Algebra Assignment: MX01021A Solutions with Step-by-Step Answers

This document presents a comprehensive solution set for an algebra assignment. It covers a range of algebraic concepts, including calculations, substitution, simplification of expressions, and solving equations. The assignment includes detailed solutions for each question, demonstrating step-by-step working and explanations. Topics addressed include basic arithmetic operations, exponent rules, like terms, and the manipulation of algebraic expressions. The solutions are presented in a clear and organized manner, providing a valuable resource for students studying algebra. The assignment covers a variety of algebraic methods, making it a useful resource for students looking to improve their understanding and problem-solving skills in algebra. The solutions provided are complete and detailed to assist students in understanding the concepts.

Running head: MATHS
Algebra
Name of the Student:
Name of the University:
Author Note:

1MATHS
Table of Contents
1. Question 1:...................................................................................................................................2
2. Question 2:...................................................................................................................................2
3. Question 3:...................................................................................................................................3
4. Question 4:...................................................................................................................................4
5. Question 5:...................................................................................................................................5
6. Question 6:...................................................................................................................................6
7. Question 7:...................................................................................................................................7
10. Question 10:.............................................................................................................................10
References:....................................................................................................................................11

2MATHS
1. Question 1:
a. 4 x−2 x – 3=−24
Answer: -24
b. 3−(−2)+5 = 10
Answer: 10
c. 16 ÷ (−4 )=−4
Answer: -4
d. 3+7 X 5=38
Answer: 38
e. 3+ 42=19
Answer: 19
f. 3 X 10+2 X – 2=26
Answer: 26
2. Question 2:
Given: p = 4, q= -3, x = -2
a. p+q=1
= 4-3
= 1
Answer: 1
b. x – 5= -7
= - 2 – 5
= -7
Answer: -7

3MATHS
c. 7 p+q=25
= 7 x 4 + (-3)
= 25
Answer: 25
d. p q – x= -10
= 4 x (-3) – (-2)
= (-12) +2
= -10
Answer: -10
e. 4 q2 =36
= 4 x (-3)2
= 36
Answer: 36
f. x3=−27
= (- 3)3
= -27
Answer: -27
3. Question 3:
a.25=32
Answer: 32
b. 3−2= 1/9
=1/32
= 1/9
Answer: 1/9

4MATHS
c. 1 00=1
Answer: 1
d. 32 X 23=72
= 9 x 8
= 72
Answer: 72
4. Question 4:
a. a6 ÷ a2 =a4
= a6-2
= a4
Answer: a4
b. 4 b6 ÷ 2b2=2 b4
= 2b6-2
= 2b4
Answer: 2b4
c. C7 X C3=C10
= C 7+3
Answer: C 10
d. 3 a−3 X 4 a4=12 a
Answer: 12 a
e. ( m3 )
4
=m12
Answer: m12
f. ( 3 a2 ) 3
=27 a6
= 3a2x 3a2 x 3a2

5MATHS
= 27 a6
Answer: 27 a6
g. a3 ÷ a−3=a6
= a3-(-3)
= a6
= a6
Answer: a6
h. d2 x d1=d2−1=d1
= d2-1
=d1
Answer: d1
i. a−2 X a−3=a−¿¿5
= a-2-3
= a-5
Answer: a-5
j. √36 a4=6 a2
Answer: 6a2
5. Question 5:
Pair of like terms:
a. a, 3a
b. 2xy, -4xy
c. 3b2, -2b2
d. 4q, 6q

6MATHS
6. Question 6:
a. 3 x+ x+5 y− y=4 x+4 y
= 3x+x+5y-y
Adding the similar elements:
= 4x+5y-y
= 4x+4y
Answer: 4x+4y
b. 6 b4 +2 b3=2 b 3(3 b +1)
Applying the exponent rule: ab+c= ab.ac
=6b.b3+2b3
=2.3b.b3 +2b3
= 2b3 (3b+1)
Answer: 2b3 (3b+1)
c. 3 p+5 q+2 r −p+ q−4 r =2 p+6 q−2r =2 p+6 q−2r
By grouping the like terms:
= 3p- p + 5q + q + 2r - 4r
= 2p + 5q + q- 2r
= 2p+ 6q- 2r
Answer: 2p+ 6q- 2r
d. x2 – x2 – 6+ 2 x2=2 x2−6
By grouping the like terms:
= x2 – x2 +2x2 – 6
= 2x2 - 6

7MATHS
Answer: 2x2 – 6
7. Question 7:
a. 2 ( b+3 ) =2 b+6
Answer: 2b + 6
b. 4 ( y−9 )=4 y−36
= 4y- 4.9
Answer: 4y- 36
c. 6 ( 4−x ) =24−6 x
= 6.4- 6x
=24- 6x
Answer: 24- 6x
d. 3 ( −2+x ) =−6+3 x
=3. (-2)+3x
= -6 +3x
Answer: -6 +3x
8. Question 8:
a. 4 ( x+2 )+6=4 x+ 14
By expanding:
= 4x+8+6
= 4x+14
Answer: 4x+14
b. 3 b+2 ( b – 3 ) =5 b−6
By expanding:

8MATHS
=3b+2b- 6
= 5b -6
Answer: 5b -6
c. 5 ( y +2 ) – 2 ( 6+ y )=3 y −2
By expanding:
=5y+10-2(6+y)
= 3y-2
Answer: 3y-2
d. 4 ( 2 x – 7 )+2 ( 5 x+ 2 )=18 x−24
By expanding and simplifying
= 8x- 28 + 2 (5x+2)
= 18x- 24
Answer: 18x- 24
9. Question 9:
a. x−6=4 ; x=10
By adding 6 to both the sides,
x- 6+6= 4+6
x=10
Answer: x=10
b. x
3 =12 ; x=36
By multiplying 3 on both the sides,
3x/3=12.3
Simplifying,

9MATHS
x=36
Answer: x=36
b. 5 y=15 ; y=3
By dividing both sides by 5,
5y/5=15/5
Simplifying,
y=3
Answer: y=3
d. 4 x+2=– 6 ; x=−2
By subtracting 2 from both the sides,
4x+2-2= -6-2
4x=-8
x=-2
Answer: -2
e. 5 y +3=24 – 2 y ; y =3
By subtracting 3 from both the sides,
5y+3-3=24-2y-3
Simplifying,
5y= - 2y+21
5y+2y= 21
7y=21
By dividing 7 on both the sides,
y=3
Answer: y=3
f. 2 ( x+4 )=11 – x ; x=1

10MATHS
By expanding and subtracting 8 from both sides:
2x+8=11-x
2x +8 - 8=11- x – 8
By adding x to both sides and simplifying,
3x=3
x=1
Answer: x=1
10. Question 10:
a. 5
8 – 1
4 = 3
8
Answer: 3/8
b. 5
8 + 1
3 = 23
24
Answer: 23/24
c. t
5 + 2
5 = t+2
5
Answer: t+2
5
d. 3
8 ÷ 9
10 = 5
12
Answer: 5/12
e. 3
4 + 2
5 ÷ 3
8 =109
60
Answer: 109/60
f. 5 x
9 X 3 f
5 = x2 f
3
Answer: x2f/3

11MATHS
g. a
10 − b
3 = a .3−b .10
30
Answer: (a.3-b.10) / 30
h. 3 c
8 ÷ 9 c
14 = 7
12
Answer: 7/12
References:
Kaput, J.J., Carraher, D.W. and Blanton, M.L. eds., 2017. Algebra in the early grades.
Routledge.
Van der Waerden, B.L., 2013. A history of algebra: from al-Khwārizmī to Emmy Noether.
Springer Science & Business Media.