value of 4c – 4d. There are several ways to go about this. One way is to multiply the terms
of the equation 2c + 3d = 17 by –3 to get –6c – 9d = –51. If you stack and add this equation
with the second equation, the result is –4d = –12, which solves to d = 3. Plug this value for
d into the equation 6c + 5d = 39 to get 6c + 15 = 39, so 6c = 24 and c = 4. Therefore, 4c –
4 d = 4(4) – 4(3) = 16 – 12 = 4. This is (C).11. A Factoring the left side of the equation x^2 + 2xy + y^2 = 64 gives (x + y)^2 = 64. Taking the
square root of both sides of the equation, we find that x + y = 8 or –8. The other equation
provides that y – x = 12, so y = x + 12. Substitute this value of y into the first equation:
either x + (x + 12) = 8, so 2x + 12 = 8, 2x = –4, and x = –2, or else or x + (x + 12) = –8, so
2 x + 12 = –8, so 2x = –20, and x = –10. Therefore, x could be either –2 or –10, and only –
10 is an option in the answers, so (A) is correct.
D Translate from English to math in bite-sized pieces. Make the price of a hot yoga lesson h
and the price of a zero gravity yoga session z. If she offers 2 hot yoga and 3 zero gravity
yoga sessions for $400, then 2h + 3z = 400. Similarly, if 4 hot yoga and 2 zero gravity yoga
sessions are $440, then 4h + 2z = 440. Now, be sure to Read the Full Question: You want to
know whether Samantha can create a package that’s greater than $800 but has fewer than 13
sessions. If you stack the two equations and then add them together, you get 6h + 5z = 880.
In other words, she can offer 6 hot yoga and 5 zero gravity yoga sessions (11 total sessions)
for $880. This satisfies her requirements, so you know the answer is “Yes”; eliminate (A)
and (B). For (C), because you don’t know the price of each lesson individually, you don’t
know yet whether 5 hot yoga and 5 zero gravity yoga sessions will be over $800; leave (C)
for now. For (D), if 6 hot yoga and 5 zero gravity yoga sessions were over $800, then
adding a zero gravity yoga session will still be over $800. Given what you already know,
(D) must be true; choose (D).
B Begin by simplifying the equation given. (3p^2 + 14p + 24) – 2(p^2 + 7p + 20) = 3p^2 + 14p +
24 – 2p^2 – 14p – 40 = p^2 – 16 = 0. Factoring the left side of the simplified equation, we
find that (p – 4)(p + 4) = 16. Solving for p, we find that p = ±4. The value of 3p + 6 must
then be either 3(–4) + 6 = –6 or 3(4) + 6 = 18. The latter value is (B).
A Taking note that i = , the expression (2 + 8i)(1 – 4i) – (3 – 2i)(6 + 4i) becomes (2 + 8
)(1 – 4 ) – (3 – 2 )(6 + 4) . Expanding, this becomes 2 – 8 + 8 –
32( )^2 – (18 + 12 – 12 – 8( )^2 ) = 2 – 32()^2 – 18 + 8( )^2 = 8( )^2 –
32( )^2 – 16. This further simplifies to –8 + 32 – 16 = 8. This is (A).
C Plug In the Answers! The answer choices aren’t in any particular order, and some numbers
appear more than once, so you don’t need to start in the middle. Instead, start with 9
because it is in three of the four choices. If x = 9, then 2 = 9 – 3. = 3, so the left side
of the equation is 2 × 3 = 6, and the right side of the equation is 9 – 3 = 6. This works, so 9
is part of the solution set; eliminate (B) because it doesn’t include 9. Next, try x = 1: 2 =
1 – 3, which solves to 2 = –2. This isn’t true, so 1 is not part of the solution set; eliminate
(D). Lastly, try x = –1:2 = –1 – 3. You cannot take the square root of a negative
number, so this doesn’t work. Eliminate (A) and choose (C).