Question

what is the least value for an integer, x, for which x, x + 5, and 2x - 15 can be the lengths of the sides of a triangle? answer: the smallest integer x is

Explanation:

Step1: Apply triangle - inequality theorem

For three side - lengths \(a\), \(b\), and \(c\) of a triangle, \(a + b>c\), \(a + c>b\), and \(b + c>a\). Here \(a=x\), \(b=x + 5\), and \(c=2x-15\).
First, consider \(a + b>c\):
\(x+(x + 5)>2x-15\)
\(2x+5>2x - 15\), which is always true for all real \(x\).
Next, consider \(a + c>b\):
\(x+(2x-15)>x + 5\)
\(3x-15>x + 5\)
\(3x-x>5 + 15\)
\(2x>20\)
\(x>10\).
Then, consider \(b + c>a\):
\((x + 5)+(2x-15)>x\)
\(3x-10>x\)
\(3x-x>10\)
\(2x>10\)
\(x>5\).

Step2: Find the smallest integer value

Since \(x>10\) from the most restrictive inequality, the smallest integer value of \(x\) is \(11\).

Answer:

\(11\)