The point (5, -3) is a solution to which system of equations?(1 point)
27 answers
There is no given system of equations in the question. Therefore, it is impossible to determine which system of equations the point (5, -3) is a solution to.
The point (5, -3) is a solution to which system of equations?(1 point)
A
y=-1/2x+6
y=3x-1
B
y=x-3
2x+y=7
C
y=2/3x+6
3y+6x-18=0
D
y=x+2
y=x+5
A
y=-1/2x+6
y=3x-1
B
y=x-3
2x+y=7
C
y=2/3x+6
3y+6x-18=0
D
y=x+2
y=x+5
To determine which system of equations the point (5, -3) is a solution to, let's substitute the x and y values of the point into each equation and see which equation(s) are satisfied.
A:
Substituting x=5 and y=-3 into the equations:
-3 = -1/2(5) + 6 [not satisfied]
-3 = 3(5) - 1 [not satisfied]
B:
Substituting x=5 and y=-3 into the equations:
-3 = 5 - 3 [satisfied]
2(5) + (-3) = 7 [satisfied]
C:
Substituting x=5 and y=-3 into the equations:
-3 = 2/3(5) + 6 [not satisfied]
3(-3) + 6(5) - 18 = 0 [not satisfied]
D:
Substituting x=5 and y=-3 into the equations:
-3 = 5 + 2 [not satisfied]
-3 = 5 + 5 [not satisfied]
Out of the given options, only system B satisfies the point (5, -3). Therefore, the point is a solution to system B:
y = x - 3
2x + y = 7
A:
Substituting x=5 and y=-3 into the equations:
-3 = -1/2(5) + 6 [not satisfied]
-3 = 3(5) - 1 [not satisfied]
B:
Substituting x=5 and y=-3 into the equations:
-3 = 5 - 3 [satisfied]
2(5) + (-3) = 7 [satisfied]
C:
Substituting x=5 and y=-3 into the equations:
-3 = 2/3(5) + 6 [not satisfied]
3(-3) + 6(5) - 18 = 0 [not satisfied]
D:
Substituting x=5 and y=-3 into the equations:
-3 = 5 + 2 [not satisfied]
-3 = 5 + 5 [not satisfied]
Out of the given options, only system B satisfies the point (5, -3). Therefore, the point is a solution to system B:
y = x - 3
2x + y = 7
How many solutions does the system of equations have?
y=−2x+12
4y+x−20=0
(1 point)
Responses
One solution
One solution
Two solutions
Two solutions
No solution
No solution
Infinitely many solutions
y=−2x+12
4y+x−20=0
(1 point)
Responses
One solution
One solution
Two solutions
Two solutions
No solution
No solution
Infinitely many solutions
To determine the number of solutions the system of equations has, we can determine if the lines represented by the equations intersect, are parallel, or coincident.
The given system of equations is:
y = -2x + 12
4y + x - 20 = 0
We can see that the first equation is in slope-intercept form (y = mx + b) where m is the slope and b is the y-intercept. The slope of the first equation is -2.
The second equation can be written as:
4y = -x + 20
y = -1/4x + 5
Comparing the slopes of the two equations, we see that one is -2 and the other is -1/4. Since the slopes are not equal, the lines represented by the equations will intersect at a single point.
Therefore, the system of equations has one solution.
The given system of equations is:
y = -2x + 12
4y + x - 20 = 0
We can see that the first equation is in slope-intercept form (y = mx + b) where m is the slope and b is the y-intercept. The slope of the first equation is -2.
The second equation can be written as:
4y = -x + 20
y = -1/4x + 5
Comparing the slopes of the two equations, we see that one is -2 and the other is -1/4. Since the slopes are not equal, the lines represented by the equations will intersect at a single point.
Therefore, the system of equations has one solution.
A company owns two dealerships, both of which sell cars and trucks. Dealership A sells a total of 225 cars and trucks. Dealership B sells twice as many cars and half as many trucks as the Dealership A, and sells a total of 300 cars and trucks.
(4 points)
a) An equation for the total cars and trucks for Dealership A: ____
b) An equation for the total cars and trucks for Dealership B: ____
c) How many cars did Dealership A sell? ____
d) Given the fact that Dealership B sold half as many trucks as Dealership A, how many trucks did Dealership B sell? ____
(4 points)
a) An equation for the total cars and trucks for Dealership A: ____
b) An equation for the total cars and trucks for Dealership B: ____
c) How many cars did Dealership A sell? ____
d) Given the fact that Dealership B sold half as many trucks as Dealership A, how many trucks did Dealership B sell? ____
a) An equation for the total cars and trucks for Dealership A:
Let C represent the number of cars sold by Dealership A and T represent the number of trucks sold by Dealership A.
