Question 1
Use substitution t solve the system of equations
5x + y = -24
x + 3y = -2
A: x = -5, y = -1
B: x = 5, y = 1
C: x = -5, y = 1
D x = 5, y = -1
Question 2:
Use substitution to solve the systems of equations. How many solutions are there?
1/2 x -1/3 y = 5
x = 2/3 y + 10
A: there are no solutions
B: there is one solution.
C: there are infinitely many solutions.
D: it is not possible to determine the number of solutions.
Question 3
y = 4.5x + 7.5
y = -3.5x - 4.5
A: x = -1.5, y = -0.75
B: x = 1.5, y = 0.75
C: x = 1.5, y = -0.75
D: x = -1.5, y = 0.75
Questions 4:
When using substitution to solve a system of equations, how can you tell when a system has no solution? Use the drop-down menus to explain your answer.
A system has no solution when the resulting equation is always. A True Or False? Next x = 4 4 = 5 4 = 4 is an example of this type of equation.
Question 5:
Tickets to a school play cost $3 for students and $8 for adults. On opening night, $1,000 was collected and 150 tickets sold. Use substitution to solve a system of equations to find how many of each kind of ticket were sold? Enter your answers in the boxes.
How many student tickets were sold?
How many adult Tickets were sold?
Please Help I have to have this done today before school is over I have exactly 2hrs left of school please and thank you for the help.
NEVER MIND I DID IT MYSELF WITH A FRIEND
what is the last one?
Question 1:
To solve the system of equations using substitution, we can start by solving one equation for one variable and substituting it into the other equation.
From the first equation, 5x + y = -24, we can rearrange it to solve for y:
y = -5x - 24
Now, we substitute this value of y into the second equation:
x + 3(-5x - 24) = -2
x - 15x - 72 = -2
-14x - 72 = -2
-14x = 70
x = -5
Now, substitute the value of x into the first equation to find y:
5(-5) + y = -24
-25 + y = -24
y = 1
Therefore, the solution to the system of equations is x = -5 and y = 1.
Answer: C: x = -5, y = 1
Question 2:
To solve the system of equations using substitution, we can start by solving one equation for one variable and substituting it into the other equation.
From the second equation, x = 2/3 y + 10, we can substitute this value of x into the first equation:
1/2 (2/3 y + 10) - 1/3 y = 5
1/3 y + 5 - 1/3 y = 5
5 = 5
We notice that both variables cancel out, and we are left with the equation 5 = 5. This means that the equation is always true, no matter the value of y. Hence, there are infinitely many solutions.
Answer: C: there are infinitely many solutions.
Question 3:
To solve the system of equations, we can equate the expressions for y:
4.5x + 7.5 = -3.5x - 4.5
Rearrange the equation:
4.5x + 3.5x = -4.5 - 7.5
8x = -12
Divide both sides by 8:
x = -12/8
x = -1.5
Substitute the value of x into one of the equations to find y:
y = -3.5(-1.5) - 4.5
y = 5.25 - 4.5
y = 0.75
Therefore, the solution to the system of equations is x = -1.5 and y = 0.75.
Answer: B: x = 1.5, y = 0.75 is incorrect, all the other options are incorrect.
Question 4:
A system has no solution when the resulting equation is always false. The statement "x = 4, 4 = 5, 4 = 4" is an example of this type of equation.
Therefore, the correct answer is True.
Question 5:
Let's use the substitution method to solve this problem. First, we can define the variables:
Let x = number of student tickets sold.
Let y = number of adult tickets sold.
From the given information, we have two equations:
3x + 8y = 1000 (equation 1)
x + y = 150 (equation 2)
We can rearrange equation 2 to solve for x:
x = 150 - y
Now, substitute this value of x into equation 1:
3(150 - y) + 8y = 1000
450 - 3y + 8y = 1000
5y = 550
y = 110
Substitute the value of y into equation 2 to find x:
x + 110 = 150
x = 40
Therefore, the number of student tickets sold is 40, and the number of adult tickets sold is 110.
Answer:
Number of student tickets sold: 40
Number of adult tickets sold: 110
Question 1:
To solve the system of equations using substitution, we can solve one equation for one variable and substitute it into the other equation.
Let's solve the first equation for x:
5x + y = -24
=> x = (-24 - y)/5
Now we substitute this value of x into the second equation:
x + 3y = -2
=> [(-24 - y)/5] + 3y = -2
To simplify the equation, we can multiply through by 5 to eliminate the fraction:
-24 - y + 15y = -10
Combining like terms:
14y - y = -10 + 24
13y = 14
y = 14/13
Now, substitute the value of y back into the first equation to solve for x:
x = (-24 - (14/13))/5
=> x = (-24*13 - 14)/ (13*5)
=> x = -140/65 = -4/3
So the solution is x = -4/3 and y = 14/13.
Checking the answer options, none of them match the solution we found. This means that none of the answer options provided are correct.
Question 2:
To solve the system of equations using substitution, we solve one equation for one variable and substitute it into the other equation.
The second equation is already solved for x:
x = 2/3 y + 10
Now we substitute this value of x into the first equation:
1/2 x - 1/3 y = 5
=> (1/2)(2/3 y + 10) - 1/3 y = 5
Simplifying the equation:
(1/3 y + 5) - 1/3 y = 5
5 = 5
In this case, the equation simplifies to 5 = 5, which means the equation is always true. This indicates that the two equations represent the same line and have infinitely many solutions.
So the answer is C: there are infinitely many solutions.
Question 3:
We have the equations:
y = 4.5x + 7.5
y = -3.5x - 4.5
To solve the system of equations, we can equate the two expressions for y and solve for x:
4.5x + 7.5 = -3.5x - 4.5
Adding 3.5x to both sides and subtracting 7.5 from both sides:
8x = -12
Dividing both sides by 8:
x = -12/8 = -3/2 = -1.5
Now substitute the value of x back into one of the original equations, for example, the first equation:
y = 4.5(-1.5) + 7.5
y = -6.75 + 7.5
y = 0.75
So the solution is x = -1.5 and y = 0.75.
Checking the answer options, we can see that option B matches the solution we found: x = 1.5 and y = 0.75.
Question 4:
A system has no solution when the resulting equation is always false. The equation x = 4 can never be true when combined with the equation 4 = 5 or 4 = 4.
Therefore, the statement "A system has no solution when the resulting equation is always false" is true.
Question 5:
Let's assume the number of student tickets sold is S and the number of adult tickets sold is A.
We are given two pieces of information:
1. Tickets sold: S + A = 150
2. Amount collected: 3S + 8A = 1000
To solve for S and A, we can use substitution or elimination method.
Let's solve this system using substitution:
From equation 1, we can express S in terms of A:
S = 150 - A
Substitute this value of S into equation 2:
3(150 - A) + 8A = 1000
450 - 3A + 8A = 1000
5A = 550
A = 110
Now substitute the value of A back into equation 1 to solve for S:
S + 110 = 150
S = 150 - 110
S = 40
So, 40 student tickets were sold and 110 adult tickets were sold.