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To determine the value of [tex]\( x \)[/tex] that makes [tex]\(\overline{KM} \parallel \overline{JN}\)[/tex], we can use the converse of the side-splitter theorem. According to this theorem, if [tex]\(\frac{JK}{KL} = \frac{JM}{MN}\)[/tex], then [tex]\(\overline{KM} \parallel \overline{JN}\)[/tex]. In this context, substituting the given expressions:
[tex]\[ \frac{JK}{KL} = \frac{x-5}{x} \quad \text{and} \quad \frac{JM}{MN} = \frac{x-3}{x+4} \][/tex]
Step-by-step solution:
1. By the converse of the side-splitter theorem, if [tex]\(\frac{JK}{KL} = \frac{x-5}{x} = \frac{JM}{MN} = \frac{x-3}{x+4}\)[/tex], then [tex]\(\overline{KM} \parallel \overline{JN}\)[/tex].
2. Substitute the expressions into the proportion:
[tex]\[ \frac{x-5}{x} = \frac{x-3}{x+4} \][/tex]
3. Cross-multiply:
[tex]\[ \text{Cross-multiply: } (x-5)(x+4) = x(x-3) \][/tex]
4. Distribute both sides:
[tex]\[ x(x) + x(4) - 5(x) - 5(4) = x(x) + x(-3) \][/tex]
5. Multiply and simplify:
[tex]\[ x^2 + 4x - 5x - 20 \sim x^2 - 3x \][/tex]
[tex]\[ x^2 - x - 20 \sim x^2 - 3x \][/tex]
6. To find [tex]\( x \)[/tex], combine and simplify the terms:
[tex]\[ x^2 - x - 20 = x^2 - 3x \][/tex]
[tex]\[ 0 = 2x - 20 \][/tex]
7. Solve for [tex]\( x \)[/tex]:
[tex]\[ 2x - 20 = 0 \][/tex]
[tex]\[ 2x = 20 \][/tex]
[tex]\[ x = 10 \][/tex]
Therefore, the value of [tex]\( x \)[/tex] that makes [tex]\(\overline{KM} \parallel \overline{JN}\)[/tex] is [tex]\( \boxed{10} \)[/tex].
[tex]\[ \frac{JK}{KL} = \frac{x-5}{x} \quad \text{and} \quad \frac{JM}{MN} = \frac{x-3}{x+4} \][/tex]
Step-by-step solution:
1. By the converse of the side-splitter theorem, if [tex]\(\frac{JK}{KL} = \frac{x-5}{x} = \frac{JM}{MN} = \frac{x-3}{x+4}\)[/tex], then [tex]\(\overline{KM} \parallel \overline{JN}\)[/tex].
2. Substitute the expressions into the proportion:
[tex]\[ \frac{x-5}{x} = \frac{x-3}{x+4} \][/tex]
3. Cross-multiply:
[tex]\[ \text{Cross-multiply: } (x-5)(x+4) = x(x-3) \][/tex]
4. Distribute both sides:
[tex]\[ x(x) + x(4) - 5(x) - 5(4) = x(x) + x(-3) \][/tex]
5. Multiply and simplify:
[tex]\[ x^2 + 4x - 5x - 20 \sim x^2 - 3x \][/tex]
[tex]\[ x^2 - x - 20 \sim x^2 - 3x \][/tex]
6. To find [tex]\( x \)[/tex], combine and simplify the terms:
[tex]\[ x^2 - x - 20 = x^2 - 3x \][/tex]
[tex]\[ 0 = 2x - 20 \][/tex]
7. Solve for [tex]\( x \)[/tex]:
[tex]\[ 2x - 20 = 0 \][/tex]
[tex]\[ 2x = 20 \][/tex]
[tex]\[ x = 10 \][/tex]
Therefore, the value of [tex]\( x \)[/tex] that makes [tex]\(\overline{KM} \parallel \overline{JN}\)[/tex] is [tex]\( \boxed{10} \)[/tex].
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