2a - 5b = 1
3a + 5b = 14
Let's solve this system using elimination. Add the equations together to cancel the b-terms, then solve for a.
2a - 5b = 1
3a + 5b = 14
+___________
5a - 0 = 15
5a = 15
a = 3
Next, plug 3 for a into either of the equations and solve for b.
2a - 5b = 1
2(3) - 5b = 1
6 - 5b = 1
- 5b = -5
b = 1
Lastly, check all work by plugging both values for their respective variables.
2a - 5b = 1
2(3) - 5(1) = 1
6 - 5 = 1 -- This is true
3a + 5b = 14
3(3) + 5(1) = 14
9 + 5 = 14 -- This is true
The solution to the system of equations is (3, 1).