Comprehensive Solutions: Math Homework on Equations and Formulas

This document presents a comprehensive solution to a math homework assignment. The solution covers a variety of algebraic problems including simplifying expressions, solving linear equations, and solving systems of equations. It includes step-by-step solutions to problems involving the manipulation of formulas, solving for variables, and applying equations to real-world scenarios. The assignment addresses topics like mobile phone plans, distance and time problems, and problems involving scoring systems. Additionally, the solutions demonstrate how to find gradients, intercepts, and formulate linear equations based on given information. The document is designed to provide students with a clear understanding of the problem-solving process and enhance their mathematical skills.

P a g e | 1
Name:
Course:
Professor:
Date:

P a g e | 2
Q1
a) a(5b-2)-5ab
Opening the brackets,
5ab-2a-5ab
Collecting like terms together,
5ab-5ab-2a
= -2a
b) 2a(3b+4)+3b(1-3a)-6b+1
Opening the brackets,
6ab+8a+3b-9ab-6b+1
Collecting like terms together,
6ab-9ab+8a+3b-6b+1
= -3ab+8a-3b+1
Q2
a) 7e-8=-120
7e=-120+8
7e=-112
=e = -112/7 = -16
b) 5(3-2f)=4f-1
= 15-10f = 4f-1
= 15+1=4f+10f
= 14f =16
Therefore f=16/14 = 8/7 = 1 1/7
c) 2(2c+4) = -{(3c+6)/4}
4(4c+8) = -(3c+6)
16c+32 = -3c-6
16c+3c = -6-32

P a g e | 3
19c =-38
C=-2
Q3
a) M=24LT(D-T)
L=6m, D=0.3m, T =5/100 = 0.05m
M = 24 x 6 x 0.05x (0.3-0.05)
= 24x 6 x 0.05 x 0.25
M = 1.8t
b) Transpose and make D subject
M =24LT (D-T)
M/24LT = D-T
D = M/24LT + T
c) Yes, the rigger can safely move the load
The load outside diameter is less than 1m
L=8m, T=0.04m and M=7.5t
Finding the maximum outside diameters D for the load which can be lifted,
7.5= 24 x 8 x 0.04(D-0.04)
D-0.04 = 7.5/(24x8x0.04)
D=0.9765625+0.04
Maximum outside diameter which can be lifted is 1.164625m. this means that the available
diameter of less than 1m will be safely lifted.
Q4
a) i. mobile plan
C= 50 +7/100m, where m is the number of sms sent per month
ii.
total monthly cost = 67.50
67.50 = 50+0.07m
67.50-50=0.07m

P a g e | 4
17.50/0,07 = m
M = 250 sms
b) i.
L= T + 26
ii.
D= 249km
249x2 = 3(sum of L and T)
249x2 = 3(T + T +26)
498=6T+78
iii.
solving for T
498=6T +78
498-78 =6T
T =70km
Q5
Let C represent correct answer, W represent wrong answers and E total earning by the student
Generally E = 10C-7W
C+W= 32
10C-7W =133
C= 32-W
Replacing C in E equation,
10(32-W)-7W = 133
320-10W-7W =133
320-133=17W
187=17W
W=11
C=32-11

P a g e | 5
C=21
Therefore the correct answers were 21
Q6
a) 2d+e=7 and 3d+2e=10
Multiplying equation 1 with 2 and equation 2 with 1
2d+e=7 /2
3d+2e=10/1
We get
4d+2e=14
3d+2e=10
Subtracting the two equations downwards
4d+2e=14
3d+2e=10
D= 4
Replacing value of d in one of the equations; 2(4)+e=7
E=7-8= -1
Therefore d = 4 and e=-1
b) Brand A=$7/kg and B=$11/kg
Let y represent kg in brand A and z kgs in brand B
Therefore; y+z=15
The total cost equation will be 7y+11z=129
Y=15-z
Replacing y in the total cost equation;
7(15-z)+11z=129
105-7z+11z=129
-7z+11z=129-105
4z=24; z=6kg

P a g e | 6
Y=15-6 = 9kg
Brand A kgs = 6kg and Brand B = 9kg
Q7
a) i.
y=1/10 x -5
Considering the standard line equation of y=mx+c; where m is the gradient;
Therefore the lines gradient will be 1/10
ii. Coordinates of y intercept means that x=0. Replacing x with 0 in the equation gives y values
as y=1/10x 0 -5. Therefore, y value = -5
Coordinates of y intercept = (0,-5)
b) y intercept of a perpendicular line
The gradient of a perpendicular is usually negative reciprocal
The negative reciprocal of 1/10 = -10. The two points which will be used to achieve this gradient
are (-1,7) and (0,c)
m = y2− y1
x2−x1
-10 = c−7
0−−7
-10(+7) = c-7
-70+7=c = -63
Therefore y intercept coordinates will be (0,-63)
c) gradient
m = y2− y1
x2−x1
m = 18−8
−3−−7 =10
4 = 5
2
d) equation of line
m = y2− y1
x2−x1
m = 1 4−−8
−5−0.5 = 22
−5.5 =−4

P a g e | 7
y=-4x+c
The y intercept will happen at point (0,c), picking one point of the line to find value of c
and using the gradient, we have
-4 = 1 4−c
−5−0
-4x-5 = 14-c
20=14-c
C=-6
The lines equation will be y = -4x-6