Therefore, the equation for the total cars and trucks sold by Dealership A can be written as:
C + T = 225
b) An equation for the total cars and trucks for Dealership B:
Since Dealership B sells twice as many cars and half as many trucks as Dealership A, if we let x represent the number of cars sold by Dealership A, then Dealership B would sell 2x cars and (1/2)(x) = (x/2) trucks.
The equation for the total cars and trucks sold by Dealership B can be written as:
2x + (x/2) = 300
Multiplying through by 2 to get rid of the fraction, we have:
4x + x = 600
Combining like terms, we get:
5x = 600
Solving for x, we find:
x = 120
c) How many cars did Dealership A sell?
From the equation we obtained for Dealership A:
C + T = 225
If we substitute x = 120 from the previous equation, we can solve for C:
120 + T = 225
Subtracting 120 from both sides, we get:
T = 225 - 120
T = 105
Therefore, Dealership A sold 120 cars.
d) Given the fact that Dealership B sold half as many trucks as Dealership A, how many trucks did Dealership B sell?
From the equation we obtained for Dealership A:
C + T = 225
We found that T = 105, so we substitute that value into the equation:
C + 105 = 225
Subtracting 105 from both sides, we get:
C = 225 - 105
C = 120
Therefore, Dealership B sold 105 trucks.
Let C represent the number of cars sold by Dealership A and T represent the number of trucks sold by Dealership A.
Therefore, the equation for the total cars and trucks sold by Dealership A can be written as:
C + T = 225
b) An equation for the total cars and trucks for Dealership B:
Since Dealership B sells twice as many cars and half as many trucks as Dealership A, if we let x represent the number of cars sold by Dealership A, then Dealership B would sell 2x cars and (1/2)(x) = (x/2) trucks.
The equation for the total cars and trucks sold by Dealership B can be written as:
2x + (x/2) = 300
Multiplying through by 2 to get rid of the fraction, we have:
4x + x = 600
Combining like terms, we get:
5x = 600
Solving for x, we find:
x = 120
c) How many cars did Dealership A sell?
From the equation we obtained for Dealership A:
C + T = 225
If we substitute x = 120 from the previous equation, we can solve for C:
120 + T = 225
Subtracting 120 from both sides, we get:
T = 225 - 120
T = 105
Therefore, Dealership A sold 120 cars.
d) Given the fact that Dealership B sold half as many trucks as Dealership A, how many trucks did Dealership B sell?
From the equation we obtained for Dealership A:
C + T = 225
We found that T = 105, so we substitute that value into the equation:
C + 105 = 225
Subtracting 105 from both sides, we get:
C = 225 - 105
C = 120
Therefore, Dealership B sold 105 trucks.
that's wrong
Apologies for the mistake. Let's correct the answers:
a) An equation for the total cars and trucks for Dealership A:
Let C represent the number of cars sold by Dealership A and T represent the number of trucks sold by Dealership A.
The equation for the total cars and trucks sold by Dealership A can be written as:
C + T = 225
b) An equation for the total cars and trucks for Dealership B:
Since Dealership B sells twice as many cars and half as many trucks as Dealership A, let C' represent the number of cars sold by Dealership B and T' represent the number of trucks sold by Dealership B.
If Dealership B sells twice as many cars and half as many trucks as Dealership A, we can write:
C' = 2C
T' = T/2
The equation for the total cars and trucks sold by Dealership B can be written as:
C' + T' = 300
Replacing the values of C' and T', we have:
2C + T/2 = 300
c) How many cars did Dealership A sell?
From equation (a), we have:
C + T = 225
d) Given the fact that Dealership B sold half as many trucks as Dealership A, how many trucks did Dealership B sell?
From equation (b), we have:
2C + T/2 = 300
To solve for the unknown variables, we need additional information or equations.
a) An equation for the total cars and trucks for Dealership A:
Let C represent the number of cars sold by Dealership A and T represent the number of trucks sold by Dealership A.
The equation for the total cars and trucks sold by Dealership A can be written as:
C + T = 225
b) An equation for the total cars and trucks for Dealership B:
Since Dealership B sells twice as many cars and half as many trucks as Dealership A, let C' represent the number of cars sold by Dealership B and T' represent the number of trucks sold by Dealership B.
If Dealership B sells twice as many cars and half as many trucks as Dealership A, we can write:
C' = 2C
T' = T/2
The equation for the total cars and trucks sold by Dealership B can be written as:
C' + T' = 300
Replacing the values of C' and T', we have:
2C + T/2 = 300
c) How many cars did Dealership A sell?
