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The solution to a system of linear equations is [tex][tex]$(-3,-3)$[/tex][/tex]. Which system of linear equations has this point as its solution?

A. [tex][tex]$x - 5y = -12$[/tex][/tex] and [tex][tex]$3x + 2y = -15$[/tex][/tex]
B. [tex][tex]$x - 5y = -12$[/tex][/tex] and [tex][tex]$3x + 2y = 15$[/tex][/tex]
C. [tex][tex]$x - 5y = 12$[/tex][/tex] and [tex][tex]$3x + 2y = -15$[/tex][/tex]
D. [tex][tex]$x - 5y = 12$[/tex][/tex] and [tex][tex]$3x + 2y = 15$[/tex][/tex]

Sagot :

GinnyAnswer
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.
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