The product of two consecutive numbers is 109 more than there sum. Find the pairs of numbers.

Question

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Answer 1

Dear Eugene,

 

   Let's try some steps here:

 

Note:   "Consecutive numbers" mean that numbers follow each other, for example, 1,2,3.  If I add one to one, I get

 

2.  If I add one to two, 2+1, I get three.

 

Remember that "product" means multiply.    "More"  means add.  "Sum" means add.

 

1.  Let X = one consecutive number.     Then X+1= 2nd consecutive number

 

2.  Take the story problem, and underline or highlight all key words and numbers.  Then change the key words to symbols, for example:

 

                                                      X(X+1) =  X  + (X+1)  +   109

The product of two consecutive numbers is 109 more than their sum.  Find the pairs of numbers.

 

 

Note:  You are looking for the first consecutive number (X) and the second, which is X+1.

 

 

3.      You need to simplify in order to see what is going on and to solve the problem.

 

                                                          X² +X =  2X  +1  +109   

   

Now this is a quadratic equation because of the "X²."  Write it in standard form:  aX² +bX +C=0                   

 

                                                          X²+ 1X -2X-1 -109=0

 

                                                         1X² + -1X-110=0

 

4.  This may be a hard problem to factor.  Therefore, I would suggest using the Quadratic Formula:

 

                                              X=  ( - (b) + /-  √((b)²- 4(a)(c)))/ 2(a)  

 

 

5.  Determine all values for "a, b," and "c."         a=  1,      b=-1,      c= -110

 

 

6.  Substitute values into the Quadratic Formula, and solve.   +/-  means plus or minus.

 

X=( - (-1) +/-  √( (-1)²- 4(1)(-110) )/ 2      =   ( 1+/-  √(1+ 440) )/2  = (1 +/-  √(441) )/2

 

 

 

The last simplification represents two answers for the quadratic:   X= (1  + 21)/2   and    X= (1- 21)/2

 

                                                                                             X= 22/2 = 11     and    X= -20/2= -10    

 

 

7.  Finally, check each answer separately by substituting into the following equation:  X(X+1) = X + (X+1) + 109.

 

If you get  132=132, then your answers are  correct.

 

     Well, Eugine, if you have any further questions, please ask them.

 

                                                                 Thank you,

 

                                                                   Susan

                             

 

 

                             

Answer 2

First, we define variables for the numbers we want to find.

 

Let the first number = x

Let the second number = x + 1

 

 

Using the variables, we can write an equation based on the problem description.

 

x(x + 1) = (x + x + 1) + 109

 

x² + x = 2x + 110

 

 

Solve for x.

 

If we subtract 2x and subtract 110 on both sides of equation, we will get a quadratic equation.

 

x² - x - 110 = 0

 

(x + 10)(x - 11) = 0

 

x = -10      and       x = 11

 

 

These values of x are one of your numbers.  Add 1 to each of these to get the other number.  Thus we have two pairs:

 

Pair 1:       -10 and -9

 

 

Pair 2:        11 and 12

 

 

But wait, we are not done yet.  We still need to verify if both pairs work, or just only one of the pairs.

 

 

Lets test the first pair by plugging them into the equation.

 

-10(-9) = (-10 + (-9)) + 109

 

90 = -19 + 109

 

90 = 90

 

This is okay.

 

 

Test the second pair.

 

11(12) = (11 + 12) + 109

 

132 = 23 + 109

 

132 = 132

 

 

This is also okay.

 

The two pairs are acceptable.

Answer 3

OK, I think the tricky part is the wording because it might seem unclear, but if we break it down carefully it is pretty straight forward.

 

Let's start with "the product of two consecutive numbers." We know that the product means multiplication, but we don't know what are the two numbers that are side by side (consecutive). This makes them variables, but with a twist...instead of x and y, we'll use x for the first number and x+1 for the other number because the other number is 1 more than the value of x. So part of the problem looks like this: (x)•(x+1).

 

The other part of the problem says "109 more than their sum" so whatever the sum of the numbers are, the product is larger than the sum by 109. In other words, you must add 109 to the sum to equal the product of both numbers which makes the other part of the equation: 109+[(x)+(x+1)]. Putting it all together as one equation: (x)•(x+1)=109+[x+(x+1)].

 

As an equation, we can combine like terms (numbers with the same "tag") as well as use the distributive property. For the product of the two numbers, multiply the x with each term of the second number, so it's now: x•x+1•x which is x²+x. Combine like terms in the second part of the equation, so now the equation is as such: x²+x=109+2x+1 or x²+x=110+2x. Let's have x be on one side of the equation by combining like terms again, so now the equation is x²-x=110.

 

If you notice, we can't solve for x as directly as usual in algebra because x2 and x can't be combined, so we must make the equation equal 0 and undo FOIL. x²-x-110=0 and undoing FOIL would mean (x± )•(x± )=0 right? Well, we know that the numbers for x must be one different in order for them to leave -x as the middle term and if multiplied together, it equals -110. Think to yourself What are the factors of -110 that have a difference of one (in terms of the number, not the sign)? Here, it would be -11 and 10 because -11•10= -110 and when FOILing the middle term becomes -1 for x: (x-11)•(x+10)= x²+10x-11x-110 or x²-x-110. So, (x-11)•(x+10)=0 leaving x=11 or x= -10 but x can't be -10 so x=11!

 

We're so close now! Going back to the problem, the two numbers are consecutive and if x is 11, then x+1 is 12. Double check to make sure it works with the problem when you replace x with 11 and x+1 with 12: (11)•(12)= 109+(11)+(12) which is 132=109+23 or 132=132!