If the sides of a triangle have measures 3x + 4, 6x – 1 and 8x + 2, find all possible values of x.
question 12 options:

a)

x > -4/3

b)

0 < x < 1/6

c)

1/6 < x < 3/11

d)

x > 0

x>-4/3

Step-by-step explanation:All sides must be greater of zero:

3x+4>0

3x> -4 Divide with 3

x> -4/3

6x-1>0

6x>1 Divide with 6

x>1/6

8x+2>0

8x> -2 Divide with 8

x> -2/8

x> -1/4

Least of that numbers is -4/3= -1.3333

x> -4/3 is solution

add -60 to 120 which will result in : 2x+3y=180

then subtract 2x: 3y=-2x+180

lastly, divide by 2: y= -2/3x+60

hope this is the answer you're looking for: )