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Solve the system of equations 3x - 5y = 12 and 12x - 20y = 28 using the elimination method.

A) (x, y) = (4, 0)
B) (x, y) = (-2, -2)
C) (x, y) = (2, 3)
D) (x, y) = (0, -4)

User Dalius
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Final answer:

To solve the system of equations 3x - 5y = 12 and 12x - 20y = 28 using the elimination method, we multiply the equations by constants to eliminate a variable and then solve for the remaining variable. However, when we perform the elimination method on this system, we find that there is no common solution.

Step-by-step explanation:

To solve the system of equations 3x - 5y = 12 and 12x - 20y = 28 using the elimination method, we need to eliminate one of the variables by multiplying one or both of the equations by a constant so that the coefficients of one of the variables in both equations become the same. Let's start by multiplying the first equation by 4 and the second equation by 1. This gives us:

12x - 20y = 48

12x - 20y = 28

Now, subtract the two equations to eliminate the variable x:

(12x - 20y) - (12x - 20y) = 48 - 28

0 = 20

Since 0 does not equal 20, these two equations do not have a common solution. Therefore, the correct answer is: There is no solution.

User Kochizufan
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