If $a:b = 1:3$ and $b$ is a positive, one-digit integer, what is the greatest possible value of $a$?

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Work Shown:

a:b = 1:3

a/b = 1/3

a*3 = b*1 ... cross multiply

3a = b

b = 3a

Since b is a one digit number, this means that the largest it can get is 9

The largest possible value of 'a' is 3

y = 2x

step-by-step explanation:

if you divide each number in the sequence, you'll see that they all have a multiple factor of 2. each number in the series, and any other numbers which may continue, will be multiplied by 2. therefore, y is the next number in the sequence, and x is the current one, so, our formula will be y = 2x.