Question

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

Answer 1

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

Answer 2

A<b<c
c - a = 2
ac = 6b-1
5 x 7 = 35 = 6x6 - 1
a = 5
b = 6
c = 7