Download MATH 1010-2: Quiz 3A

MATH 1010-2: Quiz 3A
Solution
1. (4 points) Solve the inequality and sketch the solution on the real number line:
−3x + 4 > 2x + 9
Solution:
−3x + 4 − 2x > 2x + 9 − 2x
−5x + 4 > 9
−5x + 4 − 4 > 9 − 4
−5x > 5
1
− 5 (−5x) < − 15 × 5
x < −1
b
b
-3
-2
)b
-1
b
b
b
b
0
1
2
3
2. (3 points) Solve the equation:
|x + 1| = 4
Solution:
(a) We are going to find solutions among real numbers greater than or equal to −1. In this case,
x + 1 is always positive or 0. So it’s always true that |x + 1| = x + 1. The original equation
becomes
x+1=4
whose solution is x = 3. Since 3 > −1, it’s in the area that we are finding solutions.
(b) We are going to find solutions among real numbers less than −1. In this case, x + 1 is always
negative. So it’s always true that |x + 1| = −(x + 1). The original equation becomes
−(x + 1) = 4
whose solution is x = −5. Since −5 < −1, it’s in the area that we are finding solutions.
To conclude, there are two solutions of this equation. They are 3 and −5.
3. (3 points) Plot (3,1) on a rectangular coordinate system.
y
3
2
1
b
b
b
0
(3,1)
b
b
b
b
b
1
2
3
x
MATH 1010-2: Quiz 3B
Solution
1. (4 points) Solve the inequality and sketch the solution on the real number line:
−2x + 3 > x + 9
Solution:
−2x + 3 − x > x + 9 − x
−3x + 3 > 9
−3x + 3 − 3 > 9 − 3
−3x > 6
1
− 3 (−3x) < − 13 × 6
x < −2
b
-3
)b
-2
b
b
b
b
b
-1
0
1
2
3
2. (3 points) Solve the equation:
|x + 1| = 3
Solution:
(a) We are going to find solutions among real numbers greater than or equal to −1. In this case,
x + 1 is always positive or 0. So it’s always true that |x + 1| = x + 1. The original equation
becomes
x+1=3
whose solution is x = 2. Since 2 > −1, it’s in the area that we are finding solutions.
(b) We are going to find solutions among real numbers less than −1. In this case, x + 1 is always
negative. So it’s always true that |x + 1| = −(x + 1). The original equation becomes
−(x + 1) = 3
whose solution is x = −4. Since −4 < −1, it’s in the area that we are finding solutions.
To conclude, there are two solutions of this equation. They are 2 and −4.
3. (3 points) Plot (3,2) on a rectangular coordinate system.
y
3
2
1
b
b
b
b
0
(3,2)
b
b
b
b
1
2
3
x