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Dafna1 [17]
4 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]4 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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The expanded form of 584 is:

584 = 500 + 80 + 4

Therefore:

3 · 584 = 3 · ( 500 + 80 + 4 )

Then we will use Distributive property : we can multiply a sum by muptiplying each addend separately and then add the products.

3 · 500 + 3 · 80 + 3 · 4 = 1,500 + 240 + 12 = 1,752

Finally: 3 · 584 = 1,752.

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<h2>(1, 4)</h2>

Step-by-step explanation:

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Answer:

\sf A . \sf \frac{3}{-7}

\sf E. \sf -( \frac{-3}{-7} )

This above options makes the expression true.

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Answer:

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y , 2 , 4 , 5 , 7, 9

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