Question

One number is three more than a second number. Forty nine more than the product of the 2 numbers is 89. How do you find the numbers?

Answer 1

The numbers are 5 and 8

Explanation:

let the number be n . The second number is 3 less than

the first so is (n - 3 )

49 more than the product equals 89.

hence n(n - 3 ) + 49 = 89

distribute # n^2 - 3n + 49 = 89

and so `n^2 - 3n = 89 - 49 = 40`

This is a trinomial . To solve equate to zero.

# n^2 - 3n - 40 = 0

To factorise require the factors of 40 which multiply to

give -40 and sum to -3 . Factors of 40 ± (1,40,2,20,4,10,5,8)

The pair which multiply to give - 40 and sum to -3 are 5,-8

`n^2 -3n - 40 = 0 → (n+5)(n-8) =0`

so n+5 = 0 gives n = -5

and n - 8 = 0 gives n = 8

n ≠ -5 and so n = 8 and n- 3 = 5

The 2 numbers are therefore 5 and 8