GETE 1340 Technical Mathematics for Engineers I: Homework 3 Solutions

This document presents the solutions for Homework 3 of the GETE 1340 Technical Mathematics for Engineers I course, focusing on systems of linear equations, determinants, factoring, fractions, and quadratic equations. The solutions include detailed step-by-step answers to various problems, such as solving simultaneous equations using appropriate methods, applying these concepts to real-world problems involving fuel efficiency, Kirchhoff’s law, and pump rates, as well as computer programs. Additionally, the document offers solutions related to trigonometry problems and simplifications. The problems cover a range of mathematical concepts, including solving for unknown variables, applying formulas, and interpreting results. The document is a valuable resource for students seeking to understand and solve complex mathematical problems in the context of engineering applications.

Technical Mathematics for Engineers I Spring 2019
Homework # 3
17

Question#1
a. Solution of linear simultaneous equations:
Answer:
Δ1=|29 −27
69 33 |=2820
Δ2=|10 29
40 69|=−470
Δ=|10 −27
40 33 |=1410≠0
So, i= Δ1
Δ =2820
1410 =2 and v = Δ2
Δ =−470
1410 =− 1
3
Solution: i = 2, v = - 1/3
Answer:
Δ=|0 . 42 −0 . 56
0 . 98 −1 . 4 |=−0 .0392≠0
Δ1=| 1. 26 −0 .56
−0 .28 −1 . 4 |=−1. 9208
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Δ2=|0 . 42 1. 26
0 .98 −0 .28|=−1. 3524
So, x= Δ1
Δ =−1 . 9208
−0 . 0392 =49 and y= Δ2
Δ =−1. 3524
−0 .0392 =34 .5
Solution: x = 49, y = 34.5 (Song, Wang, & Yu, 2018)
Answer:
Δ=|
2 1 1
1 −2 −1
3 3 −2
|=22≠0
Δ1=|
4 1 1
3 −2 −1
1 3 −2
|=44
Δ2=|
2 4 1
1 3 −1
3 1 −2
|=−22
Δ3=|
2 1 4
1 −2 3
3 3 1
|=22
So, x= Δ1
Δ =44
22 =2 y= Δ2
Δ =−22
22 =−1 z= Δ3
Δ =22
22 =1
Solution: x = 2, y = -1, z = 1
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Answer:
Δ=|
2 1 2
3 −2 −4
−2 3 4
|=14≠0
Δ1=|
8 1 2
5 −2 −4
−3 3 4
|=42
Δ2=|
2 8 2
3 5 −4
−2 −3 4
|=−14
Δ3=|
2 1 8
3 −2 5
−2 3 −3
|=21
So, r= Δ1
Δ =42
14 =3 s= Δ2
Δ =−14
14 =−1 t = Δ3
Δ =21
14 =3
2
Solution: r = 3, s = -1, t = 3/2
b. Let the two parts of the fuel are “x” and “y”
So, X + Y = 1
Now, at 80% efficiency X part is burnt and at 70% efficiency Y part is burned.
Hence, 0.8 * 150, 000 * X + 0.7 * 150,000 * Y = 114,000
 120000 X + 105000 Y = 114000
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Δ1=| 1 1
114000 105000|=−9000
Δ2=| 1 1
120000 114000|=−6000
Δ=| 1 1
120000 105000|=−15000≠0
X = Δ1
Δ = −9000
−15000 =0 . 6
Y = Δ2
Δ = −6000
−15000 =0 . 4
Hence, part 1 was 60% and part 2 was 40% of the total fuel amount burnt.
c. Kirchhoff’s Law problem:
Answer:
Δ=|
1 1 1
5 . 2 −3 .25 0
0 3. 25 −2. 62
|=39039
1000 ≠0
Δ1=|
0 1 1
1. 88 −3 . 25 0
−3. 35 3 . 25 −2 .62
|=1481
10000
Δ2=|
1 0 1
5. 2 1 .88 0
0 −3 . 35 −2. 62
|=−13966
625
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Δ3=|
1 1 0
5. 2 −3 . 25 1. 88
0 3 . 25 −3 .35
|=8879
400
So, i1= Δ1
Δ =1481/10000
39039/ 1000 =1481
390390 =0 . 0038
i2= Δ2
Δ =−13966/625
39039/1000 =−111728
195195 =0 . 57
i3= Δ3
Δ =8879/400
39039 /1000 =3415
6006 =0 .57
d. Computer Programs:
Answer:
A + B + C = 140
A + 3 * B + 2 * C = 236
2 * A + B + 3 * C = 304
Δ=|
1 1 1
1 3 2
2 1 3
|=3≠0
Δ1=|
140 1 1
236 3 2
304 1 3
|=204
Δ2=|
1 140 1
1 236 2
2 3. 4 3
|=72
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Δ3=|
1 1 140
1 3 236
2 1 304
|=144
A= Δ1
Δ =204
3 =68 B= Δ2
Δ =72
3 =24 C= Δ3
Δ =144
3 =48
Solution: A = 68, B = 24, C = 48 number of computer programs required.
Question#2
Answer:
x2+ y2=1
2 [ x2 + y2+ x2+ y2 ]
=> x2+ y2= 1
2 [ x2+2 xy+ y2+ x2−2 xy + y2 ]
x2+ y2=1
2 [ ( x+ y ) 2+ ( x− y ) 2 ] (Proved)
Answer:
2 ( e1−e2 )
2+2 ( e2−e3 ) 2
=2 ( e1
2−2 e1 e2+ e2
2 ) +2 ( e2
2−2 e2 e3+e3
2 )
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=2 e1
2−4 e1 e2+2 e2
2+2 e2
2−4 e2 e3+2 e3
2
=2 e1
2−4 e1 e2+ 4 e2
2−4 e2 e3+2 e3
2
=2 e1
2+ 4 e2
2+2 e3
2−4 e1 e2−4 e2 e3
Answer:
Pump 1 pumps 5000 gallon in 20 minutes
 In one minute pump 1 pumps = 5000 / 20 = 250 gallons
Pump 2 pumps 5000 gallon in 25 minutes
 In one minute pump 2 pumps = 5000 / 25 = 200 gallons
Hence, in both the pumps together pump = 250 + 200 = 450 gallons per minute.
So, 450 gallons pumped in 1 minute
 1 gallon pumped in 1 / 450 minutes
 5000 gallons pumped in = 5000 / 450 = 11.11 minutes.
17

