4a + 3a = (4 + 3)a = 7a
10a − 3 − 4a
-7a + 3a = -4a
14a - 4a + 5a - 3a = 12a
3A or 4A schools are about the size of the students. 4A schools have more students than a 3AM school
4a + 3a = (4 + 3)a = 7a
Add them together: 3a+4a-a = 6a
10a + 3a - 4a = 9a
4a+3a-4 = 10 4a+3a = 10+4 7a = 14 a = 2
"a" plus "3a" is equal to 4a. (3a+a=4a). You already have 3 "a's" and then you add another "a."
10a − 3 − 4a
-7a + 3a = -4a
3a + 11 = 7a - 13 subtract 3a from each side 3a - 3a + 11 = 7a - 3a - 13 11 = 4a - 13 add 13 to each side 11 + 13 = 4a - 13 + 13 24 = 4a divide each sides integer by 4 24/4 = (4/4)a 6 = a -------------check in original equation 3(6) + 11 = 7(6) - 13 18 + 11 = 42 - 13 29 = 29 ----------------checks out
3a + 15 = 4a Subtract 3a from both sides 15 = a So a is 15.
14a - 4a + 5a - 3a = 12a
3a-9 = 6 3a = 6+9 3a = 15 a = 15/3 a = 5 Substitute a = 5 into 4a-1. 4a-1 =4(5)-1 =20-1 =19 Therefore, the value of 4a-1 is 19.
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