Question

Directions: State whether or not (3, -2) is a solution to the following systems.
1. x-2y = 7 and 2x + 3y = 0
2.- x – y = -5 and 3x – 4y = 17
3. x + y = 1 and x – y = 6

1. (3, -2) is a solution to the system of equations x – 2y = 7 and 2x + 3y = 0, (3, -2) is not a solution to the system of equations  -x – y = -5 and 3x – 4y = 17 and (3, -2) is not a solution to the system of equations x + y = 1 and x – y = 6

### How to determine the whether or not (3, -2) is a solution to the following systems?

The systems of equations are given as:
1. x – 2y = 7 and 2x + 3y = 0
2. -x – y = -5 and 3x – 4y = 17
3. x + y = 1 and x – y = 6
Next, we substitute (3, -2) for (x, y) in the system of equations.
So, we have:
1. x – 2y = 7 and 2x + 3y = 0
3 – 2 * -2 = 7 and 2 * 3 + 3 * -2 = 0
Evaluate
7 = 7 and 0 = 0
The above equation is true
Hence, (3, -2) is a solution to the system of equations x – 2y = 7 and 2x + 3y = 0
2.- x – y = -5 and 3x – 4y = 17
-3 + 2 = -5 and 3 * 3 + 4 * 2 = 17
Evaluate
– 1 = – 5 and 1 7 = 1 7
The above equation is false
Hence, (3, -2) is not a solution to the system of equations  -x – y = -5 and 3x – 4y = 17
3. x + y = 1 and x – y = 6
3 – 2 = 1 and 3 + 2 = 6
Evaluate
1 = 1 and 5 = 6
The above equation is false
Hence, (3, -2) is not a solution to the system of equations x + y = 1 and x – y = 6