From equation (a), we have:
C + T = 225
d) Given the fact that Dealership B sold half as many trucks as Dealership A, how many trucks did Dealership B sell?
From equation (b), we have:
2C + T/2 = 300
To solve for the unknown variables, we need additional information or equations.
he sum of two numbers is 22. Three times one number increased by five is 35.(4 points)
Put responses in the correct input to answer the question. Select a response, navigate to the desired input and insert the response. Responses can be selected and inserted using the space bar, enter key, left mouse button or touchpad. Responses can also be moved by dragging with a mouse.
a) What are the equations you would use? ____ and ____
b) What are the two numbers? ____ area and ____
Put responses in the correct input to answer the question. Select a response, navigate to the desired input and insert the response. Responses can be selected and inserted using the space bar, enter key, left mouse button or touchpad. Responses can also be moved by dragging with a mouse.
a) What are the equations you would use? ____ and ____
b) What are the two numbers? ____ area and ____
a) The equations you would use are:
1) x + y = 22, where x and y are the two numbers whose sum is 22.
2) 3x + 5 = 35, where x is one of the numbers.
b) To solve the equations and find the two numbers, we can proceed as follows:
From equation 1) x + y = 22, we can solve for y:
y = 22 - x
Then, we substitute this value of y into equation 2) 3x + 5 = 35:
3x + 5 = 35
3x = 30
x = 10
Now, we can substitute the value of x into equation 1) to find the value of y:
10 + y = 22
y = 12
Therefore, the two numbers are 10 and 12.
1) x + y = 22, where x and y are the two numbers whose sum is 22.
2) 3x + 5 = 35, where x is one of the numbers.
b) To solve the equations and find the two numbers, we can proceed as follows:
From equation 1) x + y = 22, we can solve for y:
y = 22 - x
Then, we substitute this value of y into equation 2) 3x + 5 = 35:
3x + 5 = 35
3x = 30
x = 10
Now, we can substitute the value of x into equation 1) to find the value of y:
10 + y = 22
y = 12
Therefore, the two numbers are 10 and 12.
Joelle currently has 18 rocks in her collection and gains 4 each week. Lewis currently has 30 rocks in his collection and gains 3 each week. Set up a system of equations to show how many rocks each has in their collection.(6 points)
Put responses in the correct input to answer the question. Select a response, navigate to the desired input and insert the response. Responses can be selected and inserted using the space bar, enter key, left mouse button or touchpad. Responses can also be moved by dragging with a mouse.
a) Joelle has y = ____ + ____
b) Lewis has y = ____ + ____
c) After how many weeks will Joelle and Lewis have the same number of rocks in their collections? ____ weeks
d) How many rocks will Joelle and Lewis have when the amount of rocks in their collection is equal? They will each have ____ rocks in their collection.
Put responses in the correct input to answer the question. Select a response, navigate to the desired input and insert the response. Responses can be selected and inserted using the space bar, enter key, left mouse button or touchpad. Responses can also be moved by dragging with a mouse.
a) Joelle has y = ____ + ____
b) Lewis has y = ____ + ____
c) After how many weeks will Joelle and Lewis have the same number of rocks in their collections? ____ weeks
d) How many rocks will Joelle and Lewis have when the amount of rocks in their collection is equal? They will each have ____ rocks in their collection.
a) Joelle has y = 18 + 4w, where y represents the number of rocks in Joelle's collection and w represents the number of weeks.
b) Lewis has y = 30 + 3w, where y represents the number of rocks in Lewis's collection and w represents the number of weeks.
c) To find the number of weeks when Joelle and Lewis have the same number of rocks, we can set their equations equal to each other:
18 + 4w = 30 + 3w
d) To determine how many rocks Joelle and Lewis will have when the amount of rocks in their collections is equal, we can substitute the value of w from part c) into either equation.
b) Lewis has y = 30 + 3w, where y represents the number of rocks in Lewis's collection and w represents the number of weeks.
c) To find the number of weeks when Joelle and Lewis have the same number of rocks, we can set their equations equal to each other:
18 + 4w = 30 + 3w
d) To determine how many rocks Joelle and Lewis will have when the amount of rocks in their collections is equal, we can substitute the value of w from part c) into either equation.
what the answer to part c
Apologies for the oversight. To determine the number of weeks when Joelle and Lewis have the same number of rocks, we can set their equations equal to each other:
18 + 4w = 30 + 3w
To solve for w, we can subtract 3w from both sides:
w = 12
Therefore, Joelle and Lewis will have the same number of rocks in their collections after 12 weeks.
