Answer :
To find the vertices of the triangle formed by the intersections of the lines given by the equations [tex]\(3x + y = 7\)[/tex], [tex]\(3y - x = 1\)[/tex], and [tex]\(7y + x = -11\)[/tex], we need to find the points where these lines intersect. We will solve the equations pairwise to find these intersection points.
Step-by-Step Solution:
1. Intersection of the lines [tex]\(3x + y = 7\)[/tex] and [tex]\(3y - x = 1\)[/tex]:
- Write down the equations:
[tex]\[ 3x + y = 7 \quad \text{(Equation 1)} \][/tex]
[tex]\[ 3y - x = 1 \quad \text{(Equation 2)} \][/tex]
- Solve the system of equations:
- First, solve Equation 2 for [tex]\(x\)[/tex]:
[tex]\[ x = 3y - 1 \][/tex]
- Substitute [tex]\(x = 3y - 1\)[/tex] into Equation 1:
[tex]\[ 3(3y - 1) + y = 7 \][/tex]
[tex]\[ 9y - 3 + y = 7 \][/tex]
[tex]\[ 10y - 3 = 7 \][/tex]
[tex]\[ 10y = 10 \][/tex]
[tex]\[ y = 1 \][/tex]
- Substitute [tex]\(y = 1\)[/tex] back into [tex]\(x = 3y - 1\)[/tex]:
[tex]\[ x = 3(1) - 1 = 2 \][/tex]
- So, the intersection point (vertex) for these two lines is:
[tex]\[ (x, y) = (2, 1) \][/tex]
2. Intersection of the lines [tex]\(3x + y = 7\)[/tex] and [tex]\(7y + x = -11\)[/tex]:
- Write down the equations:
[tex]\[ 3x + y = 7 \quad \text{(Equation 1)} \][/tex]
[tex]\[ 7y + x = -11 \quad \text{(Equation 3)} \][/tex]
- Solve the system of equations:
- First, solve Equation 3 for [tex]\(x\)[/tex]:
[tex]\[ x = -11 - 7y \][/tex]
- Substitute [tex]\(x = -11 - 7y\)[/tex] into Equation 1:
[tex]\[ 3(-11 - 7y) + y = 7 \][/tex]
[tex]\[ -33 - 21y + y = 7 \][/tex]
[tex]\[ -33 - 20y = 7 \][/tex]
[tex]\[ -20y = 40 \][/tex]
[tex]\[ y = -2 \][/tex]
- Substitute [tex]\(y = -2\)[/tex] back into [tex]\(x = -11 - 7y\)[/tex]:
[tex]\[ x = -11 - 7(-2) = -11 + 14 = 3 \][/tex]
- So, the intersection point (vertex) for these two lines is:
[tex]\[ (x, y) = (3, -2) \][/tex]
3. Intersection of the lines [tex]\(3y - x = 1\)[/tex] and [tex]\(7y + x = -11\)[/tex]:
- Write down the equations:
[tex]\[ 3y - x = 1 \quad \text{(Equation 2)} \][/tex]
[tex]\[ 7y + x = -11 \quad \text{(Equation 3)} \][/tex]
- Solve the system of equations:
- First, add Equation 2 and Equation 3 to eliminate [tex]\(x\)[/tex]:
[tex]\[ 3y - x + 7y + x = 1 - 11 \][/tex]
[tex]\[ 10y = -10 \][/tex]
[tex]\[ y = -1 \][/tex]
- Substitute [tex]\(y = -1\)[/tex] into Equation 3:
[tex]\[ 7(-1) + x = -11 \][/tex]
[tex]\[ -7 + x = -11 \][/tex]
[tex]\[ x = -4 \][/tex]
- So, the intersection point (vertex) for these two lines is:
[tex]\[ (x, y) = (-4, -1) \][/tex]
Vertices of the Triangle:
The vertices of the triangle formed by the given lines are:
[tex]\[ (2, 1), (3, -2), \text{and} (-4, -1) \][/tex]
Step-by-Step Solution:
1. Intersection of the lines [tex]\(3x + y = 7\)[/tex] and [tex]\(3y - x = 1\)[/tex]:
- Write down the equations:
[tex]\[ 3x + y = 7 \quad \text{(Equation 1)} \][/tex]
[tex]\[ 3y - x = 1 \quad \text{(Equation 2)} \][/tex]
- Solve the system of equations:
- First, solve Equation 2 for [tex]\(x\)[/tex]:
[tex]\[ x = 3y - 1 \][/tex]
- Substitute [tex]\(x = 3y - 1\)[/tex] into Equation 1:
[tex]\[ 3(3y - 1) + y = 7 \][/tex]
[tex]\[ 9y - 3 + y = 7 \][/tex]
[tex]\[ 10y - 3 = 7 \][/tex]
[tex]\[ 10y = 10 \][/tex]
[tex]\[ y = 1 \][/tex]
- Substitute [tex]\(y = 1\)[/tex] back into [tex]\(x = 3y - 1\)[/tex]:
[tex]\[ x = 3(1) - 1 = 2 \][/tex]
- So, the intersection point (vertex) for these two lines is:
[tex]\[ (x, y) = (2, 1) \][/tex]
2. Intersection of the lines [tex]\(3x + y = 7\)[/tex] and [tex]\(7y + x = -11\)[/tex]:
- Write down the equations:
[tex]\[ 3x + y = 7 \quad \text{(Equation 1)} \][/tex]
[tex]\[ 7y + x = -11 \quad \text{(Equation 3)} \][/tex]
- Solve the system of equations:
- First, solve Equation 3 for [tex]\(x\)[/tex]:
[tex]\[ x = -11 - 7y \][/tex]
- Substitute [tex]\(x = -11 - 7y\)[/tex] into Equation 1:
[tex]\[ 3(-11 - 7y) + y = 7 \][/tex]
[tex]\[ -33 - 21y + y = 7 \][/tex]
[tex]\[ -33 - 20y = 7 \][/tex]
[tex]\[ -20y = 40 \][/tex]
[tex]\[ y = -2 \][/tex]
- Substitute [tex]\(y = -2\)[/tex] back into [tex]\(x = -11 - 7y\)[/tex]:
[tex]\[ x = -11 - 7(-2) = -11 + 14 = 3 \][/tex]
- So, the intersection point (vertex) for these two lines is:
[tex]\[ (x, y) = (3, -2) \][/tex]
3. Intersection of the lines [tex]\(3y - x = 1\)[/tex] and [tex]\(7y + x = -11\)[/tex]:
- Write down the equations:
[tex]\[ 3y - x = 1 \quad \text{(Equation 2)} \][/tex]
[tex]\[ 7y + x = -11 \quad \text{(Equation 3)} \][/tex]
- Solve the system of equations:
- First, add Equation 2 and Equation 3 to eliminate [tex]\(x\)[/tex]:
[tex]\[ 3y - x + 7y + x = 1 - 11 \][/tex]
[tex]\[ 10y = -10 \][/tex]
[tex]\[ y = -1 \][/tex]
- Substitute [tex]\(y = -1\)[/tex] into Equation 3:
[tex]\[ 7(-1) + x = -11 \][/tex]
[tex]\[ -7 + x = -11 \][/tex]
[tex]\[ x = -4 \][/tex]
- So, the intersection point (vertex) for these two lines is:
[tex]\[ (x, y) = (-4, -1) \][/tex]
Vertices of the Triangle:
The vertices of the triangle formed by the given lines are:
[tex]\[ (2, 1), (3, -2), \text{and} (-4, -1) \][/tex]
