No
It is an equation in three unknown variables.
5b = 21 + 4b 5b - 4b = 21 + 4b - 4b b = 21
a + b = 21 so a = 21 - b; 3a = 4b so 3(21 - b) = 4b ie 63 - 3b = 4b so 7b = 63 making b = 9 and a = 12
2a-6a-a=b+4b-4b -5a=5b-4b -5a=1b a=-5 b=1
No
12b2 - 8b = 0 4b(3b-2) = 0 4b = 0 and 3b-2 = 0 4b = 0 and 3b = 2 b = 0 and 2/3
b^3 - 25b^2 - 4b + 100 = 0 b = 25 or b = -2
positive 24b squared
It is an equation in three unknown variables.
This is not a difficult factorisation, as it is merely a normal factorisation with "a" stuck on the end of each number. 25a(2) - 20ab + 4b(2) (2b-5a)(2b-5a) This is: 2bx2b=4b(2) 2bx-5a=-10ab (twice) -5ax-5a=25a
2a*4b*(-3c) = ? 2 * 4 * -3 * a * b * c = ? -24abc
It is: 12b2
g=4b
2a + 4b - 52a + 4b - 52a + 4b - 52a + 4b - 5
8b - 4b = 4b
5b = 21 + 4b 5b - 4b = 21 + 4b - 4b b = 21