Algebra 2 (Consecutive Word Problem)?

Question

Find two consecutive positive integers whose product is 20 less than 13 times the smaller integer.

Answer 1

Let the first integer = x

Let the second integer = x + 1

 

 

Set up equation using the variables and info given.

 

x(x + 1) = 13x - 20

 

x² + x = 13x - 20

 

 

Subtract 13x and add 20 on both sides of equation.

 

x² - 12x + 20 = 0

 

(x - 10)(x - 2) = 0

 

x = 10        and        x = 2

 

 

Substitute these values of x into the two numbers.

 

The first integer is 2.

The second integer is 3.

 

 

Now we check these numbers to see if they make the statement true.  If not, we accept the other value of x.

 

Multiply the two numbers.

 

2 * 3 = 6

 

 

The number 6 must be 20 less than 13 times the smaller number.  The smaller number is 2.

 

We compare 6 and 2.

 

13(2) - 20 = 26 - 20 = 6

 

These numbers make the statement true.

 

 

Lets plug in x = 10 to make sure this value of x works too.

 

1 number = 10

2nd number = 11

 

 

10 * 11 = 110

 

13(10) - 20 = 130 - 20 = 110

 

This value of x makes the statement true as well.

 

 

We have 2 pairs of consecutive integers.

 

First pair:   2 and 3

Second pair:   10 and 11

Answer 2

The purpose of this problem is to build your experience at translating the words into math variables and operations.

 

For example, let S = the smaller integer

 

Then,   "13 times the smaller integer"   becomes    13*S 

   "20 less than that"  becomes   13*S - 20

   "is"   means  =

  "consecutive integers (with S being the smallest one)"    becomes     S  and (S+1)

   "the product of"    means to multiply

 

So,      S*(S+1) = 13*S - 20

 

Now, use the rules of algebra to identify the two numbers:

           S*S + S = 13*S - 20

         S^2 -12S +20 = 0

 

Using either factoring or else the Binomial formula, you get S = 2 and S = 10

 

Now check:

    For 2 and 3, is 2*3 = 13*2 - 20   ?

     For 10 and 11, is 10*11 = 13*10 - 20   ?