Answer :
To determine which linear system has the solution [tex]\( x = 5 \)[/tex] and [tex]\( y = -4 \)[/tex], we need to substitute these values into each equation and check if they satisfy the system.
### Linear System a:
1. First equation:
[tex]\[ x + 3y = 12 \][/tex]
Substituting [tex]\( x = 5 \)[/tex] and [tex]\( y = -4 \)[/tex]:
[tex]\[ 5 + 3(-4) = 5 - 12 = -7 \neq 12 \][/tex]
This equation is not satisfied.
2. Second equation:
[tex]\[ 4x - 2y = -27 \][/tex]
Substituting [tex]\( x = 5 \)[/tex] and [tex]\( y = -4 \)[/tex]:
[tex]\[ 4(5) - 2(-4) = 20 + 8 = 28 \neq -27 \][/tex]
This equation is not satisfied.
3. Third equation:
[tex]\[ 2x + 4y = 10 \][/tex]
Substituting [tex]\( x = 5 \)[/tex] and [tex]\( y = -4 \)[/tex]:
[tex]\[ 2(5) + 4(-4) = 10 - 16 = -6 \neq 10 \][/tex]
This equation is not satisfied.
Since none of the equations are satisfied, the solution [tex]\( x = 5 \)[/tex] and [tex]\( y = -4 \)[/tex] does not satisfy system a.
### Linear System b:
1. First equation:
[tex]\[ 2x + 3y = 5 \][/tex]
Substituting [tex]\( x = 5 \)[/tex] and [tex]\( y = -4 \)[/tex]:
[tex]\[ 2(5) + 3(-4) = 10 - 12 = -2 \neq 5 \][/tex]
This equation is not satisfied.
2. Second equation:
[tex]\[ 3x + y = 11 \][/tex]
Substituting [tex]\( x = 5 \)[/tex] and [tex]\( y = -4 \)[/tex]:
[tex]\[ 3(5) + (-4) = 15 - 4 = 11 \][/tex]
This equation is satisfied.
3. Third equation:
[tex]\[ -2x + y = 11 \][/tex]
Substituting [tex]\( x = 5 \)[/tex] and [tex]\( y = -4 \)[/tex]:
[tex]\[ -2(5) + (-4) = -10 - 4 = -14 \neq 11 \][/tex]
This equation is not satisfied.
4. Fourth equation:
[tex]\[ -2x + 4y = -26 \][/tex]
Substituting [tex]\( x = 5 \)[/tex] and [tex]\( y = -4 \)[/tex]:
[tex]\[ -2(5) + 4(-4) = -10 - 16 = -26 \][/tex]
This equation is satisfied.
However, since not all equations in system b are satisfied, the solution [tex]\( x = 5 \)[/tex] and [tex]\( y = -4 \)[/tex] does not satisfy system b either.
### Conclusion
The solution [tex]\( x = 5 \)[/tex] and [tex]\( y = -4 \)[/tex] does not satisfy either of the given linear systems. Therefore, neither system a nor system b has [tex]\( x = 5 \)[/tex] and [tex]\( y = -4 \)[/tex] as a solution.
### Linear System a:
1. First equation:
[tex]\[ x + 3y = 12 \][/tex]
Substituting [tex]\( x = 5 \)[/tex] and [tex]\( y = -4 \)[/tex]:
[tex]\[ 5 + 3(-4) = 5 - 12 = -7 \neq 12 \][/tex]
This equation is not satisfied.
2. Second equation:
[tex]\[ 4x - 2y = -27 \][/tex]
Substituting [tex]\( x = 5 \)[/tex] and [tex]\( y = -4 \)[/tex]:
[tex]\[ 4(5) - 2(-4) = 20 + 8 = 28 \neq -27 \][/tex]
This equation is not satisfied.
3. Third equation:
[tex]\[ 2x + 4y = 10 \][/tex]
Substituting [tex]\( x = 5 \)[/tex] and [tex]\( y = -4 \)[/tex]:
[tex]\[ 2(5) + 4(-4) = 10 - 16 = -6 \neq 10 \][/tex]
This equation is not satisfied.
Since none of the equations are satisfied, the solution [tex]\( x = 5 \)[/tex] and [tex]\( y = -4 \)[/tex] does not satisfy system a.
### Linear System b:
1. First equation:
[tex]\[ 2x + 3y = 5 \][/tex]
Substituting [tex]\( x = 5 \)[/tex] and [tex]\( y = -4 \)[/tex]:
[tex]\[ 2(5) + 3(-4) = 10 - 12 = -2 \neq 5 \][/tex]
This equation is not satisfied.
2. Second equation:
[tex]\[ 3x + y = 11 \][/tex]
Substituting [tex]\( x = 5 \)[/tex] and [tex]\( y = -4 \)[/tex]:
[tex]\[ 3(5) + (-4) = 15 - 4 = 11 \][/tex]
This equation is satisfied.
3. Third equation:
[tex]\[ -2x + y = 11 \][/tex]
Substituting [tex]\( x = 5 \)[/tex] and [tex]\( y = -4 \)[/tex]:
[tex]\[ -2(5) + (-4) = -10 - 4 = -14 \neq 11 \][/tex]
This equation is not satisfied.
4. Fourth equation:
[tex]\[ -2x + 4y = -26 \][/tex]
Substituting [tex]\( x = 5 \)[/tex] and [tex]\( y = -4 \)[/tex]:
[tex]\[ -2(5) + 4(-4) = -10 - 16 = -26 \][/tex]
This equation is satisfied.
However, since not all equations in system b are satisfied, the solution [tex]\( x = 5 \)[/tex] and [tex]\( y = -4 \)[/tex] does not satisfy system b either.
### Conclusion
The solution [tex]\( x = 5 \)[/tex] and [tex]\( y = -4 \)[/tex] does not satisfy either of the given linear systems. Therefore, neither system a nor system b has [tex]\( x = 5 \)[/tex] and [tex]\( y = -4 \)[/tex] as a solution.
