What value of [tex]$x$[/tex] would make [tex]$\overline{ KM } \parallel \overline{ JN }$[/tex]?

Complete the statements to solve for [tex][tex]$x$[/tex][/tex].

By the converse of the side-splitter theorem, if [tex]$\frac{JK}{KL} = \square$[/tex], then [tex]$\overline{ KM } \parallel \overline{ JN }$[/tex].

Substitute the expressions into the proportion:
[tex]
\frac{x-5}{x} = \frac{x-3}{x+4}.
[/tex]

Cross-multiply: [tex](x-5) \cdot (x+4) = x(x-3)[/tex].

Distribute: [tex]x(x) + x(4) - 5(x) - 5(4) = x(x) + x(-3)[/tex].

Multiply and simplify: [tex]x^2 - x - \square = x^2 - 3x[/tex].

Solve for [tex][tex]$x$[/tex][/tex]: [tex]$x = \square$[/tex].



Answer :

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].