Two equations are given below: a − 3b = 4 a = b − 2 What is the solution to the set of equations in the form (a, b)? (−2, −2) (−3, −1) (−9, −7) (−5, −3)

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I wanna teach kids the Christian truth

If you wanna reach those kids on the street

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Step-by-step explanation:

1) AB + BC = AC. Plug in for each of these, you get:

x - 3 + 7x + 12 = 6x + 23

Combine like terms:

8x + 15 = 6x + 23

Subtract 6x from both sides and subtract 15 from both sides so you get variables on one side and constants on the other.

8x - 6x = 23 - 15

2x = 8

Divide both sides by 2 to isolate the variable.

x = 8/2 = 4

x = 4.

2) Same thing to start out. AB + BC = AC.

2y + 5 + 3y - 4 = 7y - 5

Combine like terms:

5y + 1 = 7y - 5

Subtract 5y from both sides and add 5 to both sides so that you get variables on one side and constants on the other.

1 + 5 = 7y - 5y

6 = 2y

Divide both sides by 2 to isolate the variable.

y = 6/2 = 3

y = 3.

Now we must plug in our y-value. If AB = BC, then point is the mid-point.

2y + 5 = 3y - 4

2(3) + 5 = 3(3) - 4

6 + 5 = 9 - 4

11 ≠ 5

Point B is not the midpoint.

3) AH + HS = AS

5b - 1 + 12b + 7 = 8b + 9

Combine like terms:

17b + 6 = 8b + 9

Subtract 8b from both sides and 6 from both sides to get variables on one side and constants on the other:

17b - 8b = 9 - 6

9b = 3

Divide by 9 on both sides to isolate the variable:

b = 3/9 = 1/3.

Now, we have to find AS. So plug in your b-value into the AS equation.

AS = 8b + 9

8(1/3) + 9

8/3 + 9 -> 8/3 + 27/3 = 35/3 = 11 and 2/3

AS = 11 and 2/3 units