Word Problem

Question

The product of two consecutive integers is 89 more than their sum. Find the integers.

Answer 1

Hello John,

Let n = the first integer
Let n + 1 = the next consecutive integer


Now let's set up an equation, shall we? 

n(n+1) = 89 + n + (n + 1)

Let's simplify, shall we?  

n² + n = 90 + 2n

Move all of your terms to one side and do one of two things: (a) factor out for n, or (b) use the quadratic formula to solve for n. 

Let's move all of our terms to one side. 

n² - n - 90 = 0

You can use a quadratic - once again to solve this - however the way this has worked allows us to factor quite easily, doesn't it? To factor, all we need do is find two numbers that when multiplied together equal -90 and when added give us a -1, right? 

So, let's try this because my hunch is we have two such numbers, right? 

(n - 10) + (n + 9) = 0 

You now have two possible solutions when solving the above, correct? 

n - 10 = 0 
n = 10 

n + 9 = 0 
n = -9 


Now, do some substitution, to proof our work, don't we? Remember that integers can be both positive or negative.  

When n = -9

Original equation: 

n(n+1) = 89 + n + (n + 1)

-9 * (-9 + 1) = 89 - 9 - 8

81 - 9 = 89 - 9 - 8

72 = 89 - 9 -8   

72 = 72  

When n = 10:

n(n+1) = 89 + n + (n + 1)

10 * (10 +1) = 89 + 10 + 11

100 + 10 = 89 + 10 + 11

110 = 110


Therefore, you have your two solutions: {-9, -8} or {10, 11} 


Let me know if this made sense to you. Have a great day at school tomorrow. 

Answer 2

The two integers are "n" and "n + 1."
n(n + 1) = n + n + 1 + 89
n^2 + n = 2n + 90
n^2 - n - 90 = 0

 

solve for n,

 

n^2 - 10n + 9n - 90 = 0

n(n - 10) + 9(n - 10) = 0

(n-10)(n+9) =0

n = -9  and n = 10

 

so consecutive integers are

n = -9 and -9+1 = -8

n = 10 and 10 +1 = 11