Question 148699
Let x=one integer; x^2 is the square of the integer
Then x+1=the other integer; (x+1)^2 is the square of the other integer

x+(x+1)=2x+1 is the sum of the integers 10(2x+1)=10 times the sum of the integers;10(2x+1)-9=9 less than 10 times the sum of the integer. So our equation to solve is:

x^2+(x+1)^2=10(2x+1)-9 simplify
x^2+x^2+2x+1=20x+10-9
2x^2+2x+1=20x+1 subtract 20x and also 1 from each side

2x^2+2x+1-1-20x=20x-20x+1-1 collect like terms

2x^2-18x=0  divide each term by 2
x^2-9x=0
x(x-9)=0
x=0; x+1=1
and
x=9; x+1=10

CK
0^2+1^2=10(0+1)-9
1=1
and
9^2+10^2=10(19)-9
81+100=190-9
181=181

Hope this helps---ptaylor