents a line that is parallel to the line with equation y = −3x + 4?

Solution
For an equation of a line in the form y = mx + b, the constant m is the slope of the line. Thus, the line represented by y = −3x + 4 has slope −3. Lines that are parallel have the same slope. To find out which of the given equations represents a line with the same slope as the line represented by y = −3x + 4, one can rewrite each equation in the form y = mx + b, that is, solve each equation for y. Choice A, 6x + 2y = 15, can 25 be rewritten as 2y = −6x + 15 by subtracting 6x from each side of the equation. Then, dividing each side of 2y = −6x + 15 by 2 gives y = −^{6}⁄_{2}x + \(\frac{15}{2}\) = −3x + \(\frac{15}{2}\) . Therefore, this line has slope −3 and is parallel to the line represented by y = −3x + 4. (The lines are parallel, not coincident, because they have different yintercepts.)
Choices B, C, and D are incorrect and may be the result of common misunderstandings about which value in the equation of a line represents the slope of the line.
While preparing to run a marathon, Amelia created a training schedule in which the distance of her longest run every week increased by a constant amount. If Amelia’s training schedule requires that her longest run in week 4 is a distance of 8 miles and her longest run in week 16 is a distance of 26 miles, which of the following best describes how the distance Amelia runs changes between week 4 and week 16 of her training schedule?

Solution
In Amelia’s training schedule, her longest run in week 16 will be 26 miles and her longest run in week 4 will be 8 miles. Thus, Amelia increases the distance of her longest run by 18 miles over the course of 12 weeks. Since Amelia increases the distance of her longest run each week by a constant amount, the amount she increases the distance of her longest run each week is \(\frac{26−8}{164}=\frac{18}{12}=\frac{3}{2}\) = 1.5 miles.
Choices A, B, and C are incorrect because none of these training schedules would result in increasing Amelia’s longest run from 8 miles in week 4 to 26 miles in week 16. For example, choice A is incorrect because if Amelia increases the distance of her longest run by 0.5 miles each week and has her longest run of 8 miles in week 4, her longest run in week 16 would be 8 + 0.5 ∙ 12 = 14 miles, not 26 miles.
If \(\frac{ab}{b}\) = ^{3}⁄_{7} , which of the following must also be true?

Solution
The equation \(\frac{a+b}{b}\) = ^{3}⁄_{7} can be rewritten as ^{a}⁄_{b}  ^{b}⁄_{b}= ^{3}⁄_{7} or, from which it follows that ^{a}⁄_{b} − 1 = ^{3}⁄_{7}, or ^{a}⁄_{b} = ^{3}⁄_{7} + 1 = \(\frac{10}{7}\).
Choices A, C, and D are incorrect and may be the result of calculation errors in rewriting \(\frac{a − b}{b}\) = ^{3}⁄_{7}. For example, choice A may be the result of a sign error in rewriting \(\frac{a − b}{b}\) as ^{a}⁄_{b} + ^{b}⁄_{b} =^{a}⁄_{b} + 1.
3(2x + 1)(4x + 1)
Which of the following is equivalent to the expression above?

Solution
The expression 3(2x + 1)(4x + 1) can be simplified by first distributing the 3 to yield (6x + 3)(4x + 1), and then expanding to obtain 24x^{2} + 12x + 6x + 3. Combining like terms gives 24x^{2} + 18x + 3.
Choice A is incorrect and may be the result of performing the termbyterm multiplication of 3(2x + 1)(4x + 1) and treating every term as an xterm. Choice B is incorrect and may be the result of correctly finding (6x + 3)(4x + 1), but then multiplying only the first terms, (6x)(4x), and the last terms, (3)(1), but not the outer or inner terms. Choice D is incorrect and may be the result of incorrectly distributing the 3 to both terms to obtain (6x + 3)(12x + 3), and then adding 3 + 3 and 6x + 12x and incorrectly adding the exponents of x.
If f(x) = −2x + 5, what is f(−3x) equal to?

Solution
If f(x) = − 2x + 5, then one can evaluate f(−3x) by substituting −3x for every instance of x. This yields f(−3x) = −2 (−3x) + 5, which simplifies to 6x + 5.
Choices A, C, and D are incorrect and may be the result of miscalculations in the substitution or of misunderstandings of how to evaluate f(−3x).
^{x}⁄_{y} = 6
4(y + 1) = x
If (x, y) is the solution to the system of equations above, what is the value of y ?

Solution
The first equation can be rewritten as x = 6y. Substituting 6y for x in the second equation gives 4(y + 1) = 6y. The lefthand side can be rewritten as 4y + 4, giving 4y + 4 = 6y. Subtracting 4y from both sides of the equation gives 4 = 2y, or y = 2.
Choices B, C, and D are incorrect and may be the result of a computational or conceptual error when solving the system of equations.
f(x)= ^{3}⁄_{2}x + b
In the function above, b is a constant. If f(6) = 7, what is the value of f(−2) ?

Solution
Since f(x) = ^{3}⁄_{2}x + b and f(6) = 7, substituting 6 for x in f(x) = ^{3}⁄_{2}x + b gives f(6) = ^{3}⁄_{2}(6) + b = 7. Then, solving the equation ^{3}⁄_{2}(6) + b = 7 for b gives \(\frac{18}{2}\) + b = 7, or 9 + b = 7. Thus, b = 7 − 9 = −2. Substituting this value back into the original function gives f(x) =^{3}⁄_{2} x − 2; therefore, one can evaluate f( −2) by substituting −2 for x: ^{3}⁄_{2} (−2) − 2 = − ^{6}⁄_{2} − 2 = −3 − 2 = −5.
Choice B is incorrect as it is the value of b, not of f(−2). Choice C is incorrect as it is the value of f(2), not of f(−2). Choice D is incorrect as it is the value of f(6), not of f(−2).
Which of the following expressions is equal to 0 for some value of x ?

Solution
The expression x − 1 −1 will equal 0 if x − 1 = 1. This is true for x = 2 and for x = 0. For example, substituting x = 2 into the expression x − 1 − 1 and simplifying the result yields 2 − 1 − 1 = 1 − 1 = 1 − 1 = 0. Therefore, there is a value of x for which x − 1 − 1 is equal to 0.
Choice B is incorrect. By definition, the absolute value of any expression is a nonnegative number. Substituting any value for x into the expression x + 1 will yield a nonnegative number as the result. Because the sum of a nonnegative number and a positive number is positive, x + 1 + 1 will be a positive number for any value of x. Therefore, x + 1 + 1 ≠ 0 for any value of x. Choice C is incorrect. By definition, the absolute value of any expression is a nonnegative number. Substituting any value for x into the expression 1 − x will yield a nonnegative number as the result. Because the sum of a nonnegative number and a positive number is positive, 1 − x + 1 will be a positive number for any value of x. Therefore, 1 − x + 1 ≠ 0 for any value of x. Choice D is incorrect. By definition, the absolute value of any expression is a nonnegative number. Substituting any value for x into the expression x − 1 will yield a nonnegative number as the result. Because the sum of a nonnegative number and a positive number is positive, x − 1 + 1 will be a positive number for any value of x. Therefore, x − 1 + 1 ≠ 0 for any value of x.