2. Find three positive consecutive integers such that the product of the first and the third is one less than six times the second.

So consecutive integers are notated as x, x+1, x+2 since they are that distance frome ach other

such that the product (that means mulitipication) of the first and third (that means x times (x+2)) is (that means equals) one less than (minus 1) six times the second (6 times (x+1))

x times (x+2) = - 1+6 (x+1)

distribute using distributiver property which is a (b+c) = ab+ac

x (x+2) = x^2+2x

6 (x+1) = 6x+6

so we have

x^2+2x=-1+6x+6

add liket erms

-1+6x+6=6x+6-1=6x+5

x^2+2x=6x+5

make one side zeroe so we can factor

subtract 6x from both sides

x^2-4x=5

subtract 5 from both sides

x^2-4x-5=0

factor

find what 2 number multiply to get - 5 and add to get - 4

the answer is - 5 and + 1

so we put them in front of x in the factored form to get

(x-5) (x+1) = 0

set each to zero

x-5=0

x+1=0

solve for x

x-5=0

add 5 to both sides

x=5

x+1=0

subtract 1

x=-1

false since the question specified positive integers

therefor x=5

the numbers are

x, x+1, x+2

x+1=5+1=6

x+2=5+2=7

the numbers are 5,6,7

A
c - a = 2

ac = 6b-1

5 x 7 = 35 = 6x6 - 1

a = 5

b = 6

c = 7