Astudent took a system of equations, multiplied the first equation 2 by 3 and the second equation by , then added the results together. based on this, she concluded that there were no solutions. which system of equations could she have started with? a. -2x+4y=4 -3x+6y=6 b. 3x+y=12 -3x+6y=6 c.3x+6y=9 -2x-4y=4 d.2x-4y=6 -3x+6y=9

C)3x+6y=9

-2x-4y=4

Step-by-step explanation:

A student took a system of equations, multiplied the first equation 2 by 3 and the second equation by , then added the results together.

Lets check with each option

We multiply the first equation by 2  and second equation by 3 and add it.

When both x  and y terms gets cancelled then we can say there were no solutions.

A) -2x+4y=4

-3x+6y=6

Equation becomes

-4x + 8y = 8

-9x + 18y = 18

-13x + 26y = 26

B) 3x+y=12

-3x+6y=6

Multiply first equation by 2  and second equation by 3

6x + 2y = 24

-9x + 18x = 18

-3x + 20y = 42

C)3x+6y=9

-2x-4y=4

Multiply first equation by 2  and second equation by 3

6x + 12y = 18

-6x - 12y = 12

          0 = 30

Both x  and y terms becomes 0. Hence there is no solution for these system of equations.

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