Answer:
K=1 . 00−2 . 67 R+ R2
 0 .5=1. 00−2. 67 R+ R2
 0=0 .5−2. 67 R+ R2
 R2−2. 67 R+0 . 5=0
So,
R= 2 .67± √ 2. 672−4∗1∗0 . 5
2∗1 = 2. 67±2. 265
2 =0 . 202, 2. 468
Solution: R = 0.202, 2.468
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Guide for Exam 1
17

Answer 1:
i. In the triangle,
a
SinA = b
SinB = c
SinC (Z. Kalanov, 2014)
Now, angle C = 90o => Sin 90o = 1 and side c = 16 meters
So,
a
SinA = b
SinB =16
Now angle A + B + C = 180o
 A = 180o - B – C
 A = 180o - 27o – 90o = 63o

a
Sin 63o = b
Sin 27o =16
 a=16∗Sin 63o
and b=16∗Sin 27o
 a=16∗0. 891=14 .256 and b=16∗0. 454=7 . 264
Solution: A = 63o, B = 27o, C = 90o: a = 14.26, b = 7.26, c = 16
ii. In the triangle,
a
SinA = b
SinB = c
SinC
Now, angle C = 90o => Sin 90o = 1
From Pythagoras theorem, a2 +b2=c2
So, c=42. 32
So,
18
SinA =38 .3
SinB =42. 32
1
 Sin A = 18 / 42.32 = 0.425, and Sin B = 38.3 / 42.32 = 0.905
 A = Sin -1 0.425 = 25.18o, and B = Sin -1 0.905 = 64.82
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Solution: A = 25.18o, B = 64.82o, C = 90o: a = 18, b = 38.3, c = 42.32
Answer 2:
Max amplitude of 3 and period of 4 with zeros at 2, 6, 10 …
f ( x ) =3∗Cos πx
4
Max amplitude of 0.9 and period of
2 π
3
f ( x ) =3∗Cos 3 x
2
Answer 3:
Case 1: Observer in the middle: Let “x” be the distance between A and B.
Hence, Sin AOC = Sin 15o = 0.259 = AC / CO
 CO = AC / 0.259 = 8 / 0.259 = 30.89 meters.
Again, Sin BOD = Sin 23.375o = 0.397 = BD / DO
 DO = BD / 0.397 = 8 / 0.397 = 20.15 meters.
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Hence, x = CO + DO = 30.89 + 20.15 = 51.04 meters.
Case 2: Observer in one side:
Hence, Sin AOC = Sin 15o = 0.259 = AC / CO
 CO = AC / 0.259 = 8 / 0.259 = 30.89 meters.
Again, Sin BOD = Sin 23.375o = 0.397 = BD / DO
 DO = BD / 0.397 = 8 / 0.397 = 20.15 meters.
Hence, x = CO - DO = 30.89 - 20.15 = 10.74 meters.
Answer 4:
Δ=|
−2 −2 1
4 −3 2
8 −4 −3
|=−82≠0
Δ1=|
2 −2 1
−2 −3 2
13 −4 −3
|=41
17

