What is 16

What is the sum of three consecutive even integers such that six more than twice the second is 2/3 of the first increased by 3/2 of the third?

Correct answer:

s =  78

Step-by-step explanation:

$$\begin{gathered} 6+2b=2/3\cdot a+3/2\cdot c \\ b=a+2 \\ c=b+2 \\ 6+2\cdot b=2/3\cdot a+3/2\cdot c \\ b=a+2 \\ c=b+2 \\ 4a-12b+9c=36 \\ a-b=-2 \\ b-c=-2 \\ Row2-\frac{1}{4}\cdot Row1\to Row2 \\ 4a-12b+9c=36 \\ 2b-2.25c=-11 \\ b-c=-2 \\ Row3-\frac{1}{2}\cdot Row2\to Row3 \\ 4a-12b+9c=36 \\ 2b-2.25c=-11 \\ 0.125c=3.5 \\ c=\frac{3.5}{0.125}=28 \\ b=\frac{-11+2.25c}{2}=\frac{-11+2.25\cdot 28}{2}=26 \\ a=\frac{36+12b-9c}{4}=\frac{36+12\cdot 26-9\cdot 28}{4}=24 \\ a=24 \\ b=26 \\ c=28 \\ s=a+b+c=24+26+28=78 \end{gathered}$$

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