6a2
if f(x) = 3x - 10, then whatever is put (substituted) for x in the "f(x)" bit is substituted for x in the "3x - 10" bit. Thus f(2a) = 3(2a) - 10 = 6a - 10.
9a +8 -2a -3 -5a = 2a +5
Combine like terms, adding/subtracting the coefficients: (5a² + 6a + 2) + (7a² - 7a + 5) = (5+7)a² + (6-7)a + (2+5) = 12a² - a + 7 (5a² + 6a + 2) - (7a² - 7a + 5) = (5a² + 6a + 2) + (-7a² + 7a - 5) = (5-7)a² + (6+7)a + (2-5) = -2a² + 13a - 3
9a = 6a + 4 subtracting 6a from each side, 3a = 4 dividing each side by 3, a = 4/3
Since the question is 3(2a), then just write it out. 3(2a) is 2a+2a+2a or 6a.
2a*3 = 8+4a 6a = 8+4a 6a-4a = 8 2a = 8 a = 4
6a-3=8a+11 -3-11=8a-6a -14=2a a=-7
(2a - 3)(3a - 4)
9a +8 -2a -3 -5a = 2a +5
(4, 3)
-8a + 6a + 7 = 1 -2a = 1 - 7 -2a = -6 a = 6/2 a = 3 Check: -8(3) + 6(3) + 7 = 1 -24 + 18 + 7 = 1 1 = 1
I guess you mean: 2a^2+6a Well in this case you take 2a out of the bracket and then you're left with a+3 inside. 2a(a+3a)
Not sure why it is 3a2- if you mean 3*a*2, then: 3*a*2=6a (6a-2a)+5-2(a (or a2)+a)=1 4a+5-(2a+a)=1 4a+5-2a-a=1 Then take the 5 over (remember, the +5 become -5 when transposed): 4a-2a-a=1-5 Solving the equations on both sides gives: a=-4 Is this your answer?
6a2
if f(x) = 3x - 10, then whatever is put (substituted) for x in the "f(x)" bit is substituted for x in the "3x - 10" bit. Thus f(2a) = 3(2a) - 10 = 6a - 10.
If that's -10, the answer is (a + 5)(a - 2). If that's +10, it gets ugly in a hurry.