Find four different solutions of the equation $x + 2y = 6.$
(N/A) To find solutions for the linear equation $x + 2y = 6$,we can assign arbitrary values to one variable and solve for the other.
$1$. Let $x = 0$. Then $0 + 2y = 6$,which gives $2y = 6$,so $y = 3$. Thus,$(0, 3)$ is a solution.
$2$. Let $y = 0$. Then $x + 2(0) = 6$,which gives $x = 6$. Thus,$(6, 0)$ is a solution.
$3$. Let $x = 2$. Then $2 + 2y = 6$,which gives $2y = 4$,so $y = 2$. Thus,$(2, 2)$ is a solution.
$4$. Let $y = 1$. Then $x + 2(1) = 6$,which gives $x + 2 = 6$,so $x = 4$. Thus,$(4, 1)$ is a solution.
Therefore,four different solutions of the equation are $(0, 3)$,$(6, 0)$,$(2, 2)$,and $(4, 1)$.
$1$. Let $x = 0$. Then $0 + 2y = 6$,which gives $2y = 6$,so $y = 3$. Thus,$(0, 3)$ is a solution.
$2$. Let $y = 0$. Then $x + 2(0) = 6$,which gives $x = 6$. Thus,$(6, 0)$ is a solution.
$3$. Let $x = 2$. Then $2 + 2y = 6$,which gives $2y = 4$,so $y = 2$. Thus,$(2, 2)$ is a solution.
$4$. Let $y = 1$. Then $x + 2(1) = 6$,which gives $x + 2 = 6$,so $x = 4$. Thus,$(4, 1)$ is a solution.
Therefore,four different solutions of the equation are $(0, 3)$,$(6, 0)$,$(2, 2)$,and $(4, 1)$.
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