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Dafna1 [17]
3 years ago
12

A student took a system of equations, multiplied the first equation by and the second equation by , then added the results toget

her. 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
Mathematics
1 answer:
Ierofanga [76]3 years ago
3 0

Answer:

option A

-2x + 4y = 4

-3x + 6y = 6

Step-by-step explanation:

In option A, if the student multiply the first equation by 3 and the second equation by -2, then the equations become

-6x+12y = 12 and

6x-12y =-12

If she adds both equations, she will get 0 = 0

This means that the system has infinite number of solutions.

Hence, the student could have started with equations -2x + 4y = 4 and -3x + 6y = 6.

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Answer:

Step-by-step explanation:

Our equations are

y = -3x^2 + x + 12\\y = 2x^2 - 6x + 5\\y = x^2 + 7x - 11\\y = -x^2 - 8x - 16\\

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Discriminant is denoted by  D and its formula is

D=b^2-4ac\\

Where

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Properties of D: If D

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Let us start with

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