Carnegie Learning Math Series Volume 1 4th Edition Chapter 1 Linear Equations
Carnegie Learning Math Series Volume 14th Edition Chapter 1 Exercise 1.3 Solution Page 32 Problem 1 Answer
We have been given that Terry, Trudy, Tom, and Trevor have challenged their friends with this riddle.
Terry said: “If you add 150 to the number of MP3 downloads Tom has and then double that number and finally divide it by three, you have the number of MP3 downloads I have.”
Trudy said: “If you take the number of MP3 downloads Tom has, subtract 30, multiply that difference by 5, and finally, divide that product by 4, the result will be the number of MP3 downloads I have.”
Trevor said: “Well if you take twice the number of MP3 downloads Tom has, add 30, multiply the sum by 4, and finally, divide that product by 3, you will have the number of MP3 downloads I have.”
We have been asked what we need to know to determine the number of MP3 downloads each person has.
We need to know the number of MP3 downloads Tom has to find out the number of MP3 downloads each person has.
We have answered that we need to know the number of MP3 downloads Tom has to find out the number of MP3 downloads each person has.
Page 33 Problem 2 Answer
We have been given that Tom has 150 MP3 downloads.
We have to determine how many MP3 downloads Terry has.
We will find the result.
Given that, Terry has 150 MP3 downloads.
We know that,
The number of MP3 downloads Terry has is,
=2(150+x)/3
Here x=150,
=2(150+150)/3
=2(300)/3
=200
We have found that if Tom has 150 MP3 downloads, Terry will have 200 MP3 downloads.
Page 33 Problem 3 Answer
We have been given that Tom has 150 MP3 downloads.
We have to determine how many MP3 downloads Trudy has.
We will find the result.
Given that, Tom has 150 MP3 downloads.
We know that,
The number of MP3 downloads Trudy has is,
=5(x−30)/4
Here x=150,
=5(150−30)/4
=5(120)/4
=150
We have found that if Tom has 150 MP3 downloads, Trudy will have 150 MP3 downloads.
Page 33 Problem 4 Answer
We have been given that Tom has 150 MP3 downloads.
We have to determine how many MP3 downloads Trudy has.
We will find the result.
Given that, Tom has 150 MP3 downloads.
We know that,
The number of MP3 downloads Trudy has is,
=4(30+x)/3
Here x=150,
=4(30+150)/3
=4(180)/3
=240
We have found that if Tom has 150 MP3 downloads, Trevor will have 240 MP3 downloads.
Solutions For Linear Equations Exercise 1.3 In Carnegie Learning Math Series Page 33 Problem 5 Answer
Given: Statement 1: Terry said: “If you add 150 to the number of MP3 downloads Tom has and then double that number and finally divide it by three, you have the number of MP3 downloads I have.”
Statement 2: Trudy said: “If you take the number of MP3 downloads Tom has, subtract 30, multiply that difference by 5, and finally divide that product by 4, the result will be the number of MP3 downloads I have.”
Statement 3: Trevor said: “Well, if you take twice the number of MP3 downloads Tom has, add 30, multiply the sum by 4, and finally divide that product by 3, you will have the number of MP3 downloads I have.”
Let x-number of MP3 downloads Terry is having.
y-number of MP3 downloads Trudy is having.
z-number of MP3 downloads Tom is having.
w-number of MP3 downloads Trevor is having.
According to first statement equation is
150+2z/3=x
→ 2z+450/3=x
According to second statement equation is
5(z−30)/4=y
→5z−150/4=y
According to third statement equation is
4(2z+30)/3=w
→8z+120/3=w
Now if Terry and Trevor have the same number of MP3 downloads then
2z+450/3=8z+120/3
→6z=550
→z=55
Tom has 55 MP3 downloads
Page 33 Problem 6 Answer
It is given that Terry and Trevor have the same number of MP3 downloads
y-number of MP3 downloads Trudy is having.
According to second statement equation is
5(z−30)/4=y
→5z−150/4=y
Since z=55
y=5(55)−150/4
→y=31.25
→y=31
Trudy has 31 MP3 downloads.
