document.write( "Question 91017: Find four consecutive intergers such that twice the sum of the two greater intergers exceeds three times the first by 91. \n" ); document.write( "
Algebra.Com's Answer #66068 by checkley75(3666)![]() ![]() ![]() You can put this solution on YOUR website! THE 4 CONSECUTIVE NUMBERS ARE X, X+1, X+2 & X+3 \n" ); document.write( "2(X+2+X+3)=3(X+X+1)+91 \n" ); document.write( "2(2X+5)=3(2X+1)+91 \n" ); document.write( "4X+10=6X+3+91 \n" ); document.write( "4X-6X=94-10 \n" ); document.write( "-2X=84 \n" ); document.write( "x=84/-2 \n" ); document.write( "x=-42 ANSWER FOR THE SMALLEST NUMBER. \n" ); document.write( "-42+1=-41 \n" ); document.write( "-42+2=-40 \n" ); document.write( "-42+3=-39 \n" ); document.write( "PROOF \n" ); document.write( "2(-40-39)=3(-42-41)+91 \n" ); document.write( "2*-79=3*-83+91 \n" ); document.write( "-158=-249+91 \n" ); document.write( "-158=-158 \n" ); document.write( " |