Question

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if the sides of a triangle have the following lengths, find a range of possible values for x

1. ( pq = 2x + 13 ), ( qr = 10x - 2 ), ( pr = x + 27 )

range of x - values:

1. ( lm = 6x + 3 ), ( mn = 4x - 17 ), ( ln = x + 40 )

range of x - values:

1. ( cd = x - 4 ), ( de = 3x + 21 ), ( ce = 6x - 13 )

range of x - values:

1. ( jk = 5x + 17 ), ( kl = x - 1 ), ( jl = 7x - 9 )

Explanation:

Problem 1: $PQ=2x+13$, $QR=10x-2$, $PR=x+27$

Step1: Apply Triangle Inequality 1

Sum of $PQ+QR>PR$

$$\begin{align} (2x+13)+(10x-2) &> x+27 \\ 12x+11 &> x+27 \\ 11x &> 16 \\ x &> \frac{16}{11} \end{align}$$

Step2: Apply Triangle Inequality 2

Sum of $PQ+PR>QR$

$$\begin{align} (2x+13)+(x+27) &> 10x-2 \\ 3x+40 &> 10x-2 \\ -7x &> -42 \\ x &< 6 \end{align}$$

Step3: Apply Triangle Inequality 3

Sum of $QR+PR>PQ$

$$\begin{align} (10x-2)+(x+27) &> 2x+13 \\ 11x+25 &> 2x+13 \\ 9x &> -12 \\ x &> -\frac{4}{3} \end{align}$$

Step4: Take strictest bounds

The tightest constraints are $x>\frac{16}{11}$ and $x<6$.

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Problem 2: $LM=6x+3$, $MN=4x-17$, $LN=x+40$

Step1: Ensure side lengths are positive

First, $MN=4x-17>0 \implies x>\frac{17}{4}$

Step2: Apply Triangle Inequality 1

Sum of $LM+MN>LN$

$$\begin{align} (6x+3)+(4x-17) &> x+40 \\ 10x-14 &> x+40 \\ 9x &> 54 \\ x &> 6 \end{align}$$

Step3: Apply Triangle Inequality 2

Sum of $LM+LN>MN$

$$\begin{align} (6x+3)+(x+40) &> 4x-17 \\ 7x+43 &> 4x-17 \\ 3x &> -60 \\ x &> -20 \end{align}$$

Step4: Apply Triangle Inequality 3

Sum of $MN+LN>LM$

$$\begin{align} (4x-17)+(x+40) &> 6x+3 \\ 5x+23 &> 6x+3 \\ -x &> -20 \\ x &< 20 \end{align}$$

Step5: Take strictest bounds

The tightest constraints are $x>6$ and $x<20$.

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Problem 3: $CD=x-4$, $DE=3x+21$, $CE=6x-13$

Step1: Ensure side lengths are positive

$CD=x-4>0 \implies x>4$; $CE=6x-13>0 \implies x>\frac{13}{6}$

Step2: Apply Triangle Inequality 1

Sum of $CD+DE>CE$

$$\begin{align} (x-4)+(3x+21) &> 6x-13 \\ 4x+17 &> 6x-13 \\ -2x &> -30 \\ x &< 15 \end{align}$$

Step3: Apply Triangle Inequality 2

Sum of $CD+CE>DE$

$$\begin{align} (x-4)+(6x-13) &> 3x+21 \\ 7x-17 &> 3x+21 \\ 4x &> 38 \\ x &> \frac{19}{2}=9.5 \end{align}$$

Step4: Apply Triangle Inequality 3

Sum of $DE+CE>CD$

$$\begin{align} (3x+21)+(6x-13) &> x-4 \\ 9x+8 &> x-4 \\ 8x &> -12 \\ x &> -\frac{3}{2} \end{align}$$

Step5: Take strictest bounds

The tightest constraints are $x>\frac{19}{2}$ and $x<15$.

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Problem 4: $JK=5x+17$, $KL=x-1$, $JL=7x-9$

Step1: Ensure side lengths are positive

$KL=x-1>0 \implies x>1$; $JL=7x-9>0 \implies x>\frac{9}{7}$

Step2: Apply Triangle Inequality 1

Sum of $JK+KL>JL$

$$\begin{align} (5x+17)+(x-1) &> 7x-9 \\ 6x+16 &> 7x-9 \\ -x &> -25 \\ x &< 25 \end{align}$$

Step3: Apply Triangle Inequality 2

Sum of $JK+JL>KL$

$$\begin{align} (5x+17)+(7x-9) &> x-1 \\ 12x+8 &> x-1 \\ 11x &> -9 \\ x &> -\frac{9}{11} \end{align}$$

Step4: Apply Triangle Inequality 3

Sum of $KL+JL>JK$

$$\begin{align} (x-1)+(7x-9) &> 5x+17 \\ 8x-10 &> 5x+17 \\ 3x &> 27 \\ x &> 9 \end{align}$$

Step5: Take strictest bounds

The tightest constraints are $x>9$ and $x<25$.

Answer:

1. $\frac{16}{11} < x < 6$
2. $6 < x < 20$
3. $\frac{19}{2} < x < 15$
4. $9 < x < 25$