Question about integers

Question

A positive integer is 3 more than twice a smaller positive integer. If the product of the two integers is equal to six times the larger, then find the integers.

Answer 1

Hi Haley.

 

Let's call S your smaller number and L your larger number, and craft algebraic equations that match what your statements claim.

 

"A positive integer is 3 more than twice the smaller integer."

 

That means:

 

L = 2S + 3

 

"The product of both integers is 6x the larger."

 

That means:

 

S*L = 6L

 

Now you have two equations, two unknowns.

 

L = 2S + 3

S*L = 6L

 

It may be very tempting to cancel out the L's in the second equation, and in this case you are absolutely allowed to do so. The only time you can't cancel out variables on both sides is if they have a possibility of being 0 (since everyone knows you can't divide by 0). Since the question tells you both integers are positive, they can't be 0. So feel free to cancel them out.

 

S = 6

 

If S = 6, then

 

L = 2*6 + 3

 

L = 15

 

The other way to do this without canceling the zeros is to substitute equation 1 into equation 2.

 

S*(2S + 3) = 6*(2S + 3)

 

2S^2 + 3S = 12S + 18

 

2S^2 - 9S - 18 = 0

 

Solve this as you would any other quadratic. I'm going to use the quadratic formula.

 

S = (9 + √(81+144)) / 4

 

S = (9 + √225)/4

 

S = (9 + 15)/4

 

S = (9+15)/4, (9-15)/4

 

S = 6, -3/2

 

The problem claimed it's a positive integer, so you can throw away -3/2.

 

S = 6, meaning

 

L = 2S + 3

 

L = 15

 

So either way you do it, your smaller integer is 6 and your larger integer is 15.

 

Hope this helps.

Answer 2

X and y are the two integers 

x = 2y + 3

x*y = 6x 

 

after selecting random numbers I tried 6 for y to see if the product and 6x would be the same, so:

 

2(6) + 3 = x          15*6 = 6(15)

12 + 3 = x             90 = 90

15 = x

 

 

X=15

y = 6