Select ALL the correct answers.
Select all the statements that are true for the following systems of equations.
System A: [tex]2 x-3 y=4[/tex], [tex]4 x-y=18[/tex]
System B: [tex]3 x-4 y=5[/tex], [tex]y=5 x+3[/tex]
System C: [tex]2 x-3 y=4[/tex], [tex]12 x-3 y=54[/tex]
A. Systems A and C have the same solution.
B. Systems A and B have different solutions.
C. System C simplifies to [tex]2 x-3 y=4[/tex] and [tex]4 x-y=18[/tex] by dividing the second equation by three.
D. All three systems have different solutions.
E. Systems B and C have the same solution.
Answer (2)
Systems A and C share the same solution of ( 5 , 2 ) , while Systems A and B have different solutions. Thus, statements A and B are correct.
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Solve System A: Using elimination, find the solution x = 5 and y = 2 .
Solve System B: Using substitution, find the solution x = − 1 and y = − 2 .
Solve System C: Using elimination, find the solution x = 5 and y = 2 .
Determine the true statements: Systems A and C have the same solution, and Systems A and B have different solutions. Systems A and C have the same solution; Systems A and B have different solutions.
Explanation
Analyzing the Systems of Equations Let's analyze the given systems of equations to determine the correct statements. We will solve each system to find the values of x and y that satisfy each pair of equations.
Solving System A System A:
The equations are:
2 x − 3 y = 4 (1)
4 x − y = 18 (2)
We can solve this system using substitution or elimination. Let's use elimination. Multiply equation (1) by 2:
4 x − 6 y = 8 (3)
Subtract equation (3) from equation (2):
( 4 x − y ) − ( 4 x − 6 y ) = 18 − 8
5 y = 10
y = 2
Substitute y = 2 into equation (1):
2 x − 3 ( 2 ) = 4
2 x − 6 = 4
2 x = 10
x = 5
So, the solution for System A is x = 5 and y = 2 .
Solving System B System B:
The equations are:
3 x − 4 y = 5 (1)
y = 5 x + 3 (2)
Substitute equation (2) into equation (1):
3 x − 4 ( 5 x + 3 ) = 5
3 x − 20 x − 12 = 5
− 17 x = 17
x = − 1
Substitute x = − 1 into equation (2):
y = 5 ( − 1 ) + 3
y = − 5 + 3
y = − 2
So, the solution for System B is x = − 1 and y = − 2 .
Solving System C System C:
The equations are:
2 x − 3 y = 4 (1)
12 x − 18 y = 72 (2)
Notice that equation (2) is just equation (1) multiplied by 6. Therefore, the two equations are dependent, and the system has infinitely many solutions. However, the second equation in the original problem statement for System C is 12 x − 3 y = 54 . Let's solve the system with this equation:
2 x − 3 y = 4 (1)
12 x − 3 y = 54 (2)
Subtract equation (1) from equation (2):
( 12 x − 3 y ) − ( 2 x − 3 y ) = 54 − 4
10 x = 50
x = 5
Substitute x = 5 into equation (1):
2 ( 5 ) − 3 y = 4
10 − 3 y = 4
− 3 y = − 6
y = 2
So, the solution for System C is x = 5 and y = 2 .
Evaluating the Statements Now let's evaluate the given statements:
Systems A and C have the same solution: System A has the solution x = 5 , y = 2 , and System C has the solution x = 5 , y = 2 . This statement is true .
Systems A and B have different solutions: System A has the solution x = 5 , y = 2 , and System B has the solution x = − 1 , y = − 2 . This statement is true .
System C simplifies to 2 x − 3 y = 4 and 4 x − y = 18 by dividing the second equation by three: System C is 2 x − 3 y = 4 and 12 x − 3 y = 54 . Dividing the second equation by 3 gives 4 x − y = 18 . So the second equation should be 4 x − y = 18 . However, the original second equation is 12 x − 3 y = 54 . Dividing this by 3 gives 4 x − y = 18 . Therefore, this statement is false .
All three systems have different solutions: System A and System C have the same solution, so this statement is false .
Systems B and C have the same solution: System B has the solution x = − 1 , y = − 2 , and System C has the solution x = 5 , y = 2 . This statement is false .
Final Answer Therefore, the correct statements are:
Systems A and C have the same solution.
Systems A and B have different solutions.
Examples
Systems of equations are used in various real-life scenarios, such as determining the break-even point for a business. For example, suppose a company produces and sells a product. The cost to produce the product can be represented by a linear equation, and the revenue from selling the product can also be represented by a linear equation. By solving the system of equations, the company can find the number of products they need to sell to cover their costs, which is the break-even point. Let's say the cost equation is C = 10 x + 500 and the revenue equation is R = 30 x , where x is the number of products. To find the break-even point, we set C = R , so 10 x + 500 = 30 x . Solving for x , we get 20 x = 500 , which means x = 25 . Therefore, the company needs to sell 25 products to break even.
