Is (-2,6)a solution to the system of these liner equations x+ 2y=10 and 3x+y=0? why
To determine if (-2, 6) is a solution to the system of linear equations, we substitute the values of x and y into each equation and check if the equations hold true.
For the first equation:
x + 2y = 10
Substituting x = -2 and y = 6:
(-2) + 2(6) = 10
-2 + 12 = 10
10 = 10
This equation is true, so (-2, 6) satisfies the first equation.
For the second equation:
3x + y = 0
Substituting x = -2 and y = 6:
3(-2) + 6 = 0
-6 + 6 = 0
0 = 0
This equation is also true, so (-2, 6) satisfies the second equation.
Since (-2, 6) satisfies both equations, it is a solution to the system of linear equations.
To check if the point (-2, 6) is a solution to the system of linear equations x + 2y = 10 and 3x + y = 0, we need to substitute the values of x and y into both equations and check if the equations are satisfied.
For the first equation x + 2y = 10:
Substituting x = -2 and y = 6:
-2 + 2(6) = 10
-2 + 12 = 10
10 = 10
The equation is satisfied.
For the second equation 3x + y = 0:
Substituting x = -2 and y = 6:
3(-2) + 6 = 0
-6 + 6 = 0
0 = 0
The equation is satisfied.
Since both equations are satisfied, the point (-2, 6) is indeed a solution to the system of linear equations.
To determine whether (-2, 6) is a solution to the system of linear equations, we need to substitute the values of x and y from (-2, 6) into both equations and check if the resulting equations are true.
1. x + 2y = 10
Substitute x = -2 and y = 6 into the equation:
-2 + 2(6) = 10
-2 + 12 = 10
10 = 10
The equation is true for the given values of x and y.
2. 3x + y = 0
Substitute x = -2 and y = 6 into the equation:
3(-2) + 6 = 0
-6 + 6 = 0
0 = 0
The equation is also true for the given values of x and y.
As both equations hold true for x = -2 and y = 6, we can conclude that (-2, 6) is indeed a solution to the system of linear equations.