Westonci.ca offers fast, accurate answers to your questions. Join our community and get the insights you need now. Explore a wealth of knowledge from professionals across various disciplines on our comprehensive Q&A platform. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.
Sagot :
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.
Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.
