a) plug in (0,3) for each equation & see if it works:
(1/3)x + y = 3
(1/3)0 + 3 = 3
0+ 3 = 3 so yes (0,3) works for the 1st equation.
2y = (-2/3)x+10
2(3) = (-2/3)0+10
6 = 0+10 so it doesn't work for the 2nd equation.
So no, (0,3) is not a solution to the system of equations.
b) 1/3x + y = 3
We can rewrite that as y = 3 - (1/3)x
2y = (-2/3)x+10
2(3 - (1/3)x) = (-2/3)x + 10
6 - (2/3)x = -(2/3)x + 10
0 = 0
There is no solution. These lines are actually parallel lines.
Here's another way to show that:
1/3x + y = 3 which is the same as y = 3- (1/3)x
2y = (-2/3)x + 10
With the 2nd equation, divide everything by 2:
y = -1/3x + 5
They have the same slope (-1/3) but different y intercepts so they are parallel lines!!!
c) Yes, a system of 2 linear could have infinite number of solutions. If the two linear equations have the same slope & y intercept (meaning they are on top of each other, the same exact equations when you graph them), they would have infinite solutions. You can tell by looking at the equations in pt slope form. Is the slope the same? Is the intercept the same? If both are YES, then it has infinite solutions. You can tell by graphing bc it would be the same line for both.
Here's a very simple example:
y = x + 2
2y = 2x + 4
If you divide the second equaton by 2 it becomes:
y = x + 2 - - - this is the same as the 1st equation (same slope which is 1, same y intercept which is 2), so when you graph these lines, it's the same line & therefore infinite solutions.