Algebra 1 Student Journal 1st Edition Chapter 2 Solving Linear Inequalities
Page 33 Essential Question Answer
If we want to add or subtract from one side of the equation, we must perform the same operation to the other side of the equation.
This ensures that the inequality is still true. When solving inequalities by subtracting, our goal is to have the variable on its own.
Adding or subtracting will not change the directions of the < or > than signs.
For using addition or subtraction to solve an inequality we need to perform the same action on both sides of the equation.
Read and Learn More Big Ideas Math Algebra 1 Student Journal 1st Edition Solutions
Page 33 Exploration 1 Answer
Given: \(P=\frac{8.4 Y+100 C+330 T-200 N}{A}\) and Y is the total length of all completed passes (in Yards), C is Completed passes, T is passes resulting in a Touchdown, N is intercepted passes, A is Attempted passes, M is incomplete passes
To Find Whether T<C is true or not.
Evaluate the question to get the answer.
From the given we get,
A=C+N+M…….(1)
C=T+Nt……..(2)
Now, using equation (2) we get,
T<C
T<T+Nt
C>T
Thus, T<C.
This equation is valid given that all the variables are greater than or equal to zero.
Hence, T<C is true because this equation is valid given that all the variables are greater than or equal to zero.
Given: \(P=\frac{8.4 Y+100 C+33 U I^{\prime}-200 N}{A}\)and Y is the total length of all completed passes (in Yards), C is Completed passes,T is passed resulting in a Touchdown, N is intercepted passes, A is Attempted passes, M is incomplete passes.
To Find Whether C+N≤A is true or not.
Evaluate the question to get the answer.
From the given we get,
A=C+N+M ……..(1)
C=T+Nt……..(2)
Now, using equation (1) we get
C+N≤A
C+N≤C+N+M
A≥C+N
Thus, C+N≤A.
This equation is valid given that all the variables are greater than or equal to zero.
This equation is valid given that all the variables are greater than or equal to zero C+N≤A
is true.
Given: \(P=\frac{8.4 Y+100 C+33 U I^{\prime}-200 N}{A}\) and Y is the total length of all completed passes (in Yards), C is Completed passes, T is passes resulting in a Touchdown, N is intercepted passes, A is Attempted passes, M is incomplete passes.
To Find Is N<A true.
Evaluate the question to get the answer.
From the given we get,
A=C+N+M…….(1)
C=T+Nt………(2)
Now, using equation(1) we get
N<A
N<C+N+M
A>N
Thus, N<A.
This equation is valid given that all the variables are greater than or equal to zero.
This equation is valid given that all the variables are greater than or equal to zero N<A is true.
Given: \(P=\frac{8.4 Y+100 C+33 U I^{\prime}-200 N}{A}\) and Y is the total length of all completed passes (in Yards), C is Completed passes, T is passes resulting in a Touchdown, N is intercepted passes, A is Attempted passes, M is incomplete passes.
To Find Is A−C≥M true.
Evaluate the question to get the answer.
From the given we get,
A=C+N+M……..(1)
C=T+Nt………(2)
Now, using equation(1) we get
A−C≥M
(C+N+M)−C≥M
N+M≥M
Thus, A−C≥M.
This equation is valid given that all the variables are greater than or equal to zero.
A−C ≥M is true.
Page 34 Exercise 3 Answer
To determine if an inequality is legitimate, we utilize addition and subtraction.
In addition, we solve inequalities using addition and subtraction in the same way as we solve linear equations.
We use addition and subtraction to solve an inequality the same way we solve an equality.
Page 34 Exercise 4 Answer
To Find Solve the inequality x+3<4
Evaluate the question to get the answer.
Consider the following inequality
x+3<4
=x+3−3<4−3
=x<1
After solving x+3<4 we get x<1.
To Find Solve the inequality x−3≥5
Evaluate the question to get the answer.
Consider the inequality
x−3≥5
=x−3+3≥5+3
=x≥8
After solving x−3≥5 we get, x≥8.
To Find Solve the inequality 4>x−2
Evaluate the question to get the answer.
From the given we get,
4<x−2
4+2<x
x>6
After solving 4>x−2 we get x>6.
To Find Solve the inequality −2≤x+1
Evaluate the question to get the answer.
From the given we get,
−2≤x+1
−2−1≤x+1−1
x≥−3
After solving −2≤x+1 we get x≥−3.
Page 36 Exercise 1 Answer
To Find Solve the inequality x−3<−4 and graph the solution.
Evaluate the question to get the answer.
From the given, we get that,
x−3<−4
x−3+3<−4+3
x<−1
Now, graphing the solution we get
After solving x−3<−4 we get x<−1 and the graph is
Page 36 Exercise 2 Answer
To Find Solve the inequality −3>−3+h and graph the solution.
Evaluate the question to get the answer.
From the given we get
−3>−3+h
−3+3>−3+h+3
0>h
Now, graphing the solution we get
After solving −3>−3+h we get h<0 and the required graph is
Page 37 Exercise 3 Answer
To Find Solve the inequality s−(−1)≥2 and graph the solution.
Evaluate the question to get the answer.
From the given we get,
s−(−1)≥2
s+1≥2
s+1−1≥2−1
s≥1
Now, graphing the solution we get
After solving s−(−1)≥2 we get s≥1 and the required graph is
Page 37 Exercise 5 Answer
To Find Solve the inequality 12≤4c−3c+10 and graph the solution.
Evaluate the question to get the answer.
From the given we get,
12≤4c−3c+10
12≤c+10
12−10≤c+10−10
2≤c
Now, graphing the solution we get
After solving 12≤4c−3c+10 we get c≥2 and the required graph is
Page 37 Exercise 6 Answer
To Find Solve the inequality 15−7p+8p>15−2
and graph the solution.
Evaluate the question to get the answer.
From the given we get,
15−7p+8p>15−2
15+p>13
15+p−15>13−15
p>−2
Now, graphing the solution we get
After solving 15−7p+8p>15−2 we get p>−2 and the required graph is
Page 37 Exercise 7 Answer
Given: The amount of money to be spend is $15 and amount of groceries in the cart is$12.25
To Find Write an inequality that represents how much more money m that can be spend.
Evaluate the question to get the answer.
Let ,S=15 , C=12.25
Now, from the given we get the inequality as
S≥C+m……(1)
S≥C+m is the inequality that represents how much more money m can be spend.
Given: The amount of money to be spend is $15 and the amount of groceries in the cart is$12.25
To Find: Solve the inequality.
Evaluate the question to get the answer.
Given: $15 to spend on groceries and $12.25 worth of groceries already in your cart.
Let, S=15 , C=12.25
Substituting the values in we get
S≥C+m
15≥12.25+m
15−12.25≥12.25+m−12.25
2.75≥m
After solving the inequality we get m≤2.75.