QUESTION EXPLANATIONS

1. (B) — First, we move all the terms containing x to one side to get 6 = 5x – 2x – x. Then, we simplify the equation to 6 = 2x and solve for x. The correct answer is (B, x = 3.

2. (B) — When we add c and d, it is equivalent to adding a^2 + 3a + 1" and –4a + 5, which gives us c + d = a^2 + 3a + 1 + (–4a + 5) = a^2 + 3a + 1 – 4a + 5 = a ^2 – a + 6.

3. (A) — We can eliminate (C) and (D) right away because the y values of graph are always greater than the linear function, and (C) and (D) are equations where y values are always lesser than the linear function. We can see that the right arm of the graphed function has a slope of 1 and would extend below the x-axis to have a y-intercept of –2, which is described by x – 2 in answer option (A). Furthermore, if we expand the absolute value into two inequalities, we see that y ≥ x – 2 and y ≥ –x + 2. The second inequality describes the left arm of the function, having a slope of –1 and a y-intercept of 2. The correct answer is (A).

4. (C) — We can solve this problem by writing a system of equations. We know that Sophie and Jazmin start with the same amount of money to invest, so s = j, where s and j are the amount of money Sophie and Jazmin start with, respectively. If Jazmin borrows $15,000 from Sophie (j + 15000), she will have twice as much as Sophie does after Sophie has lent Jazmin $15,000, so j + 15000 = 2(s – 15000). We can then substitute our first equation (s = j) into the second to find how much money Jazmin had originally: j + 15000 = 2(j – 15000). Expanding the equation, we get j + 15000 = 2j – 30000. We can then move all the terms containing j to one side and solve: 15000 + 30000 = 2j – j, j = 45000.

5. (B) — We can eliminate (C) and (D) right away because these equations describe linear functions. Between (A) and (B), we are looking for a parabola that has been shifted left by 5, which is done by adding a positive constant to x before the square, giving us (B) as the correct answer. Alternatively, you can see that the root of A is + 5 and the root of B is –5 just by looking at the equations.

6. (A) — Luca’s rent, $1195, is a constant and does change depending on the amount of electricity he uses. Therefore, x should not affect this number, so we can eliminate (B) and (D). Next, Luca pays 10 cents per kWh for electricity, meaning he pays $0.1 times x kWh, giving us 1195 + 0.1x.

7. (B) — We first find the slope of 2y + 6x = 5, which means finding the coefficient of x. Isolating y and simplifying, we get y = –3x + 5/2. We can then look for an answer option that also has a slope of –3. We can do this by isolating y in each answer option, but we can first scan the answers for instances where y can be easily isolated. If we do this, we can quickly see that in option B, when y becomes positive, 3x becomes negative, thus giving us a slope of –3.

8. (A) — Angle CEF = angle BDF because line BD || line CE. Angle CEF = angle ACE because they are alternate interior angles. These equalities mean that angle ACE, which is supplementary to angle x, is equal to 60 degrees. Line AF and line BD intersect to create vertical angles, meaning angle y = 180 degrees - angle CAF - angle ABD. Angle ABD = angle BDF because they are alternate interior angles, y = 180 degrees - 30 degrees - 60 degrees = 90 degrees. Therefore, x - y = 120 degrees - 90 degrees = 30 degrees.

9. (D) — The given equation can be broken down into a system of 2 equations to solve for x and y. We will use 8x + y = 36 and 8x + y = 2y +4x, but you may choose to use a different set of equations. Isolating y in the first equation gives us y = 36 – 8x. We can then simplify the second equation to get 4x = y, then plug in the first equation 4x = (36 – 8x). Solving for x, we get x = 3. We can plug in the value of x to any equation and solve for y, giving us y = 12. Therefore, x + y = 3 + 12 = 15.

10. (C) — Because the quadratic equation is written in standard form, we can look at the transformations applied to the equation to visualize the parabola. Having a negative a means the parabola is reflected about the x-axis, and having a y-intercept of b means the maximum of this parabola will be shifted up to where the minimum of the first parabola is. All of this means the resulting parabola will cross the x-axis at 2 points, giving it 2 roots.

11. (B) — Within the first bracket in the numerator is a difference of squares, so we can factor this: x^2 - 1 = (x + 1)(x – 1). We now have ((x + 1)(x - 1)(x-1)/(x + 1) and can cancel out x + 1 in the numerator and denominator, leaving us with (x – 1)(x – 1), which is the same as (x – 1)^2.

12. (D) — Because h (t) = 8t – t^2 models the height of the ball, h(t) = 0 when the ball is on the ground. To solve for the times at which the ball is on the ground, we are essentially looking for the roots of this quadratic equation. Factoring h(t), we get t(8 – t). Therefore, the ball is on the ground when t = 0 (which is when it was kicked) and when t = 8 seconds.

13. (D) — We’ll start with the quadratic equation right away to find the solution because none of the answer options are integers. Plugging all the terms into the equation, we get . When we simplify the terms under the radical, we find that we get a negative number, –4. There is therefore no real solution for this equation.

14. (A) — We can set up an equation for this word problem. If we let x = the negative number, we’ll first square x, then decrease it by 14, which gives us x^2 – 14. What we have now is equal to 5 times the original number, so x^2 – 14 = 5x. When we solve for x we get x = 7 or –2. Recall that this number should be negative, so we take the reciprocal of –2 to find the final answer, which is –1/2.

15. (B) — To solve for the area of this irregular shape, we‘―ll first have to visualize it in terms of a shape we know how to find the area for. We can visualize a square surrounding a garden with sides the same length as the diameter of the garden. When we subtract the area of the garden from that square, what we are left with is the same area as the shaded region between the gardens. We can therefore find the area of the shaded region by subtracting the area of the garden from the area of the square around it. To find the length of each side of the square, work backwards from the area of the garden to find the garden‘―s diameter. A = ¦Πr^2, ¦Π = ¦Πr2, ¦Π/¦Π = r^2, 1 = r^2, 1 = r. We know then that d = 2r = 2(1) = 2. The area of the square is therefore = 2 ‘Α 2 = 4. When we subtract the area of the garden from the area of the square, we find that the area of the shaded region is 4 ¨C ¦Π.

16. (1) — When we expand we get

17. (1/5 < or = x < or = 1 ) — If |3x -1| 2x, then -2x 3x -1 2x. We can separate this into -2x 3x - 1 and 3x - 1 2x and solve for x, giving us x = 1/5 for the first equation and x = 1 for the second equation.

18. (7) — We know that slope = so we can plug in the slope and coordinates given by the question: -2/3 = (1 - 5)/(a - 1) and solve for a, giving us a = 7.

19. (3) — We can set up the equation for the area of a rectangle and plug in the given area and coordinates. We can use the x coordinates to find length and the y coordinates to find width. Substituting these into A = lw, we get A = . When we plug in our known values, we get 24 = (7 – 1)(m – (–1)). Solving for this equation, we get m = 3.

20. (18) — To evaluate the second equation, we solve for x in the first equation and plug this value in to the second equation. Manipulating the first equation into the general quadratic equation form, we get 0 = x^2 + 6x + 9. From here we can factor the equation to get (x + 3)(x + 3) and solve for the root, which is –3. Plugging x = –3 into the second equation, we get an answer of 18.