IDNLearn.com: Your destination for reliable and timely answers to any question. Get accurate and timely answers to your queries from our extensive network of experienced professionals.
Sagot :
To find which of the given points are solutions to the system of equations:
[tex]\[ \begin{array}{l} 2x + y = 5 \\ 3y = 15 - 6x \end{array} \][/tex]
we will substitute each of the points [tex]\((x, y)\)[/tex] into both equations and see if both equations are satisfied.
### Checking the point [tex]\((6, -7)\)[/tex]:
1. Substitute [tex]\(x = 6\)[/tex] and [tex]\(y = -7\)[/tex] into the first equation [tex]\(2x + y = 5\)[/tex]:
[tex]\[ 2(6) + (-7) = 12 - 7 = 5 \quad \text{(True)} \][/tex]
2. Substitute [tex]\(x = 6\)[/tex] and [tex]\(y = -7\)[/tex] into the second equation [tex]\(3y = 15 - 6x\)[/tex]:
[tex]\[ 3(-7) = 15 - 6(6) \implies -21 = 15 - 36 \implies -21 = -21 \quad \text{(True)} \][/tex]
Since both equations are satisfied, [tex]\((6, -7)\)[/tex] is a solution.
### Checking the point [tex]\((2, 1)\)[/tex]:
1. Substitute [tex]\(x = 2\)[/tex] and [tex]\(y = 1\)[/tex] into the first equation [tex]\(2x + y = 5\)[/tex]:
[tex]\[ 2(2) + 1 = 4 + 1 = 5 \quad \text{(True)} \][/tex]
2. Substitute [tex]\(x = 2\)[/tex] and [tex]\(y = 1\)[/tex] into the second equation [tex]\(3y = 15 - 6x\)[/tex]:
[tex]\[ 3(1) = 15 - 6(2) \implies 3 = 15 - 12 \implies 3 = 3 \quad \text{(True)} \][/tex]
Since both equations are satisfied, [tex]\((2, 1)\)[/tex] is a solution.
### Checking the point [tex]\((-2, -9)\)[/tex]:
1. Substitute [tex]\(x = -2\)[/tex] and [tex]\(y = -9\)[/tex] into the first equation [tex]\(2x + y = 5\)[/tex]:
[tex]\[ 2(-2) + (-9) = -4 - 9 = -13 \quad \text{(False)} \][/tex]
Since the first equation is not satisfied, we don't need to check the second equation. [tex]\((-2, -9)\)[/tex] is not a solution.
### Checking the point [tex]\((-4, 13)\)[/tex]:
1. Substitute [tex]\(x = -4\)[/tex] and [tex]\(y = 13\)[/tex] into the first equation [tex]\(2x + y = 5\)[/tex]:
[tex]\[ 2(-4) + 13 = -8 + 13 = 5 \quad \text{(True)} \][/tex]
2. Substitute [tex]\(x = -4\)[/tex] and [tex]\(y = 13\)[/tex] into the second equation [tex]\(3y = 15 - 6x\)[/tex]:
[tex]\[ 3(13) = 15 - 6(-4) \implies 39 = 15 + 24 \implies 39 = 39 \quad \text{(True)} \][/tex]
Since both equations are satisfied, [tex]\((-4, 13)\)[/tex] is a solution.
### Conclusion:
The points that are solutions to the given system of equations are:
[tex]\[ (6, -7), (2, 1), \text{and} (-4, 13) \][/tex]
[tex]\[ \begin{array}{l} 2x + y = 5 \\ 3y = 15 - 6x \end{array} \][/tex]
we will substitute each of the points [tex]\((x, y)\)[/tex] into both equations and see if both equations are satisfied.
### Checking the point [tex]\((6, -7)\)[/tex]:
1. Substitute [tex]\(x = 6\)[/tex] and [tex]\(y = -7\)[/tex] into the first equation [tex]\(2x + y = 5\)[/tex]:
[tex]\[ 2(6) + (-7) = 12 - 7 = 5 \quad \text{(True)} \][/tex]
2. Substitute [tex]\(x = 6\)[/tex] and [tex]\(y = -7\)[/tex] into the second equation [tex]\(3y = 15 - 6x\)[/tex]:
[tex]\[ 3(-7) = 15 - 6(6) \implies -21 = 15 - 36 \implies -21 = -21 \quad \text{(True)} \][/tex]
Since both equations are satisfied, [tex]\((6, -7)\)[/tex] is a solution.
### Checking the point [tex]\((2, 1)\)[/tex]:
1. Substitute [tex]\(x = 2\)[/tex] and [tex]\(y = 1\)[/tex] into the first equation [tex]\(2x + y = 5\)[/tex]:
[tex]\[ 2(2) + 1 = 4 + 1 = 5 \quad \text{(True)} \][/tex]
2. Substitute [tex]\(x = 2\)[/tex] and [tex]\(y = 1\)[/tex] into the second equation [tex]\(3y = 15 - 6x\)[/tex]:
[tex]\[ 3(1) = 15 - 6(2) \implies 3 = 15 - 12 \implies 3 = 3 \quad \text{(True)} \][/tex]
Since both equations are satisfied, [tex]\((2, 1)\)[/tex] is a solution.
### Checking the point [tex]\((-2, -9)\)[/tex]:
1. Substitute [tex]\(x = -2\)[/tex] and [tex]\(y = -9\)[/tex] into the first equation [tex]\(2x + y = 5\)[/tex]:
[tex]\[ 2(-2) + (-9) = -4 - 9 = -13 \quad \text{(False)} \][/tex]
Since the first equation is not satisfied, we don't need to check the second equation. [tex]\((-2, -9)\)[/tex] is not a solution.
### Checking the point [tex]\((-4, 13)\)[/tex]:
1. Substitute [tex]\(x = -4\)[/tex] and [tex]\(y = 13\)[/tex] into the first equation [tex]\(2x + y = 5\)[/tex]:
[tex]\[ 2(-4) + 13 = -8 + 13 = 5 \quad \text{(True)} \][/tex]
2. Substitute [tex]\(x = -4\)[/tex] and [tex]\(y = 13\)[/tex] into the second equation [tex]\(3y = 15 - 6x\)[/tex]:
[tex]\[ 3(13) = 15 - 6(-4) \implies 39 = 15 + 24 \implies 39 = 39 \quad \text{(True)} \][/tex]
Since both equations are satisfied, [tex]\((-4, 13)\)[/tex] is a solution.
### Conclusion:
The points that are solutions to the given system of equations are:
[tex]\[ (6, -7), (2, 1), \text{and} (-4, 13) \][/tex]
We appreciate your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. IDNLearn.com is dedicated to providing accurate answers. Thank you for visiting, and see you next time for more solutions.