Δ2=|
−2 2 1
4 −2 2
8 13 −3
|=164
Δ3=|
−2 −2 2
4 −3 −2
8 −4 13
|=246
So, u= Δ1
Δ =41
−82 =− 1
2 v = Δ2
Δ =164
−82 =−2 w= Δ3
Δ =246
−82 =−3
Solution: u = -0.5, v = -2, w = -3
Answer 5:
Simplifications:
3 x +1
6+ x + x +2
30+ 5 x =3 x +1
6+x + x +2
5 ( 6+ x )

3 x +1
6+x + x +2
30+ 5 x =5 ( 3 x +1 ) + x+2
5 ( 6+ x ) =16 x +7
5 x+ 30
6
y 4+ 2 y3−8 y2 + y +1
y2−2 y − y + 4
y2+ 4 y = 6
y 4+2 y3−8 y2 + y +1
y2−2 y − y+ 4
y ( y + 4 )
= 6
y4 +2 y3−8 y2 + y +1
y2−2 y − 1
y
= 6
y2 ( y2+2 y−8 ) + y +1
y ( y−2 ) − 1
y
17

= 6
y2 ( y−2 ) ( y +4 ) + y +1
y ( y−2 ) − 1
y
= 6+ y ( y +4 ) ( y+1 ) − y ( y−2 ) ( y+ 4 )
y2 ( y−2 ) ( y +4 )
= 6+ y ( y2+5 y + 4 ) − y ( y2+2 y−8 )
y2 ( y −2 ) ( y +4 )
= 6+ ( y3 +5 y2+ 4 y ) − ( y3+2 y2−8 )
y2 ( y−2 ) ( y +4 )
= 6+ ( y3 +5 y2+ 4 y ) − ( y3+2 y2−8 )
y2 ( y−2 ) ( y +4 )
= 6+3 y2 +4 y + 8
y2 ( y−2 ) ( y +4 )
= 3 y2+4 y+14
y2 ( y−2 ) ( y +4 )
Answer 6:
Let X = First type (ton) and Y = second type (ton)
X + Y = 42 ……………..(I)
18 X + 30 Y = 1050…….(II)
Solving (I) and (II), we get
Δ1=|42 1
1050 30|=210
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Δ2=| 1 42
18 1050|=294
Δ=| 1 1
18 30|=12≠0
X = Δ1
Δ =210
12 =17 . 5
Y = Δ2
Δ =294
12 =24 . 5
Solution: First type = X = 17.5 tons, Second type = Y = 24.5 tons
Answer 7:
x2−x−12=0
 x2−4 x +3 x−12=0
 x ( x−4 ) +3 ( x−4 ) =0
 ( x+ 3 )∗( x−4 ) =0
 x = -3 or x = 4
Answer 8:
−2 SinA +1=0
 SinA =1
2

A=Sin−1 1
2 =−330o , 30o , 390o . ..
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Answer 9:
Sin2 A−Sin A+1=0
Let Sin A = x
The equation reduces to x2−x +1=0
So,
x= 1± √ 1−4
2 = 1±i √ 3
2
So, Sin A has no real solution. Hence, angle A has no real solution.
References
Song, G. J., Wang, Q. W., & Yu, S. W. (2018). Cramer’s rule for a system of quaternion
matrix equations with applications. Applied Mathematics and Computation, 336,
490-499.
Z. Kalanov, T. (2014). On the System Analysis of the Foundations of Trigonometry. Pure
And Applied Mathematics Journal, 3(2), 26. doi: 10.11648/j.pamj.20140302.12
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