Multiplying each side of \(\frac{t+5}{t-5}\) = 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.

The question states that \(\sqrt{x-a}\) = x − 4 and that a = 2, so substituting 2 for a in the equation yields \(\sqrt{x-2}\) = x − 4. To solve for x, square each side of the equation, which gives (\(\sqrt{x-2}\))² = (x − 4)², or x − 2 = (x − 4)². Then, expanding (x − 4)² yields x − 2 = x² −8x + 16, or 0 = x² − 9x + 18.Factoring the right-hand side gives 0 = (x − 3)(x − 6), and so x = 3 or x = 6. However, for x = 3, the original equation becomes \(\sqrt{3-2}\) = 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 \(\sqrt{6-2}\) = 6 − 4, which yields √4 = 2, or 2 = 2. Since this is true, the solution set of \(\sqrt{x-2}\) = 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.

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 = −⁶⁄₂x + \(\frac{15}{2}\) = −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 y-intercepts.)

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.

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 \(\frac{26−8}{16-4}=\frac{18}{12}=\frac{3}{2}\) = 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.

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.

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 24x² + 12x + 6x + 3. Combining like terms gives 24x² + 18x + 3.

Choice A is incorrect and may be the result of performing the term-by-term multiplication of 3(2x + 1)(4x + 1) and treating every term as an x-term. 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, 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).

The first equation can be rewritten as x = 6y. Substituting 6y for x in the second equation gives 4(y + 1) = 6y. The left-hand 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.

Since f(x) = ³⁄₂x + b and f(6) = 7, substituting 6 for x in f(x) = ³⁄₂x + 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).

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.