Identify the point of intersection listed below each system of linear equations that is the solution of that system.
2x+5y=10
x-3y=-6
(5,0);(0,2); (3,1)
2x+5y=10
x-3y=-6
Multiply the 2nd equation by -2 so that the x's cancel.
2x+5y=10
-2x+6y=12
Add the lines together.
11y=22
Solve for y, replace that in either original equation to solve for x.
To find the point of intersection for each system of linear equations, we can substitute the given points into the equations and see which points satisfy both equations.
1. Point (5,0):
Substituting x = 5 and y = 0 into the equations:
2(5) + 5(0) = 10 => 10 = 10 (satisfied)
5 - 3(0) = -6 => 5 = -6 (not satisfied)
Therefore, the point (5,0) is not the solution for the system of equations.
2. Point (0,2):
Substituting x = 0 and y = 2 into the equations:
2(0) + 5(2) = 10 => 10 = 10 (satisfied)
0 - 3(2) = -6 => -6 = -6 (satisfied)
Therefore, the point (0,2) is the solution for the system of equations.
3. Point (3,1):
Substituting x = 3 and y = 1 into the equations:
2(3) + 5(1) = 10 => 11 = 10 (not satisfied)
3 - 3(1) = -6 => 0 = -6 (not satisfied)
Therefore, the point (3,1) is not the solution for the system of equations.
Hence, the point of intersection listed under the given system of linear equations that is the solution is (0,2).
To find the point of intersection for each system of linear equations, we need to solve the equations simultaneously.
Let's solve the first system of linear equations:
1. Start with the equations:
2x + 5y = 10 ---(equation 1)
x - 3y = -6 ---(equation 2)
2. We will solve this system of equations using the method of substitution. Solve equation 2 for x:
x = 3y - 6
3. Substitute the value of x from equation 2 into equation 1:
2(3y - 6) + 5y = 10
6y - 12 + 5y = 10
11y - 12 = 10
4. Add 12 to both sides of the equation:
11y = 22
5. Divide both sides of the equation by 11:
y = 2
6. Substitute the value of y back into equation 2 to find x:
x = 3(2) - 6
x = 0
Therefore, the point of intersection for the first system of equations is (0, 2).
Now let's solve the second system of linear equations:
1. Start with the equations:
2x + 5y = 10 ---(equation 1)
x - 3y = -6 ---(equation 2)
2. We will solve this system of equations using the method of substitution. Solve equation 2 for x:
x = 3y - 6
3. Substitute the value of x from equation 2 into equation 1:
2(3y - 6) + 5y = 10
6y - 12 + 5y = 10
11y - 12 = 10
4. Add 12 to both sides of the equation:
11y = 22
5. Divide both sides of the equation by 11:
y = 2
6. Substitute the value of y back into equation 2 to find x:
x = 3(2) - 6
x = 0
Therefore, the point of intersection for the second system of equations is (0, 2).
To find the point of intersection for the third system of linear equations, we need to use the same steps again. However, instead of going through the process, let's directly substitute the given points into the equations and check which one satisfies both equations.
Let's substitute each of the given points into the equations and see which one satisfies both:
For point (5, 0):
Substituting into equation 1: 2(5) + 5(0) = 10, which is true.
Substituting into equation 2: 5 - 3(0) = -6, which is false.
For point (0, 2):
Substituting into equation 1: 2(0) + 5(2) = 10, which is true.
Substituting into equation 2: 0 - 3(2) = -6, which is true.
For point (3, 1):
Substituting into equation 1: 2(3) + 5(1) = 10, which is true.
Substituting into equation 2: 3 - 3(1) = -6, which is false.
Therefore, the point of intersection for the third system of equations is (0, 2) only.
In summary:
- The point of intersection for the first system of equations is (0, 2).
- The point of intersection for the second system of equations is (0, 2).
- The point of intersection for the third system of equations is (0, 2) only.