Use an algebraic solution to find four?

Question

consecutive positive integers such that the product of the first and fourth is four less than twice the first multiplied by the fourth.

Answer 1

x, x+1, x+2, x+3
so that
(x)(x+3) +4= 2{(x)(x+3)}

s a v v y

the trick in Algebra - is learning how to put words into motion

consecutive - starting point, +1, +2, +3
x, x+1, x+2, x+3
product of the first and fourth
first is x
fourth is x+3
so x times x+3
is four less - - - - - -so add four to this to equal what's coming

than twice = 2x{..........}
the first (x) multiplied by the fourth (x+3)

so that comes out to
(x)(x+3) +4= 2{(x)(x+3)}
do we need to solve it?
x^2 + 3x +4 = 2x^2 + 6x
subtract x^2 from both sides
3x + 4 = x^2 + 6x
subtract 3x from both sides
4 = x^2 + 3x
subtract 4 from both sides
x^2 + 3x -4 = 0
(x-1)(X+4) = 0

x = 1, -4

since we wanted a positive integer = 1

does it fit
1 x 4 + 4 = 2(1x4)
yup!

we have a winner!!!!!!!!!

Answer 2

consecutive positive integers such that the product of the first and fourth is four less than twice the first multiplied by the fourth.

let the consecutive positive integer numbers be :

n; n+1; n+2; n+3
n(n+3) = 2n(n+3) -4
n^2 +3n -2n^2 -6n =-4
-n^2 -3n +4 =0
n^2 +3n -4 =0
(n-1)(n+4) =0
n =1 is the solution

the 4 numbers are

1 ; 2 ; 3 ; 4

Answer 3

x^2+3x = 2x^2+6x-4

x^2+3x-4 = 0

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

x = -4 or x = 1

Only positive are accepted.

x = 1

The integers are: 1, 2, 3, 4

Answer 4

i dont know