Carnegie Learning Math Series 4th Edition Exercise 1.3 Solutions Page 33 Problem 7 Answer
Terry and Trevor have the same number of MP3 downloads
According to first statement equation is
150+2z/3=x
→ 2z+450/3=x
According to third statement equation is
4(2z+30)/3=w
→8z+120/3=w
Since z=55
1.x=2(55)+450/3
→x=186.67
→x=187
2.w=8(55)+120/3
→w=186.67
→w=187
Answer:
Terry and Trevor has same number of MP3 downloads=187
Page 34 Problem 8 Answer
Given: Statement 1: Terry said: “If you add 150 to the number of MP3 downloads Tom has and then double that number and finally divide it by three, you have the number of MP3 downloads I have.”
Statement 2: Trudy said: “If you take the number of MP3 downloads Tom has, subtract 30, multiply that difference by 5, and finally divide that product by 4, the result will be the number of MP3 downloads I have.”
Statement 3: Trevor said: “Well, if you take twice the number of MP3 downloads Tom has, add 30, multiply the sum by 4, and finally divide that product by 3, you will have the number of MP3 downloads I have.”
If the sum of Trudy’s and Trevor’s MP3 downloads is 39 more than Terry has then equation becomes:
5z−150/4+8z+120/3
=39+2z+450/3
→39z=1338
→z=57.384615
→z=57
Answer:
Tom has 57 MP3 downloads
Linear Equations Solutions Chapter 1 Exercise 1.3 Carnegie Learning Math Series Page 34 Problem 9 Answer
Given: Sum of Trudy’s and Trevor’s MP3 downloads is 39 more than the number of download Terry has
Since z=57
y=5(57)−150/4
→y=33.75
→y=34
Answer:
Trudy has 34 MP3 downloads
Page 34 Problem 10 Answer
Given: Sum of Trudy’s and Trevor’s MP3 downloads is 39 more than the number of download Terry has
Since z=57
w=8(57)+120/3
→w=192
Answer:
Trevor has 192 MP3 downloads
Page 34 Problem 11 Answer
Given: Sum of Trudy’s and Trevor’s MP3 downloads is 39 more than the number of download Terry has
Since z=57
x= 2(57)+450/3
→x=188
Terry has 188 MP3 downloads
Step-By-Step Solutions For Carnegie Learning Math Series Chapter 1 Exercise 1.3 Page 35 Problem 12 Answer
Given equation is
−7(3x+6)/3=7
We have to find out the value of the variable x.
Given equation is
−7(3x+6)/3=7
Multiply both sides by 3
−7(3x+6)/3×3=7×3
−7(3x+6)=7×3
Divide both sides by 7
−7(3x+6)/7
=7×3/7
−(3x+6)=3 [ Cancel out 7]
−3x−6=3
Add 6 both sides
−3x−6+6=3+6
−3x=9
Divide both sides by −3
−3x/−3
=9/−3
x=−3
To check solution plug x=−3 of the left sides of the given equation
L.HS=−7(3×(−3)+6)/3
=−7(−9+6)/3
=−7×(−3)/3
=7×3/3
=3
=RHS
The solution for the equation is x=−3
Carnegie Learning Math Series Chapter 1 Exercise 1.3 Free Solutions Page 36 Problem 13 Answer
Given equation:
−3(−2x−5)/4=−5(3x+5)+5/4
We have to find out the value of x and substitute the value of the variable in the equation to check whether the value x
satisfies the equation or not.
Given equation is −3(−2x−5)/4=5/4−20(3x+5)/4
Multiply both sides by 4
−3(−2x−5)/4×4=5/4×4−20(3x+5)/4×4
−3(−2x−5)=5−20(3x+5)
Simplify the above equation
6x+15=5−60x−100
Add both sides by 60x and −15
6x+60x+15−15=5−60x−100+60x−15
66x=−110
Divide both sides by 66
x=−5/3
Plug the value of x in the L.H.S of the given equation
L.H.S=−3(−2×−5/3−5)/4
=−3(10/3−5)/4
=−3(10−15/3)/4
=−3(−5/3)/4
=5/4
Plug the value of x in the R.H.S of the given equation
R.HS=5/4−20(3(−5/3)+5)/4
=5/4−20(−5+5)/4
=5/4−0/4
=5/4
Hence L.HS=RHS
The value of x is correct.
Solution for the equation is x=−5/3