Select all the systems of equations that have exactly one solution. Group of answer choices {y=3x+1y=-3x+7 ( I know this one is wrong) {y=3x+1y=x+1 {y=3x+1y=3x+7 {x+y=102x+2y=20 {x+y=10x+y=12

1) Exactly one solution

2) Exactly one solution

3) No solution

4) Infinite many solution

5) No solution

Step-by-step explanation:

As given

1)

y = 3x + 1     (1)

y = -3x + 7   (2)

Put the value of y in equation (1) , we get

-3x + 7 = 3x + 1

⇒7 - 1 = 3x + 3x

⇒6 = 6x

⇒x = 1

Now as

y = -3x + 7

⇒y = -3(1) + 7 = -3 + 7 = 4

∴ we get

x = 1, y = 4

Unique Solution ( Exactly one solution )

2)

y = 3x+ 1      (3)

y = x+1         (4)

Put the value of y in equation (3), we get

x + 1 = 3x + 1

⇒1 - 1 = 3x - x

⇒0 = 2x

⇒x = 0

Now, as

y = 3x + 1

⇒y = 3(0) + 1 = 1

∴ we get

x = 0, y = 1

Unique solution ( Exactly one solution )

3)

y = 3x + 1        (5)

y = 3x + 7       (6)

Put the value of y in equation (5), we get

3x + 7 = 3x + 1

⇒7 - 1 = 3x - 3x

⇒6 = 0

No Solution

4)

x+y=10        (7)

2x+2y=20  (8)

Divide equation (8) by 2, we get

x + y = 10    (9)

⇒y = 10 - x

Put the value of y in equation (7), we get

x + 10 - x = 10

⇒10 = 10

Infinite many solutions

5)

x + y = 10     (10)

x + y = 12     (11)

⇒y = 12 - x

Put the value of y in equation (10) , we get

x + 12 - x = 10

⇒12 = 10

No solution

i dont know sorry

step-by-step explanation:

. + λrh() e 1 1 2 4 4 4 4 4 3 4 4 4 4 4 + th−1(− s(x)2 dx ω + λrh−1() e 2 1 2 4 4

step-by-step explanation: