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Let a, b, and c be integers that satisfy 2a+3b=52, 3b+c=41, and bc=60. Find a+b+ c.

 May 11, 2022
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Hi, hope this helps:
                                                         \(2a+3b=52 \)         (1)

                                                         \(3b+c=41\)           (2)

                                                            \(bc=60\)                (3)

Consider (2) and (3):
From (3): \(c=\dfrac{60}{b}\), which when we substitute into (2) gives:

\(3b+\dfrac{60}{b}=41\) Multiply by b

\(3b^2-41b+60=0\)

This can be factorised or you can use the Quadratic formula.

\(3b^2-41b+60=0 \iff (3b-5)(b-12)=0\)

Given that a,b,c are integers, then we only accept \(b=12\)

Now from (3): \(12*c=60 \implies c=5\)

Then from (1): \(2a+3(12)=52 \implies a=8\)

So, \(a+b+c=8+12+5=25\)

 May 11, 2022

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