MATH 1010 Sec. 3 Midterm 1 Solution
March 10, 2011
(Total: 100 points + extra credits: 5 points)
You have to show all the work on the space given. Please be neat! If you need more
space, you can use extra sheets of paper provided. (Please put your name on all the
sheets and label question numbers.) Not showing your work gets 0 point. Where
appropriate, clearly indicate your final answer by circling it. You are not allowed to
get help from your textbook, class notes, calculators, other students, or any other form
of outside aid. Do not forget to turn off your cell phone. If you have any question,
please ask the instructor. Good luck!
1. (Total: 10 points) Solve the following equations
(a) (5 points) 3(2x − 1) + 5 − x = 2x − (x + 3)
(b) (5 points) |2x + 3| = |x − 1|
(a)
6x − 3 + 5 − x = 2x − x − 3
5x + 2 = x − 3
4x = −5
5
x=−
4
(b) 2x + 3 = x − 1 or 2x + 3 = −x + 1. From the first equation we get x = −4. From the
second equation we get x = − 32 . Therefore, the solutions are x = −4 and x = − 23
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2. (5 points) Determine whether the two lines are parallel, perpendicular, or neither.
(a) L1 : y − 5x = 1
(b) L2 : 2y = 10x + 3
Notice that L1 can be written as y = 5x + 1 and L2 can be written as y = 5x + 32 . Since
the slope of L1 is m1 = 5 and the slope of L2 is m2 = 5, we have m1 = m2 . Therefore the
lines are parallel.
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3. (5 points) Simplify the expression 2x2 + 3y − 5x2 + y + 4xy by combining like terms.
2x2 + 3y − 5x2 + y + 4xy = (2 − 5)x2 + (3 + 1)y + 4xy = −3x2 + 4y + 4xy
4. (5 points) Find the distance between the points (2, 4) and (−1, 3).
d=
p
√
√
(−1 − 2)2 + (3 − 4)2 = 9 + 1 = 10
5. (Total: 10 points) Factor the following polynomials
(a) (5 points) x4 − y 4
(b) (5 points) x2 y − 25y
(a) x4 − y 4 = (x2 + y 2 )(x2 − y 2 ) = (x2 + y 2 )(x + y)(x − y)
(b) x2 y − 25y = y(x2 − 25) = y(x + 5)(x − 5)
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6. (Total: 10 points) Solve the following linear inequalities
(a) (5 points) −2x + 3 > −x − 1
(b) (5 points) x + 2 ≤ 2x + 1
(a)
2x − 3 < x + 1
x<4
Solution = {x | x < 4}
(b)
x + 2 ≤ 2x + 1
1≤x
Solution = {x | x ≥ 1}
7. (5 points) Simplify
10x6 y 2
5x5 y −3
10x6 y 2
= 2x6−5 y 2+3 = 2xy 5
5x5 y −3
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8. (10 points) Sketch the graph of the line given by 3y − 6x = 2. Then find the slope of the
line, the x-intercept, and the y-intercept.
The line is y = 2x + 32 .
Slope: 2.
x-intercept: (− 13 , 0).
y-intercept: (0, 32 ).
y
x
5
9. (Total: 10 points) Let
f (x) =
3x + 1, if x < 1
x − 4, if x ≥ 1
Find each value of the function.
(a) (5 points) f (−1) = 3(−1) + 1 = −3 + 1 = −2
(b) (5 points) f (2) + f (0) = (2 − 4) + (3(0) + 1) = −2 + 1 = −1
10. (10 points) Solve the following system of linear equations
−x + 2y = 5
2x − 3y = 0
Summing Equation 1 and Equation 2 we get y = 10. Substituting the value of y in the
first equation we have −x + 20 = 5. Hence x = 15. The solution is (15, 10).
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11. (10 points) Solve the following system of linear

 x − 2y + z
2x − 3y − z

2y + z
The matrix is
equations
= 1
= −1
= 8


..
 1 −2 1 . 1 


.
 2 −3 −1 .. −1 


..
0 2
1 . 8
Substituting R2 with R2 − 2R1 we get


..
 1 −2 1 . 1 


.
 0 1 −3 .. −3 


..
0 2
1 . 8
Substituting R3 with R3 − 2R2 we get


..
1
−2
1
.
1




..
 0 1 −3 . −3 


..
0 0
7 . 14
Dividing R3 by 7 we get


..
 1 −2 1 . 1 


.
 0 1 −3 .. −3 


.
0 0
1 .. 2
The system is

1
 x − 2y + z =
y − 3z = −3

z =
2
Substituting z = 2 in the second equation we get y = 3. Substituting z = 2 and y = 3 in
the first equation we get x = 5. The solution is (5, 3, 2).
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12. (10 points) Sketch the graph of the following

 2x + y
x

y
system of linear inequalities
< 1
> 1
≤ −2
The lines are y = −2x + 1, x = 1, and y = −2.
y
y=−2x+1
x
y=−2
x=1
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13. (Extra problem: 5 points) Find the determinant of the following matrix


1 −1 3
 2 −2 1 
0 −3 0


1 −1 3
−2 1
det  2 −2 1  = (1)
−3 0
0 −3 0
− (−1)
2 1
0 0
+ (3)
2 −2
0 −3
=
= [(−2)(0) − (−3)(1)] + [(2)(0) − (0)(1)] + (3)[(2)(−3) − (0)(−2)] = 3 + 0 − 18 = −15
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–END–
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