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