3 (2a 5) and 6a+15 Verify informally that the two expressions are equivalent using substitution. (Let a = 4)

Since a = 4, just plug in 4 every time you see a, and solve normally.

3 (2a + 5)

3 (2 • 4 + 5)

3 (8 + 5)

3 (12) = 36

6a + 15

6 • 4 + 15

24 + 15 = 39

Those expressions are not equivalent.