Question

hw 7: relationships in triangles
if the sides of the triangle have the following lengths, find the range of the possible values for x.
jk = 5x + 17, kl = x - 1, jl = 7x - 9
range of values
< x <

Explanation:

Step1: Apply Triangle Inequality (Sum of two sides > third side)

First, we need to ensure all side lengths are positive. So, for \( KL = x - 1 \), we have \( x - 1>0 \), which gives \( x > 1 \). For \( JK = 5x + 17 \), it's always positive for real \( x \). For \( JL = 7x - 9 \), we have \( 7x - 9>0 \), so \( 7x>9 \) and \( x>\frac{9}{7}\approx1.2857 \). Now, apply the triangle inequalities:

1. \( JK + KL>JL \):

\( (5x + 17)+(x - 1)>7x - 9 \)
Simplify left side: \( 6x + 16>7x - 9 \)
Subtract \( 6x \) from both sides: \( 16>x - 9 \)
Add 9 to both sides: \( 25>x \) or \( x < 25 \)

1. \( JK + JL>KL \):

\( (5x + 17)+(7x - 9)>x - 1 \)
Simplify left side: \( 12x + 8>x - 1 \)
Subtract \( x \) from both sides: \( 11x + 8>-1 \)
Subtract 8 from both sides: \( 11x>-9 \)
Divide by 11: \( x>-\frac{9}{11}\) (but we already have \( x>\frac{9}{7}\) from positive side, so this is redundant)

1. \( KL + JL>JK \):

\( (x - 1)+(7x - 9)>5x + 17 \)
Simplify left side: \( 8x - 10>5x + 17 \)
Subtract \( 5x \) from both sides: \( 3x - 10>17 \)
Add 10 to both sides: \( 3x>27 \)
Divide by 3: \( x>9 \)

Step2: Combine all valid inequalities

From positive sides, the strictest is \( x>\frac{9}{7}\), but from the third inequality \( x>9 \), and from the first inequality \( x < 25 \). Also, we need to check if the side lengths are positive, but \( x>9 \) already satisfies \( x>\frac{9}{7}\) and \( x > 1 \). So combining \( x>9 \) and \( x < 25 \).

Answer:

\( 9 < x < 25 \)