Question 98746
Let x = the first even integer
x+2= second even integer
x+4= third even integer
So we have;
5(x+4)=2(x+x+2)+42
5x+20=2x+2x+4+42
5x+20=4x+46
5x-4x=46-20
x=26
Now that we know x we can get the others;
x=26; 1st integer
x+2=28; 2nd integer
x+4=30; 3rd integer
:)