The product of the first and third of three consecutive integers is 3 more than 3 times the second integer. find the integers.

Let:

x = the middle integer

Then the 3 consecutive integers are:

x - 1, x, x + 1

The product of the first and the third integer:

(x - 1)*(x + 1) = x^2 - 1

is equal to:

3 more than 3 times the second integer:

3x + 3

So:

x^2 - 1 = 3x + 3

Solve:

x^2 - 3x - 4 = 0

(x-4)(x+1) = 0

x = -1 and 4

Since x is the middle integer, we have two sets of solutions:

3,4,5 and -2,-1,0

c. (0.6,-1.4)

step-by-step explanation:

hope this

option 3 . (5x + 1) - (-10x + 6)

step-by-step explanation:

option 1 :

(4x + 3) + (-9x + 4)

or 4x+3 -9x + 4

or -5x + 7

option 2

(x - 2) - (6x - 9)

or x-2-6x + 9

or -5x + 7

option 3

(5x+ + 6)

or 5x + 1 + 10x -6

or 15x -5

option 4

-4x+2x+7-3x

-7x + 2x + 7

-5x + 7

hence only option 3 did   not satisfy the given simplify expression.