Question 551834
"The sum of three time a number and 7 more than the number is the difference between -11 and twice the number" translates to {{{3n + n + 7=-11-2n}}}



{{{3n+n+7=-11-2n}}} Start with the given equation.



{{{4n+7=-11-2n}}} Combine like terms on the left side.



{{{4n=-11-2n-7}}} Subtract {{{7}}} from both sides.



{{{4n+2n=-11-7}}} Add {{{2n}}} to both sides.



{{{6n=-11-7}}} Combine like terms on the left side.



{{{6n=-18}}} Combine like terms on the right side.



{{{n=(-18)/(6)}}} Divide both sides by {{{6}}} to isolate {{{n}}}.



{{{n=-3}}} Reduce.



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Answer:


So the solution is {{{n=-3}}}