Part A: The pair of equations for lines A and B have one solution. This can be determined by observing the graph and seeing that the two lines intersect at one point.
Part B: The solution to the equations of lines A and B is the point of intersection on the graph, which is approximately (3.67, 2.67). This can be calculated by finding the coordinates of the point where lines A and B intersect. Since this point lies on both lines, it satisfies both equations simultaneously.
The graph shows two lines, A and B.
A graph is shown with x- and y-axes labeled from 0 to 6 at increments of 1. A straight line labeled A joins the ordered pair 2, 6 with the ordered pair 5, 0. Another straight line labeled B joins the ordered pair 0, 2 with the ordered pair 6, 6.
Part A: How many solutions does the pair of equations for lines A and B have? Explain your answer. (5 points)
Part B: What is the solution to the equations of lines A and B? Explain your answer. (5 points)
5 answers
A pair of equations is shown below:
y = 2x − 1
y = 4x − 5
Part A: In your own words, explain how you can solve the pair of equations graphically. Write the slope and y-intercept for each equation that you will plot on the graph to solve the equations. How can you solve it by substitution or elimination? If you decide to solve by elimination, by which number would you multiply the first equation to eliminate x? (6 points)
Part B: What is the solution to the pair of equations? (4 points)
y = 2x − 1
y = 4x − 5
Part A: In your own words, explain how you can solve the pair of equations graphically. Write the slope and y-intercept for each equation that you will plot on the graph to solve the equations. How can you solve it by substitution or elimination? If you decide to solve by elimination, by which number would you multiply the first equation to eliminate x? (6 points)
Part B: What is the solution to the pair of equations? (4 points)
Part A: To solve the pair of equations graphically, you would plot the two lines on a graph where the x-axis represents the value of x and the y-axis represents the value of y. To plot each equation, you would use the slope-intercept form of the equation, y = mx + b, where m is the slope and b is the y-intercept. For the equation y = 2x - 1, the slope is 2 and the y-intercept is -1. For the equation y = 4x - 5, the slope is 4 and the y-intercept is -5.
To solve the equations by substitution, you would solve for one variable (y in this case) in one of the equations and then substitute it into the other equation. To solve by elimination, you would multiply one or both equations by a constant to make the coefficients of one of the variables the same (x or y) so that when you add or subtract the equations, one variable is eliminated.
If you decide to solve by elimination in this case, you would multiply the first equation by 2 to eliminate x. This would make the first equation become y = 4x - 2, which would match the coefficient of x in the second equation, making elimination possible.
Part B: The solution to the pair of equations is x = 1 and y = 1. This can be found by substituting x = 1 into either equation to solve for y, or by substituting y = 1 into either equation to solve for x.
To solve the equations by substitution, you would solve for one variable (y in this case) in one of the equations and then substitute it into the other equation. To solve by elimination, you would multiply one or both equations by a constant to make the coefficients of one of the variables the same (x or y) so that when you add or subtract the equations, one variable is eliminated.
If you decide to solve by elimination in this case, you would multiply the first equation by 2 to eliminate x. This would make the first equation become y = 4x - 2, which would match the coefficient of x in the second equation, making elimination possible.
Part B: The solution to the pair of equations is x = 1 and y = 1. This can be found by substituting x = 1 into either equation to solve for y, or by substituting y = 1 into either equation to solve for x.
Josh and his friends bought vanilla wafers for $4 per packet and chocolate wafers for $1 per packet at a carnival. They spent a total of $45 to buy a total of 27 packets of wafers of the two varieties.
Part A: Write a system of equations that can be solved to find the number of packets of vanilla wafers and the number of packets of chocolate wafers that Josh and his friends bought at the carnival. Define the variables used in the equations. (5 points)
Part B: How many packets of chocolate wafers and vanilla wafers did they buy? Explain how you got the answer and why you selected a particular method to get the answer. (5 points)
Part A: Write a system of equations that can be solved to find the number of packets of vanilla wafers and the number of packets of chocolate wafers that Josh and his friends bought at the carnival. Define the variables used in the equations. (5 points)
Part B: How many packets of chocolate wafers and vanilla wafers did they buy? Explain how you got the answer and why you selected a particular method to get the answer. (5 points)
Part A:
Let x be the number of packets of vanilla wafers and y be the number of packets of chocolate wafers bought.
The cost equation is $4x + $1y = $45
The total number of packets equation x + y = 27
Part B:
To solve this system of equations, the substitution method could be used.
From the total number of packets equation, you can isolate x to get x = 27 - y. Substitute this into the cost equation to get $4(27 - y) + $1y = $45.
Solving for y, you get y = 13. Substitute this back into the equation x = 27 - y to get x = 14.
Therefore, Josh and his friends bought 14 packets of vanilla wafers and 13 packets of chocolate wafers at the carnival. This method was selected because it is straightforward and easier to manage given the equations provided.
Let x be the number of packets of vanilla wafers and y be the number of packets of chocolate wafers bought.
The cost equation is $4x + $1y = $45
The total number of packets equation x + y = 27
Part B:
To solve this system of equations, the substitution method could be used.
From the total number of packets equation, you can isolate x to get x = 27 - y. Substitute this into the cost equation to get $4(27 - y) + $1y = $45.
Solving for y, you get y = 13. Substitute this back into the equation x = 27 - y to get x = 14.
Therefore, Josh and his friends bought 14 packets of vanilla wafers and 13 packets of chocolate wafers at the carnival. This method was selected because it is straightforward and easier to manage given the equations provided.