Answer :
To determine which system of linear equations has the point [tex]\((-3, -3)\)[/tex] as its solution, we need to substitute [tex]\((x, y) = (-3, -3)\)[/tex] into each pair of equations. We will then verify if both equations in any system are satisfied by this point.
Let's analyze each system one by one:
### System 1:
[tex]\[ x - 5y = -12 \][/tex]
[tex]\[ 3x + 2y = -15 \][/tex]
Substitute [tex]\((x, y) = (-3, -3)\)[/tex]:
1. For the first equation:
[tex]\[ -3 - 5(-3) = -3 + 15 = 12 \neq -12 \][/tex]
The point [tex]\((-3, -3)\)[/tex] does not satisfy the first equation. Therefore, this system cannot be the answer.
### System 2:
[tex]\[ x - 5y = -12 \][/tex]
[tex]\[ 3x + 2y = 15 \][/tex]
Substitute [tex]\((x, y) = (-3, -3)\)[/tex]:
1. For the first equation:
[tex]\[ -3 - 5(-3) = -3 + 15 = 12 \neq -12 \][/tex]
The point [tex]\((-3, -3)\)[/tex] does not satisfy the first equation. Therefore, this system cannot be the answer.
### System 3:
[tex]\[ x - 5y = 12 \][/tex]
[tex]\[ 3x + 2y = -15 \][/tex]
Substitute [tex]\((x, y) = (-3, -3)\)[/tex]:
1. For the first equation:
[tex]\[ -3 - 5(-3) = -3 + 15 = 12 \][/tex]
The point [tex]\((-3, -3)\)[/tex] satisfies the first equation.
2. For the second equation:
[tex]\[ 3(-3) + 2(-3) = -9 - 6 = -15 \][/tex]
The point [tex]\((-3, -3)\)[/tex] satisfies the second equation.
Since the point [tex]\((-3, -3)\)[/tex] satisfies both equations, this system could be the correct one.
### System 4:
[tex]\[ x - 5y = 12 \][/tex]
[tex]\[ 3x + 2y = 15 \][/tex]
Substitute [tex]\((x, y) = (-3, -3)\)[/tex]:
1. For the first equation:
[tex]\[ -3 - 5(-3) = -3 + 15 = 12 \][/tex]
The point [tex]\((-3, -3)\)[/tex] satisfies the first equation.
2. For the second equation:
[tex]\[ 3(-3) + 2(-3) = -9 - 6 = -15 \neq 15 \][/tex]
The point [tex]\((-3, -3)\)[/tex] does not satisfy the second equation. Therefore, this system cannot be the answer.
After substituting the point [tex]\((-3, -3)\)[/tex] into each system of equations, we find that the only system that satisfies both equations is:
[tex]\[ \boxed{ x - 5y = 12 \text{ and } 3x + 2y = -15 } \][/tex]
This corresponds to the third system of linear equations.
Let's analyze each system one by one:
### System 1:
[tex]\[ x - 5y = -12 \][/tex]
[tex]\[ 3x + 2y = -15 \][/tex]
Substitute [tex]\((x, y) = (-3, -3)\)[/tex]:
1. For the first equation:
[tex]\[ -3 - 5(-3) = -3 + 15 = 12 \neq -12 \][/tex]
The point [tex]\((-3, -3)\)[/tex] does not satisfy the first equation. Therefore, this system cannot be the answer.
### System 2:
[tex]\[ x - 5y = -12 \][/tex]
[tex]\[ 3x + 2y = 15 \][/tex]
Substitute [tex]\((x, y) = (-3, -3)\)[/tex]:
1. For the first equation:
[tex]\[ -3 - 5(-3) = -3 + 15 = 12 \neq -12 \][/tex]
The point [tex]\((-3, -3)\)[/tex] does not satisfy the first equation. Therefore, this system cannot be the answer.
### System 3:
[tex]\[ x - 5y = 12 \][/tex]
[tex]\[ 3x + 2y = -15 \][/tex]
Substitute [tex]\((x, y) = (-3, -3)\)[/tex]:
1. For the first equation:
[tex]\[ -3 - 5(-3) = -3 + 15 = 12 \][/tex]
The point [tex]\((-3, -3)\)[/tex] satisfies the first equation.
2. For the second equation:
[tex]\[ 3(-3) + 2(-3) = -9 - 6 = -15 \][/tex]
The point [tex]\((-3, -3)\)[/tex] satisfies the second equation.
Since the point [tex]\((-3, -3)\)[/tex] satisfies both equations, this system could be the correct one.
### System 4:
[tex]\[ x - 5y = 12 \][/tex]
[tex]\[ 3x + 2y = 15 \][/tex]
Substitute [tex]\((x, y) = (-3, -3)\)[/tex]:
1. For the first equation:
[tex]\[ -3 - 5(-3) = -3 + 15 = 12 \][/tex]
The point [tex]\((-3, -3)\)[/tex] satisfies the first equation.
2. For the second equation:
[tex]\[ 3(-3) + 2(-3) = -9 - 6 = -15 \neq 15 \][/tex]
The point [tex]\((-3, -3)\)[/tex] does not satisfy the second equation. Therefore, this system cannot be the answer.
After substituting the point [tex]\((-3, -3)\)[/tex] into each system of equations, we find that the only system that satisfies both equations is:
[tex]\[ \boxed{ x - 5y = 12 \text{ and } 3x + 2y = -15 } \][/tex]
This corresponds to the third system of linear equations.
