+983 Hey bigguy1989!
To solve for A and B, we plug in our two values for x and y to set up the systems of equations.
When x = 1 and y = 3: (1,3), we get: 3A = B + 1
When x = 5 and y = 13: (5,13), we get: 13A = 5B + 1
We obtain the systems:
3A = B + 1
13A = 5B + 1
To solve the systems, we multiply the top one by 5, and we get:
3A = B + 1 \(\Rightarrow\) 15A = 5B + 5
Now we can subtract first equation from second equation so the B's cancel out.
15A - 13A = 5B - 5B + 5 - 1
2A = 4
A = 2 \(\Rightarrow\) B = 5
The line becomes 2y = 5x + 1
In standard form: 2y - 5x = 1
In slope-intercept: y = 2.5x + 0.5
I hope this helped,
Gavin
