For this type of problem, we need to substitute each ordered pair to the equations.
If the ordered pair satisfied the equation, then it is the solution for that specific equation.
Let's start with the first equation :
3x + 2y = 6
(0, -6) => 3(0) + 2(-6)=6
0 - 12 = 6
-12 = 6 (False!)
(0, 3) => 3(0) + 2(3) = 6
0 + 6 = 6
6 = 6 (True!)
Therefore the solution for the equation 3x + 2y = 6 is (0, 3)
For the second equation :
-5x + y = -10
(0, -6) => -5(0) + (-6) = -10
0 - 6 = -10
-6 = -10 (False!)
We will not go for (0, 3) since it is already a solution for the first equation.
Next is :
(4, -1) => -5(4) + (-1) = -10
-20 - 1 = -10
-21 = -10 (False!)
(1, -5) => -5(1) + (-5) = -10
-5 - 5 = -10
-10 = -10 (True!)
Therefore the solution for the equation -5x + y = -10 is (1, -5)
For the third equation :
x - 4y = 8
(0, -6) => (0) - 4(-6) = 8
0 + 24 = 8
24 = 8 (False!)
Next is :
(4, -1) => (4) - 4(-1) = 8
4 + 4 = 8
8 = 8 (True!)
Therefore the solution for the equation x - 4y = 8 is (4, -1)
For the last equation :
-6x - 5y = 30
The only option left is :
(0, -6) => -6(0) - 5(-6) = 30
0 + 30 =30
30 = 30 (True!)
Therefore the solution for the equation -6x - 5y = 30 is (0, -6)
Summarizing the answers :
-6x - 5y = 30 => (0, -6)
3x + 2y = 6 => (0, 3)
x - 4y = 8 => (4, -1)
-5x + y = -10 => (1, -5)
