Question

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

a)

x > -4/3


b)

0 < x < 1/6


c)

1/6 < x < 3/11


d)

x > 0

Answer 1

c)

1/6 < x < 3/11

Explanation:

Answer 2

`x>(3)/(11)`

Explanation:

If a, b and c are three sides of a triangle, then

`a+b>c\\ \\a+c>b\\ \\b+c>a`

In your case,

`a=3x+4\\ \\b=6x-1\\ \\c=8x+2`

Thus,

1.

`3x+4+6x-1>8x+2\\ \\9x+3>8x+2\\ \\9x-8x>2-3\\ \\x>-1`

2.

`3x+4+8x+2>6x-1\\ \\11x+6>6x-1\\ \\11x-6x>-1-6\\ \\5x>-7\\ \\x>-1.4`

3.

`6x-1+8x+2>3x+4\\ \\14x+1>3x+4\\ \\14x-3x>4-1\\ \\11x>3\\ \\x>(3)/(11)`

Thus,

`x>(3)/(11)`

Note that a>0, b>0 and c>0, so

1.

`3x+4>0\\ \\3x>-4\\ \\x>-(4)/(3)`

2.

`6x-1>0\\ \\6x>1\\ \\x>(1)/(6)`

3.

`8x+2>0\\ \\8x>-2\\ \\x>-(1)/(4)`

Thus,

`x>(1)/(6)`

As result,

`x>(3)/(11)`