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OverLord2011 [107]
3 years ago
11

Need help again ...these questions are hard

Mathematics
2 answers:
kherson [118]3 years ago
6 0
The answer is
(0, 5)
(1, 3)
OLga [1]3 years ago
4 0

Answer:

Number 1 and 4

Step-by-step explanation:

To do this substitute the different values of x and y into the equations and see if they come out true or not.

1. (4, -3)

Put x = 4 and y = -3 into both equations:

2x + y = 5 --> 2(4) - 3 = 5 --> 8 - 3 = 5 --> 5=5 (true)

6x + 3y = 15 --> 6(4) + 3(-3) = 15 --> 24 - 9 = 15 --> 15 = 15 (true

So number 1 works

2. (2, 4)

Put x = 2 and y = 4 into both equations:

2x + y = 5 --> 2(2) + 4 = 5 --> 4 + 4 = 5 --> 8 = 5 (NOT true)

Number 2 does not work

3. (1, 3)

Put x = 1 and y = 3 into both equations:

2x + y = 5 --> 2(1) + 3 = 5 --> 2+ 3 = 5 --> 5 = 5 (true)

6x + 3y = 15 --> 6(1) + 3(9) = 15 --> 6 + 27 = 15 --> 33 = 15 (not true)

Number 3 does not work

4. (0, 5)

Put x = 0 and y = 5 into both equations:

2x + y = 5 --> 2(0) + 5 = 5 --> 5 = 5 (true)

6x + 3y = 15 --> 6(0) + 3(5) = 15 --> 15 = 15 (true)

Number 4 does work

Number 1 and number 4 are the answers.

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The sum of all three numbers is 12. The sum of twice the first number, 4 times the second number, and 5 times the third number i
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Answer:

The numbers are 7, 4 and 1

Step-by-step explanation:

Let the numbers =x, y, z.

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3 years ago
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