If the sides of a triangle have measures 3x + 4, 6x – 1 and 8x + 2, find all possible values of x.a) x > -4/3b) 0 < x < 1/6c) 1/6 < x < 3/11d) x > 0
Accepted Solution
A:
Answer:[tex]x>\dfrac{3}{11}[/tex]Step-by-step explanation:If a, b and c are three sides of a triangle, then [tex]a+b>c\\ \\a+c>b\\ \\b+c>a[/tex]In your case,[tex]a=3x+4\\ \\b=6x-1\\ \\c=8x+2[/tex]Thus,1. [tex]3x+4+6x-1>8x+2\\ \\9x+3>8x+2\\ \\9x-8x>2-3\\ \\x>-1[/tex]2. [tex]3x+4+8x+2>6x-1\\ \\11x+6>6x-1\\ \\11x-6x>-1-6\\ \\5x>-7\\ \\x>-1.4[/tex]3. [tex]6x-1+8x+2>3x+4\\ \\14x+1>3x+4\\ \\14x-3x>4-1\\ \\11x>3\\ \\x>\dfrac{3}{11}[/tex]Thus,[tex]x>\dfrac{3}{11}[/tex]Note that a>0, b>0 and c>0, so1. [tex]3x+4>0\\ \\3x>-4\\ \\x>-\dfrac{4}{3}[/tex]2.[tex]6x-1>0\\ \\6x>1\\ \\x>\dfrac{1}{6}[/tex]3. [tex]8x+2>0\\ \\8x>-2\\ \\x>-\dfrac{1}{4}[/tex]Thus,[tex]x>\dfrac{1}{6}[/tex]As result,[tex]x>\dfrac{3}{11}[/tex]