Using the Bowtie method, you find that \(\frac{y+x}{xy}\)=6.By multiplying both sides of the second equation by -\(\frac{y+x}{xy}\), you find that \(\frac{1}{6}=\frac{xy}{z+y}\).Flipping both fractions yields \(\frac{z+y}{xy}\) =6, and thus,\(\frac{y+x}{xy}=\frac{z+y}{zy}\). Inspecting the two fractions, you may realize that z must equal x. Alternatively, by applying the Bowtie again, you obtain (y + x)(zy) = (z + y)(xy), and thus zy² + xyz = xyz + xy², meaning zy² = xy², or z = x, so the answer is choice (C).

Translate what you are told into algebra: 2 ≤ x + y ≤ 5 and 0 ≤ 2x + 3y ≤ 40.By manipulating the inequalities, you find that 2 − y ≤ x ≤ 5 – y and –2x ≤ 3y ≤ 40 − 2x. Substitute the endpoints of the values of x into the second inequality: –2(2 – y) ≤ 3y ≤ 40 − 2(2 − y) and –2(5 − y) ≤ 3y ≤ 40 − 2(5 − y).By solving for y, the first range yields –4 ≤ y ≤ 36, and the second yields – 10 ≤ y ≤ 30, meaning –10 ≤ y ≤ 36 overall.

Translate what you are told into algebra: ^x⁄_y =2 and x = \(\frac{75}{100}\times z=\frac{3}{4}z\). Notice that the problem then tells you that z is ³⁄₄ larger than w, not z is ³⁄₄ of w.That means \(z=w+\left ( \frac{3}{4}w \right )=\left ( 1+\frac{3}{4} \right )w=\frac{7}{4}w\). At this point, you can either Plug In or do algebra.To Plug In, choose 4 for w, so z = 7, and \(x=\frac{3}{4}\times 7=\frac{21}{4}\). If =2, then x = 2y, so\(\frac{21}{4}\) =2y; y= \(\frac{21}{4}=2y\). Now plug \(\frac{21}{8}\) into the answer choices for y to see which hits your target number, w = 4. Only choice (C) does. Alternatively, to do algebra, combine the first two equations you translated into algebra: 2y= ³⁄₄z; z= ⁸⁄₃.Combining this equation with the one you derived above, it follows that (⁸⁄₃)y= ⁷⁄₄w, and w= \(\frac{32}{21}\)y; the answer is choice (C).

Know your common quadratics.You are asked for the value of x + y.Because x² − y² = (x + y)(x − y) = 24, if the value of (x − y) is found, (x + y) can be figured out. Manipulate the equation 0.5(y − x) = –1 to get (y − x) = –2. Multiplying both sides by –1, you get x – y = 2. Substitute this value into the equation (x + y)(x – y) = 24 and then divide both sides of the equation by 2 to get (x + y) = 12.The answer is choice (D).

If a range of values for a can be found, then the range of values for a² can be found. Start by testing the end values of b, –3 and 1. Plug in –3 for b in the first given inequality then solve for a.You find that –4 < a < 7. If b = 1, 0 < a <11; b could be any integer in the range –3 ≤ b ≤ 1, this means –4 < a < 11 overall.Remember to take the last step, though!The question is looking for the range of a², not a; a² is always positive (i.e., 0 < a²).Because a < 11, a² < 121.This means 0 < a² < 121; the answer is choice (D).

The problem contains the phrase “must be,” so try to find numbers to plug in that disprove four of the five answer choices. If a = 4 and ab = –12, then b = –3; eliminate choices (B), (D), and (E). Now determine whether b must be an integer. If a = 4 and ab = –2, then b = -¹⁄₂; eliminate choice (C), and you’re left with choice (A), the correct answer.

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First, factor the quadratic: (2x + 2)(x – 3) = 0. Next, solve for each solution (root). If 2x + 2 = 0, then x = –1. If x – 3 = 0, then x = 3.The sum of the roots is 2, and twice their sum is 4.The factors of 4 are 1, 2, 4, and the product of those factors is 8, the final answer.

Use the information in the problem to write the equations x + y = 27 and ^x⁄₄ + x = y.You can combine these two equations to say that ^x⁄₄ = x + x = 27. When you solve this new equation you find that x = 12.You can then go back to your first equation and find that y = 15.The positive difference between 15 and 12 is 3, so select choice (A).

You know that \(\frac{xy}{z^{2}}\), so solve this one by plugging in values for z, x, and y. If x = 12, then z = 36 and y = 144;\(\frac{xy}{z^{2}}\).The correct answer is choice (B).

First, plug in 1 for the number of boxes of chocolate chip cookies. In Quantity A, if Sally buys 1 box of chocolate chip cookies, then she buys 2(1) + 3 = 5 boxes of oatmeal cookies, for a total of 6 boxes, or 36 cookies. In Quantity B, if Sally buys 1 box of chocolate chip cookies, then she buys 3(1) – 2 = 1 box of oatmeal cookies, for a total of 2 boxes, or 12 cookies. Quantity A is greater, so eliminate choices (B) and (C). Now plug in 10 boxes of chocolate chip cookies: Quantity A is now 2(10) + 3 = 23 boxes of oatmeal cookies, for a total of 33 boxes, or 198 cookies, and Quantity B is now 3(10) – 2 = 28 boxes of oatmeal cookies, for a total of 38 boxes, or 228 cookies. Quantity B is now greater, so eliminate choice (A), and you’re left with choice (D).