Here you have a system of 2 equations and 2 unknowns. An easy way to solve this type is to isolate one of the variables.
12a + 2b = 8
2b = 8 - 12a
b =

b = 4 - 6a
Now plug 4 - 6a into equation 1 to solve for a.
a + 5(4 - 6a) = 19
a + 20 - 30a = 19
-29a = -1
a = 1/29 (answer)
Now plug a into the equation for b
b= 4 - 6(1/29)
b= 110/29 (answer)
<span>10 < –2b or 2b + 3 > 11 represents 2 different sets of numbers.
</span><span>10 < –2b can be reduced by dividing both sides by -2; you must then reverse the direction of the < sign: -5 > b, which is the interval b < -5: (-infinity, -5).
</span>2b + 3 > 11 reduces to 2b > 8, which in turn reduces to b > 4: (4, infinity).
It may be helpful to graph these sets.
If you really do mean "<span>10 < –2b or 2b + 3 > 11," then the "solution" is made up of two sub-intervals: b < -5 and b > 4.
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4 and 5 are right, but 2 isn't because consonants are the opposite of vowels. So, you wouldn't use a or e.