<PRE><B>Question 162820</B>
Find three consecutive even integers
x, (x+2), (x+4)
:
 such that the sum of three times the first integer, one-third the second integer, and eight times the third integer is -58
3x + {{{1/3}}}(x+2) + 8(x+4) = -58
:
3x + {{{1/3}}}x + {{{2/3}}} + 8x + 32 = -58 
:
11{{{1/3}}}x + 32{{{2/3}}} = -58
:
11{{{1/3}}}x = -58 - 32{{{2/3}}}
:
11{{{1/3}}}x = -90{{{2/3}}} 
Multiply equation by 3
34x = -272
x = {{{(-272)/34}}}
x = -8
:
The three numbers: -8, -6, -4
:
You can check solution in the original equation</PRE>