(-3, 0) is a solution to given equation
(-6, -1) is a solution to given equation
Solution:
Given that equation is:
[tex]y = \frac{1}{3}x + 1[/tex]
(-3, 0)
Substitute x = -3 and y = 0 in given equation
[tex]0 = \frac{1}{3} \times -3 + 1\\\\0 = -1 + 1\\\\0 = 0[/tex]
Thus (-3, 0) is a solution to given equation
(-9, -1)
Substitute x = -9 and y = -1 in given equation
[tex]-1 = \frac{1}{3} \times -9 + 1\\\\-1 = -3 + 1\\\\-1 \neq -2[/tex]
Thus (-9, -1) is not a solution to given equation
(-6, -1)
Substitute x = -6 and y = -1 in given equation
[tex]-1 = \frac{1}{3} \times - 6 + 1\\\\-1 = -2 + 1\\\\-1 = -1[/tex]
Thus (-6, -1) is a solution to given equation
