Answer:
81, 82, 83, 84.
Step-by-step explanation:
Let x represent 1st integer. Then the next three consecutive integers would be [tex]x+1,x+2\text{ and }x+3[/tex].
We have been given that twice the sum of the two greater integers exceeds three times the first by 91. We can represent our given information in an equation as:
[tex]2(x+2+x+3)=3x+91[/tex]
Let us solve for x.
[tex]2(2x+5)=3x+91[/tex]
[tex]4x+10=3x+91[/tex]
[tex]4x-3x+10=3x-3x+91[/tex]
[tex]x+10=91[/tex]
[tex]x+10-10=91-10[/tex]
[tex]x=81[/tex]
Therefore, the 1st integer would be 81.
The next three consecutive integers would be:
[tex]81+1=82\\81+2=83\\81+3=84[/tex]
Therefore, our required four consecutive integers are 81, 82, 83, 84.
