Arrange the systems of equations that have a single solution in Increasing order of the x-values in their solutions. 2x + y = 10
(x-3y=-2
x + 2y = 5
2x + y = 4
x + 3y = 5
6x - y = 11
2x + y = 10
- 6x - 3y = -2
HELP

Complete question is;

Arrange the systems of equations that have a single solution in Increasing order of the x-values in their solutions.

1) 2x + y = 10

x - 3y = -2

2) x + 2y = 5

2x + y = 4

3) x + 3y = 5

6x - y = 11

4) 2x + y = 10

-6x - 3y = -2

Arranging the systems in increasing value of x gives;

System 2, system 3, system 1

Step-by-step explanation:

1) 2x + y = 10 - - - (eq 1)

x - 3y = -2 - - - (eq 2)

Multiply eq 1 by 3 to get;

6x + 3y = 30

Add it to eq 2 to get;

7x = 28

x = 28/7

x = 4

2) x + 2y = 5 - - - (eq 1)

2x + y = 4 - - - (eq 2)

Multiply eq 2 by 2 to get;

4x + 2y = 8

Subtract eq 1 from it to get;

3x = 3

x = 3/3

x = 1

3) x + 3y = 5 - - - (eq 1)

6x - y = 11 - - - (eq 2)

Multiply eq 2 by 3 to get;

18x - 3y = 33

Add to eq(1) to get;

19x = 38

x = 38/19

x = 2

4) 2x + y = 10 - - - (eq 1)

-6x - 3y = -2 - - - (eq 2)

Make y the subject in eq 1 to get;

y = 10 - 2x

Put in eq (2);

-6x - 3(10 - 2x) = -2

-6x - 30 + 6x = -2

0 = 0

Thus,no solution exists.

Arranging the systems in increasing value of x gives;

System 2, system 3, system 1

2 x^2 -3x+p

sum of roots = 3/2

let other root be x

3+ x = 3/2

x= 3/2 - 3 = -3/2

product of roots = p/2

3× -3/2 = p/2

-9 = p

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step-by-step explanation:

sry but any answer choices