18 + 4w = 30 + 3w
To solve for w, we can subtract 3w from both sides:
w = 12
Therefore, Joelle and Lewis will have the same number of rocks in their collections after 12 weeks.
what the answer to part d
To determine how many rocks Joelle and Lewis will have when the amount of rocks in their collections is equal, we can substitute the value of w from part c) (w = 12) into either equation.
Let's substitute w = 12 into Joelle's equation:
y = 18 + 4w
y = 18 + 4(12)
y = 18 + 48
y = 66
Therefore, when the amount of rocks in their collections is equal, both Joelle and Lewis will have 66 rocks in their collection.
Let's substitute w = 12 into Joelle's equation:
y = 18 + 4w
y = 18 + 4(12)
y = 18 + 48
y = 66
Therefore, when the amount of rocks in their collections is equal, both Joelle and Lewis will have 66 rocks in their collection.
Solve the system of equations.
3x+2y=2
−2x+y=8
(1 point)
Responses
(14, -20)
(14, -20)
(4, -2)
(4, -2)
(-2, 4)
(-2, 4)
(-4, 2)
3x+2y=2
−2x+y=8
(1 point)
Responses
(14, -20)
(14, -20)
(4, -2)
(4, -2)
(-2, 4)
(-2, 4)
(-4, 2)
To solve the system of equations:
Equation 1: 3x + 2y = 2
Equation 2: -2x + y = 8
We can use the method of substitution or elimination to solve the system. Let's use the method of elimination:
Multiply equation 2 by 2:
2*(-2x + y) = 2*8
-4x + 2y = 16
Now, we can add equation 1 and the modified equation 2:
(3x + 2y) + (-4x + 2y) = 2 + 16
-x + 4y = 18
Now, let's multiply equation 2 by 3:
3*(-2x + y) = 3*8
-6x + 3y = 24
Now, we can add equation 1 and the modified equation 3:
(3x + 2y) + (-6x + 3y) = 2 + 24
-3x + 5y = 26
We now have a system of equations:
-x + 4y = 18
-3x + 5y = 26
To solve this system, let's multiply the first equation by 3:
3*(-x + 4y) = 3*18
-3x + 12y = 54
Now, we can subtract this equation from the second equation:
(-3x + 5y) - (-3x + 12y) = 26 - 54
-3x + 5y + 3x - 12y = -28
-7y = -28
y = 4
We can substitute y = 4 into the first equation to solve for x:
-x + 4(4) = 18
-x + 16 = 18
-x = 2
x = -2
Therefore, the solution to the system of equations is x = -2 and y = 4. The correct response is (-2, 4).
Equation 1: 3x + 2y = 2
Equation 2: -2x + y = 8
We can use the method of substitution or elimination to solve the system. Let's use the method of elimination:
Multiply equation 2 by 2:
2*(-2x + y) = 2*8
-4x + 2y = 16
Now, we can add equation 1 and the modified equation 2:
(3x + 2y) + (-4x + 2y) = 2 + 16
-x + 4y = 18
Now, let's multiply equation 2 by 3:
3*(-2x + y) = 3*8
-6x + 3y = 24
Now, we can add equation 1 and the modified equation 3:
(3x + 2y) + (-6x + 3y) = 2 + 24
-3x + 5y = 26
We now have a system of equations:
-x + 4y = 18
-3x + 5y = 26
To solve this system, let's multiply the first equation by 3:
3*(-x + 4y) = 3*18
-3x + 12y = 54
Now, we can subtract this equation from the second equation:
(-3x + 5y) - (-3x + 12y) = 26 - 54
-3x + 5y + 3x - 12y = -28
-7y = -28
y = 4
We can substitute y = 4 into the first equation to solve for x:
-x + 4(4) = 18
-x + 16 = 18
-x = 2
x = -2
Therefore, the solution to the system of equations is x = -2 and y = 4. The correct response is (-2, 4).
Not including tax, a total of 19 pens and markers cost $11.50. The pens cost $0.25 each, and the markers cost $0.75 each. Write the system of equations that could be used to solve for the number of pens, p, and the number of markers, m, bought.(4 points)
Put responses in the correct input to answer the question. Select a response, navigate to the desired input and insert the response. Responses can be selected and inserted using the space bar, enter key, left mouse button or touchpad. Responses can also be moved by dragging with a mouse.
p + m = Response area
Response area p + Response area m = Response area
Put responses in the correct input to answer the question. Select a response, navigate to the desired input and insert the response. Responses can be selected and inserted using the space bar, enter key, left mouse button or touchpad. Responses can also be moved by dragging with a mouse.
p + m = Response area
Response area p + Response area m = Response area
To write the system of equations that could be used to solve for the number of pens, p, and the number of markers, m, bought, we can use the given information.
