Question 249047
PROBLEM 1

Let n = the first integer, so n+1 is the next integer.  Now translate the word problem into an algebraic expresion.  So, (n+(n+1))^2=361.  Take the square root of both sides of the equation, so, n+(n+1)=19 and n+(n+1)=-19.

Case 1 yields the following linear equation 2n+1=19.  Solving this linear equation results in the value n = (19-1)/2=9. The pair is 9 and 10 

Case 2 yields the folowing linear equation n+(n+1)=-19.  Solving this linear equation results in the value for n = (-19-1)/2=-10.  The pair is then -10 
and -9.

Problem 2 

Let n = the first even integer.  So n = 2k for some k. The next even integer is then n+2 = 2k+2.  Therefore, the words of the problem can be translated into the following algebraic equation  2k*(2k+2)=288.  Expanding this equation we have 4k^2+4k-288=0.  Dividing both sides by 4 produces k^2+k-72=0.  This equation can be solved for k by factoring. k^2+k-72=(k+9)(k-8)=0.  Therefore, 
k=-9 and k=8.  So,  the first pair of integers is n=2*k = -18 and the second integer is -16.

The second pair of integers is 2*8=16 and 18.

It is important to recognize the pairs of solutions for each question.

Good luck with your study of math.

Larry