Find four consecutive even integers so that the sum of the first three is 2 more than twice the fourth

Answer:The numbers are : -20,-18,-16,-14Step-by-step explanation:Let 1st even integer be [tex]=x[/tex]So, next consecutive even integer would be [tex]=x+2[/tex]3rd consecutive even integer would be [tex]=x+4[/tex]4th consecutive even integer would be [tex]=x+6[/tex]Sum of first three integers would be [tex]=x+x+2+x+4=3x+6[/tex]It says sum of the first three is 2 more than twice the fourth,So we have [tex]3x+6=2+4(x+6)[/tex]using distribution to simplify.[tex]3x+6=2+(4\times x)+(4\times6)[/tex][tex]3x+6=2+4x+24[/tex][tex]3x+6=4x+26[/tex]Subtracting [tex]4x[/tex] from both sides[tex]3x+6-4x=4x+26-4x[/tex][tex]-x+6=26[/tex]Subtracting 6 from both sides[tex]-x+6-6=26-6[/tex][tex]-x=20[/tex]∴ [tex]x=-20[/tex]So first integer =-202nd integer  [tex]=-20+2=-18[/tex]3rd integer [tex]=-18+2=-16[/tex]4th integer  [tex]=-16+2=-14[/tex]So, the numbers are : -20,-18,-16,-14