Let's assign variables:
p = the number of pens
m = the number of markers
Based on the given information:
The cost of a pen is $0.25, so the total cost of the pens is 0.25p.
The cost of a marker is $0.75, so the total cost of the markers is 0.75m.
The total cost of 19 pens and markers is $11.50, so we can set up the equation:
0.25p + 0.75m = 11.50
The total number of pens and markers is 19, so we can set up the equation:
p + m = 19
The system of equations could be written as:
0.25p + 0.75m = 11.50
p + m = 19
Let's assign variables:
p = the number of pens
m = the number of markers
Based on the given information:
The cost of a pen is $0.25, so the total cost of the pens is 0.25p.
The cost of a marker is $0.75, so the total cost of the markers is 0.75m.
The total cost of 19 pens and markers is $11.50, so we can set up the equation:
0.25p + 0.75m = 11.50
The total number of pens and markers is 19, so we can set up the equation:
p + m = 19
The system of equations could be written as:
0.25p + 0.75m = 11.50
p + m = 19
Joyce wants to go to the zoo with her friend over spring break. You can pay $45 to get into the zoo each time, or join a membership for $50 and then only pay $20 each visit to the zoo. How many times will Joyce need to visit the zoo to justify the expense of the membership fee?(1 point)
Responses
10
10
2
2
3
3
5
Responses
10
10
2
2
3
3
5
To determine how many times Joyce needs to visit the zoo to justify the expense of the membership fee, we need to compare the cost of individual visits to the cost of having a membership.
Let's assume Joyce needs to visit the zoo x times.
If she pays $45 per visit without a membership, the total cost would be 45x.
If she gets a membership for $50 and then pays $20 per visit, the total cost would be 50 + 20x.
To justify the expense of the membership fee, the cost of having a membership should be less than or equal to the cost of individual visits.
So, we can set up the inequality:
50 + 20x ≤ 45x
Let's solve this inequality:
50 ≤ 25x
Dividing both sides of the inequality by 25:
2 ≤ x
Since x represents the number of times Joyce needs to visit the zoo, she would need to visit at least 2 times to justify the expense of the membership fee.
Therefore, the correct response is 2.
Let's assume Joyce needs to visit the zoo x times.
If she pays $45 per visit without a membership, the total cost would be 45x.
If she gets a membership for $50 and then pays $20 per visit, the total cost would be 50 + 20x.
To justify the expense of the membership fee, the cost of having a membership should be less than or equal to the cost of individual visits.
So, we can set up the inequality:
50 + 20x ≤ 45x
Let's solve this inequality:
50 ≤ 25x
Dividing both sides of the inequality by 25:
2 ≤ x
Since x represents the number of times Joyce needs to visit the zoo, she would need to visit at least 2 times to justify the expense of the membership fee.
Therefore, the correct response is 2.
Two systems of equations are shown:
4x+y=−1
x+y=2
and
8x+2y=−2
−4x+cy=−8
What would the value of c need to be to get the same solution for both systems?
(1 point)
Responses
-4
-4
2
2
-2
-2
4
4
1
4x+y=−1
x+y=2
and
8x+2y=−2
−4x+cy=−8
What would the value of c need to be to get the same solution for both systems?
(1 point)
Responses
-4
-4
2
2
-2
-2
4
4
1
To get the same solution for both systems of equations, we need to find the value of c such that the second system becomes equivalent to the first system.
Let's solve the first system of equations:
4x + y = -1 ...(1)
x + y = 2 ...(2
Let's solve the first system of equations:
4x + y = -1 ...(1)
x + y = 2 ...(2
So what's the answer.
Susie determined that the system of equations below has infinitely many solutions.
The first line goes through the points (2, -13) and (-2, -1).
The second line goes through the points (5, -22) and (-3, 2).
Is she correct?
(1 point)
Responses
No, the lines are parallel.
No, the lines are parallel.
Yes, Susie is correct.
Yes, Susie is correct.
No, there is no point of intersection.
No, there is no point of intersection.
She may or may not be correct as the lines possibly cross.
The first line goes through the points (2, -13) and (-2, -1).
The second line goes through the points (5, -22) and (-3, 2).
Is she correct?
(1 point)
Responses
No, the lines are parallel.
No, the lines are parallel.
Yes, Susie is correct.
Yes, Susie is correct.
No, there is no point of intersection.
No, there is no point of intersection.
She may or may not be correct as the lines possibly cross.