To determine which ordered pairs [tex]\((x, y)\)[/tex] are solutions to the linear equation [tex]\(2x + 3y = 6\)[/tex], we need to substitute each pair into the equation and check if the equation holds true.
Let's evaluate each option one by one:
A. [tex]\((0, 2)\)[/tex]:
Substitute [tex]\(x = 0\)[/tex] and [tex]\(y = 2\)[/tex]:
[tex]\[2(0) + 3(2) = 0 + 6 = 6\][/tex]
Since [tex]\(6 = 6\)[/tex], the pair [tex]\((0, 2)\)[/tex] is a solution.
B. [tex]\((0, 6)\)[/tex]:
Substitute [tex]\(x = 0\)[/tex] and [tex]\(y = 6\)[/tex]:
[tex]\[2(0) + 3(6) = 0 + 18 = 18\][/tex]
Since [tex]\(18 \ne 6\)[/tex], the pair [tex]\((0, 6)\)[/tex] is not a solution.
C. [tex]\((2, 3)\)[/tex]:
Substitute [tex]\(x = 2\)[/tex] and [tex]\(y = 3\)[/tex]:
[tex]\[2(2) + 3(3) = 4 + 9 = 13\][/tex]
Since [tex]\(13 \ne 6\)[/tex], the pair [tex]\((2, 3)\)[/tex] is not a solution.
D. [tex]\((3, -2)\)[/tex]:
Substitute [tex]\(x = 3\)[/tex] and [tex]\(y = -2\)[/tex]:
[tex]\[2(3) + 3(-2) = 6 - 6 = 0\][/tex]
Since [tex]\(0 \ne 6\)[/tex], the pair [tex]\((3, -2)\)[/tex] is not a solution.
E. [tex]\((3, 0)\)[/tex]:
Substitute [tex]\(x = 3\)[/tex] and [tex]\(y = 0\)[/tex]:
[tex]\[2(3) + 3(0) = 6 + 0 = 6\][/tex]
Since [tex]\(6 = 6\)[/tex], the pair [tex]\((3, 0)\)[/tex] is a solution.
F. [tex]\((6, -2)\)[/tex]:
Substitute [tex]\(x = 6\)[/tex] and [tex]\(y = -2\)[/tex]:
[tex]\[2(6) + 3(-2) = 12 - 6 = 6\][/tex]
Since [tex]\(6 = 6\)[/tex], the pair [tex]\((6, -2)\)[/tex] is a solution.
So, the ordered pairs that satisfy the equation [tex]\(2x + 3y = 6\)[/tex] are:
- [tex]\((0, 2)\)[/tex]
- [tex]\((3, 0)\)[/tex]
- [tex]\((6, -2)\)[/tex]
Thus, the correct choices are:
A. [tex]\((0, 2)\)[/tex]
E. [tex]\((3, 0)\)[/tex]
F. [tex]\((6, -2)\)[/tex]
