If a, b and c are positive integers for which (a, b,c) is the only ordered triple such that 6a +9b + 20c = 61, then what is
a+b+c?

If a, b, and c are all positive integers and 6a + 9b + 20c = 61, then right away you know c < 3, because if c = 3, then 20c = 60, and there are no solutions for a and b.

So consider two cases:

• If c = 2, then we're left with 6a + 9b = 21.

• If c = 1, then 6a + 9b = 41.

We have gcd(6, 9) = 3, but 41 is prime and 3 does not divide 41. So we throw out the second case.

This leaves us with 6a + 9b = 21. Notice that if a = b = 1, then 6a + 9b = 15. Add 6 to both sides to get 21. So a = 2, b = 1, and c = 2, and hence a + b + c = 5.

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step-by-step explanation:

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?