5a + 11
-8a + 3a = (-8 + 3)a = (5)a = 5a
9a +8 -2a -3 -5a = 2a +5
It is: 7a+1 when simplified
The three integers, since they are consecutive, can be listed as a, a+1, and a+2. Twice the first is 2a. Three times the third is 3(a+2) = 3a+6. First make a formula of the information given: 2a+(3a+6)= -24 Next, solve the formula: 5a + 6 = -24 Subtract 6 from each side. 5a = -30 Divide each side by 5. a = -6 The three consecutive numbers are -6, -5, and -4.
If 3a+5=5a+3, then: 3a+5 - 3a = 5a+3 - 3a 5=2a+3 5 - 3=2a+3 - 3 2=2a a=1
5a - 5 = 7 + 2a ( subtract 2a from both sides) 3a - 5 = 7 (add 5 to both sides) 3a = 12 ( divide both sides by 3) a = 4
3a + 4b - 5 - 1a + 7b + 3 - 1b 2a + 10b - 5 + 3 2a + 10b - 2
(6a3 - 19a2 + 15a)/(2a - 3) = a(6a2 - 19a + 15)/(2a - 3) = a(2a - 3)(3a -5)/(2a -3) = a(3a - 5)
5a + 11
depends what 'a' =
2a + 4b - 52a + 4b - 52a + 4b - 52a + 4b - 5
(3a+4)/12 - (5/3) = (2a-1)/2Multiply each side by 12:(3a+4) - (20) = 6 (2a-1)Eliminate the parentheses:3a + 4 - 20 = 12a -63a -16 = 12a -6Add 6 to each side:3a -10 = 12aSubtract 3a from each side:-10 = 9aDivide each side by 9:-10/9 = a
= 3 X 2 = 6 = a X a = a2 (a square) = 6a2 + 5 Answer
The problem 3a-1 and 2a-5 is greater by 7. This is a math problem.
-2a + 5 = 1 -2a = -5 + 1 -2a = -4 a = -4/-2 a = 2
2a-5=-3a5a-5=05a=5a=0