Hello there!
In order to solve this system of equations, you would first need to make one equation have a negative variable.
2a + 3b = -1
(3a + 5b = -2) -1
2a + 3b = -1
-3a - 5b = 2
Now, we have to modify the equations for it to be able to cancel out a variable. To do this, you would need to look at the least common multiple.
2: 2, 4, 6, 8, 10
3: 3, 6, 9, 12, 15
One least common multiple would be 6 so let's modify the equation!
(2a + 3b = -1) 3
(-3a - 5b = 2) 2
6a + 9b = -3
-6a - 10b = 4
Now, let's cancel out the variable and solve for b.
9b = -3
-10b = 4
-1b = 1
b = -1
In order to solve for a, just plug in the b into one equation and solve for a.
2a + 3(-1) = -1
2a - 3 = -1
2a = 2
a = 1
The answer would be a = 1 and b = -1
