1. Which ordered pair is not a solution of y = –12x – 3?
(–3, 33)
(8, –98)
(–9, 105)
(3, –39)*
2. Find three solutions of the equation.
y = 2x – 6
(–2, –10), (1, –4), (2, –1)
(–2, –10), (1, –4), (0, –6)*
(0, –6), (3, –2), (–2, –10)
(1, –4), (0, –6), (–1, –9)
3. Which pair of points has a positive slope?
(–2, 9), (1, –9)*
(–2, 5), (2, –11)
(–2, –15), (4, 21)
(–2, 13), (1, –11
4. The ordered pairs (2, –21) and (5, –45) are solutions to which of the following equations?
y = –8x – 5
y = –8x + 5
y = 8x – 5
y = 8x+ 5*
5. Which is a rule that describes the translation of a point from (–5, 4) to (–1, 2)?
(x, y) (x – 4, y – 2)
(x, y) (x + 4, y – 2)*
(x, y) (x + 4, y + 2)
(x, y) (x – 4, y + 2)
6. ΔJKL has vertices J(10, 5), K(2, 8), and L(4, 9). It is translated right 5 units and up 4 units. What are the
coordinates of J’, K’, and L’?
J'(15, 9), K'(12, 7), L'(9, 13)
J'(15, 9), K'(7, 12), L'(9, 13)
J'(14, 10), K'(6, 13), L'(10, 12)*
J'(15, 5), K'(7, 8), L'(9, 9)
7. The point M(x, y) is reflected over the x-axis. Use arrow notation to describe the original point and its
reflection.
(x, y) (x, y)*
(x, y) (–x, y)
(x, y) (–x, –y)
(x, y) (x, –y)
#1. Nope: -12(3)-3 = -36-3 = -39
#2 ok
#3 Nope: as x increases, y decreases
#4 Nope: 8*2+5 = 16+5 = 21
#5 ok
#6 Nope J(10, 5) -> (15,9)
#7 Nope: you didn't change anything!
1. (3, -39) is not a solution of y = –12x – 3 because when we substitute x = 3 into the equation, we get y = -39, not -12(3) - 3 = -39.
2. The three solutions of the equation y = 2x - 6 are (-2, -10), (1, -4), and (0, -6). These are determined by plugging in different values of x and solving for y.
3. The pair of points (–2, 9) and (1, –9) has a positive slope because when we calculate the difference in y-coordinates and divide it by the difference in x-coordinates, we get a positive value.
4. The ordered pairs (2, –21) and (5, –45) are solutions to the equation y = -8x - 5. By plugging in the x and y values, we find that they satisfy the equation.
5. The rule that describes the translation of the point from (-5, 4) to (-1, 2) is (x, y) -> (x + 4, y - 2). This means that we add 4 to the x-coordinate and subtract 2 from the y-coordinate to get the new coordinates.
6. The coordinates of J', K', and L' after the translation are J'(14, 10), K'(6, 13), and L'(10, 12). The x-coordinates have been increased by 5 units and the y-coordinates by 4 units.
7. The point M(x, y) when reflected over the x-axis remains unchanged, so the original point and its reflection are (x, y).
1. The ordered pair (3, -39) is not a solution of the equation y = -12x - 3.
2. Three solutions of the equation y = 2x - 6 are:
- (-2, -10)
- (1, -4)
- (0, -6)
3. The pair of points (–2, 9) and (1, –9) has a positive slope.
4. The ordered pairs (2, -21) and (5, -45) are solutions to the equation y = 8x + 5.
5. The rule that describes the translation of a point from (-5, 4) to (-1, 2) is: (x, y) → (x + 4, y - 2).
6. The coordinates of the vertices of the translated triangle are:
- J' (15, 9)
- K' (7, 12)
- L' (9, 13)
7. The point M(x, y) is reflected over the x-axis. The original point and its reflection are the same: (x, y) → (x, y).
1. To find the pair that is not a solution of the equation y = –12x – 3, we substitute the x and y values of each pair into the equation and check if the equation holds true.
For (–3, 33):
y = –12x – 3
33 = –12(-3) – 3
33 = 36 - 3
33 = 33
For (8, –98):
y = –12x – 3
-98 = –12(8) – 3
-98 = -96 - 3
-98 = -99
For (–9, 105):
y = –12x – 3
105 = –12(-9) – 3
105 = 108 - 3
105 = 105
For (3, –39):
y = –12x – 3
-39 = –12(3) – 3
-39 = -36 - 3
-39 = -39
Therefore, the pair (3, -39) is not a solution of the equation.
2. To find three solutions of the equation y = 2x - 6, we can choose any three different values for x and substitute them into the equation to find the corresponding y values.
For x = -2:
y = 2(-2) - 6
y = -4 - 6
y = -10
For x = 1:
y = 2(1) - 6
y = 2 - 6
y = -4
For x = 0:
y = 2(0) - 6
y = 0 - 6
y = -6
Therefore, three solutions of the equation are (-2, -10), (1, -4), and (0, -6).
3. To determine which pair of points has a positive slope, we need to calculate the slope between the two points using the formula (y2 - y1) / (x2 - x1). If the result is positive, the slope is positive.
For (–2, 9) and (1, –9):
Slope = (–9 - 9) / (1 - (-2))
Slope = (-18) / 3
Slope = -6
For (–2, 5) and (2, –11):
Slope = (–11 - 5) / (2 - (-2))
Slope = (-16) / 4
Slope = -4
For (–2, –15) and (4, 21):
Slope = (21 - (-15)) / (4 - (-2))
Slope = 36 / 6
Slope = 6
For (–2, 13) and (1, –11):
Slope = (–11 - 13) / (1 - (-2))
Slope = (-24) / 3
Slope = -8
Therefore, the pair (–2, 9) and (1, –9) has a positive slope.
4. To determine which equation the ordered pairs (2, -21) and (5, -45) are solutions to, we substitute the x and y values of each pair into each equation and check if the equation holds true.
For y = –8x – 5:
-21 = -8(2) - 5
-21 = -16 - 5
-21 = -21
For y = –8x + 5:
-45 = -8(5) + 5
-45 = -40 + 5
-45 = -45
For y = 8x – 5:
-21 = 8(2) - 5
-21 = 16 - 5
-21 = -21
For y = 8x + 5:
-45 = 8(5) + 5
-45 = 40 + 5
-45 = -45
Therefore, the ordered pairs (2, -21) and (5, -45) are solutions to the equation y = –8x + 5.
5. To find the rule that describes the translation of a point from (–5, 4) to (–1, 2), we need to determine the change in x and y values between the two points.
Change in x = -1 - (-5) = -1 + 5 = 4
Change in y = 2 - 4 = -2
Therefore, the rule that describes the translation is (x, y) -> (x + 4, y - 2).
6. To find the coordinates of the vertices J', K', and L' after translating ΔJKL right 5 units and up 4 units, we simply add 5 to the x-coordinate and 4 to the y-coordinate of each vertex.
For J(10, 5):
J'(10 + 5, 5 + 4) = J'(15, 9)
For K(2, 8):
K'(2 + 5, 8 + 4) = K'(7, 12)
For L(4, 9):
L'(4 + 5, 9 + 4) = L'(9, 13)
Therefore, the coordinates of J', K', and L' are (15, 9), (7, 12), and (9, 13).
7. When a point M(x, y) is reflected over the x-axis, the y-value changes sign while the x-value remains the same. In arrow notation, the reflection of (x, y) over the x-axis is denoted as (x, -y).
Therefore, the original point and its reflection are described by (x, y) and (x, -y) respectively.