1/6x< 1/3(4x+3) me figure this out

Simplify 1/6x to x/6

x/6 < 1/3(4x + 3)

Simplify 1/3(4x + 3) to 4x + 3/3

x/6 < 4x + 3/3

Simplify 4x + 3/3 to 1 + 4x/3

x/6 < 1 + 4x/3

Multiply both sies by 6 (The LCM of 6,3)

x < 6 + 8x

Subtract 8x from both sides

x - 8x < 6

Divide both sides by -7

x > -6/7

4.0 ft

step-by-step explanation:

4.0 ft

answer: integers are just whole numbers: positive, negative, and zero.

i = {-∞, -2, -1, 0, 1, 2, 3, ∞}

let x = 1st integer

let y = 2nd integer

x = 3y + 17   {equation 1}

xy = -24         {equation 2}

substitute 3y+17 in place of x n equation 2

and solve for y.

(3y+17)y = -24

3y2 + 17y = -24

3y2 + 17y + 24 = 0

from quadratic equation:

y = [-17±√(172-4(3)(24))]/(2(3))

y = (-17 ±√1)/6

y = (-17+1)/6   or     y = (-17-1)/6

y = -16/6           or     y = -18/6

y = -2 2/3         or   y = -3

since we are dealing with

y = -3

x = 3y+17 = -9+17 = 8

the integers are -3, 8

step-by-step explanation: