-4 + 10 + 6a - 2a + 3
9 + 4a
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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