MATH A105 Intermediate Algebra
Exam 1 Key
Instructions
1. Do NOT write your answers on these sheets. Nothing written on the test papers will be graded.
2. Do NOT write your name on any of your answer sheets.
3. Please begin each section of questions on a new sheet of paper.
4. Do not write problems side by side.
5. Do not staple test papers.
6. Limited credit will be given for incomplete or incorrect justification.
Questions
1. Absolute Values
Find all solutions for each of the following. (5 each)
(a) |(5x − 2) − 8| <
1
10 .
|(5x − 2) − 8| <
|5x − 10| <
1
5x − 10 < 10
.
1
10 .
1
10 .
x
<
>
>
>
>
>
>
1
10 .
1
− 10
.
1
− 10
+10.
1
− 10
+ 100
10 .
99
− 10 .
1 99
5 · 10 .
99
50 .
2x−7 2x−7 3 + 2 = 9.
+ 2−2 = 9−2.
3
2x−7 = 7.
3
2x−7
= 7.
− 2x−7
3
3 3 · 2x−7
= 3 · 7.
−3 · − 2x−7
3
3
2x − 7 = 21.
2x − 7
2x − 7+7 = 21+7.
2x − 7+7
2x = 28.
2x
2x
28
2x
=
.
2
2
2
x = 14.
x
=
=
=
=
=
=
=
7.
−3 · 7.
−21.
−21+7.
−14.
−14
2 .
−7.
5x − 10+10 <
5x <
5x <
5x
<
5
x <
1
10 +10.
1
100
10 + 10 .
101
10 .
1 101
5 10 .
101
50 .
−(5x − 10)
5x − 10
5x − 10+10
5x
5x
5x
5
99 101
,
50 50
+ 2 = 9.
(b) 2x−7
3
1
(c) |x + 4| + |x − 2| = 8.
|x + 4| + |x − 2| =
8.
x + 4 + |x − 2| =
8.
x+4+x−2
=
8.
2x + 2
=
8.
2x =
6.
x =
3.
x + 4 − (x − 2)
=
8.
x+4−x+2
=
8.
6
=
8.
−(x + 4) + |x − 2| =
8.
−x − 4 + |x − 2| =
8.
−x − 4 + x − 2
=
8.
−6
=
8.
−x − 4 − (x − 2)
=
8.
−x − 4 − x + 2
=
8.
−2x − 2
=
8.
2
−2x =
10.
x =
−5.
2. Other Solving
Find all solutions for each of the following. (5 each)
(a) 10x2 − x − 21 = 0.
1
2
10
10
5
1
3
21
21
7
(5x − 3)(2x + 7) =
10x + 35x − 6x − 21 =
2
10x2 + 29x − 21.
Not correct
(5x − 7)(2x + 3)
=
10x + 15x − 14x − 21
=
2
10x2 + x − 21.
Not correct
(5x + 7)(2x − 3)
=
10x − 15x + 14x − 21
=
2
2
10x − x − 21.
Correct
5x + 7
=
0.
x
=
−7/5.
2x − 3
=
0.
x
=
3/2.
(b) x3 − 3x2 + 2x = 0.
(c) T −
3.5(5500−200)
1000
x3 − 3x2 + 2x
=
0.
x(x2 − 3x + 2)
=
0.
x(x − 1)(x − 2)
=
0.
x
=
0.
x−1
=
0.
x
=
1.
x−2
0.
x
=
2.
> 50.
T−
3.5(5500 − 200)
1000
3.5(5300)
T−
1000
T − 3.5(5.3)
T − 18.55
T
3
> 50.
> 50.
> 50.
> 50.
> 68.55.
(d)
470
50+460
=
447
T +460
470
50 + 460
470
510
=
=
470
(T + 460)
510
470
(T + 460)
510
T + 460
T
T
4
=
=
447
.
T + 460
447
.
T + 460
447
(T + 460).
T + 460
447.
510
.
470
510
− 460.
= 447
470
≈ 25.
=
447
3. Applications
(a) (4) Pressure drops one inch of mercury (inHg) per 1000 feet of altitude gained. If the pressure at 200 ft
MSL is 30.11. What is the pressure at 12,500 ft MSL?
30.11 −
12500 − 200
≈ 17.81
1000
(b) (4) Total resistance of two resistors in parallel is given by the formula below. If you need a total resistance
of 40/13 and have one resistor with resistance 8 how much resistance must the other resistor have?
1
1
1
=
+
.
Rtotal
R1
R2
1
40/13
13
40
13 1
−
40 8
5
13
−
40 40
8
40
1
5
R2
=
=
=
=
=
=
=
1
+
8
1
+
8
1
.
R2
1
.
R2
1
.
R2
1
.
R2
5.
1
.
R2
1
.
R2
(c) Temperature drops 3.5◦ F per 1000 feet altitude gained. This equation calculates the altitude at which
the temperature drops to 32◦ F.
A
72 − (3.5)
= 32.
1000
i. What does the 72◦ represent?
It is the temperature at the surface–the starting temperature.
ii. In what units is A? E.g., inches, miles, kilometers
A must be in feet because the ratio is in feet.
iii. Modify the equation to calculate the altitude at which the temperature is 40◦ .
72 − (3.5)
5
A
= 40.
1000