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

The sum of the lengths of any two sides of a triangle must be greater than the length of the third side. So we have three inequalities:

1. \(x+(x + 5)>2x-15\), which simplifies to \(2x+5>2x - 15\), and \(5>-15\) (this is always true).
2. \(x+(2x - 15)>x + 5\).

Simplify the left - hand side: \(3x-15>x + 5\).
Subtract \(x\) from both sides: \(3x-x-15>x-x + 5\), \(2x-15>5\).
Add 15 to both sides: \(2x-15 + 15>5+15\), \(2x>20\), then \(x > 10\).

1. \((x + 5)+(2x-15)>x\).

Simplify the left - hand side: \(3x-10>x\).
Subtract \(x\) from both sides: \(3x-x-10>x-x\), \(2x-10>0\).
Add 10 to both sides: \(2x-10 + 10>0+10\), \(2x>10\), then \(x>5\).

Step2: Find the smallest integer value of \(x\)

Since \(x>10\) from the second inequality (which gives the most restrictive condition), the smallest integer value of \(x\) that satisfies all the conditions is \(x = 11\).

Answer:

11