Need help again ...these questions are hard

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.

answer: answer

step-by-step explanation:

the number of possible outcomes that do not show a 1 on the first spinner and show the number 4 on the second spinner is 2.

spinner 1 spinner 2

1                     1

1                     2

1                     3

1                     4

2                   1

2                   2

2                   3

2                   4   = not 1, and 4

3                   1

3                   2

3                   3

3                   4 = not 1, and 4

only 2 outcomes have passed the given condition.

step-by-step explanation: