One positive number exceeds another by 6, and their product is 91. What are the numbers?

Answer 1

Let x, y be our positive numbers. We know they are both positive so, x,y > 0 (This is important).

So, taking one positive number to be x, we know this exceeds the other positive number y by 6.

This means that x is (or equals) 6 more than y or y+6 = x (eq 1).

We also know that their product is equal to 91 product implies multiplication so, x*y = 91 (eq 2).

We now have a system of equations.

Since we know y+6 = x from (eq 1) we can plug that into our second equation x*y = 91 to get

y² + 6y - 91 = 0 Then, factor:

From the negative before the 91, we can say we are going to have the form of

(y - _ )(y +_ ) =0 because only a negative multiplied by a positive number will give us a negative sign there.

If we look at the factors of 91 : 1 and 91, 7 and 13

13-7 = 6 which is the b coefficient in the polynomial y² + 6y - 91 = 0

So,

(y + 13)(y - 7)=0 is our factorization.

We can then solve for x and y.

Answer 2

A couple ways to solve this one.

1) Find all pairs of #s that multiply to 91

1, 91

7, 13

Which pair differs by 6? It's "7 & 13"

2) Let x = one number and y = the other number

We know one number is 6 more than the other, so

x = y + 6

We also know the product of the two numbers is 91, so

(Note: * means "times")

x * y = 91

Replace "x" in the second equation with "y + 6" (from the first equation)

(y + 6) * y = 91

Distribute the y, and subtract the 91

y² + 6y - 91 = 0

Factor

(y - 7)(y + 13)=0

Set each factor equal to 0 and solve

y - 7 = 0

y=7 ---> this is the correct answer as we know the # must be positive

y + 13 = 0

y = -13 --> this is incorrect, as we are told the # must be positiive

Therefore, y=7

In which case x = y + 6 = 7 + 6 = 13

The two numbers are 7 & 13