Question

what value of x would make (overline{km} parallel overline{jn})?
complete the statements to solve for x.
by the converse of the side-splitter theorem, if ( \frac{jk}{kl} = \boldsymbol{square} ), then ( overline{km} parallel overline{jn} ).
substitute the expressions into the proportion: ( \frac{x - 5}{x} = \frac{x - 3}{x + 4} ).
cross-multiply: ( (x - 5)(\boldsymbol{square}) = x(x - 3) ).
distribute: ( x(x) + x(4) - 5(x) - 5(4) = x(x) + x(-3) ).
multiply and simplify: ( x^2 - x - \boldsymbol{square} = x^2 - 3x ).
solve for ( x ): ( x = \boldsymbol{square} ) (options: 0.5, 4.5, 10).
diagram: triangle ( ljn ) with ( k ) on ( jl ), ( m ) on ( ln ). ( jl ) segments: ( jk = x - 5 ), ( kl = x ). ( ln ) segments: ( lm = x + 4 ), ( mn = x - 3 ). ( km ) is a segment connecting ( k ) and ( m ).

Explanation:

Step1: Recall Side - Splitter Theorem

The converse of the side - splitter theorem states that if a line divides two sides of a triangle proportionally, then it is parallel to the third side. So, if \(\frac{JK}{KL}=\frac{JM}{MN}\), then \(\overline{KM}\parallel\overline{JN}\). From the diagram, \(JK = x - 5\), \(KL=x\), \(JM=x - 3\), \(MN=x + 4\)? Wait, no, looking at the proportion given \(\frac{x - 5}{x}=\frac{x - 3}{x + 4}\), so the ratio should be \(\frac{JK}{KL}=\frac{JM}{MN}\) (or \(\frac{JL}{LK}=\frac{JN}{NM}\), but from the proportion \(\frac{x - 5}{x}=\frac{x - 3}{x + 4}\), the first blank is \(\frac{JM}{MN}\) or \(\frac{x - 3}{x + 4}\) (the ratio of the other two sides). But let's focus on solving the equation \(\frac{x-5}{x}=\frac{x - 3}{x + 4}\).

Step2: Cross - Multiply

Cross - multiplying the proportion \(\frac{x-5}{x}=\frac{x - 3}{x + 4}\) gives \((x - 5)(x + 4)=x(x - 3)\). So the second blank is \(x + 4\).

Step3: Distribute and Simplify Left - Hand Side

Distribute \((x - 5)(x + 4)\):
\[

$$\begin{align*} (x-5)(x + 4)&=x\times x+x\times4-5\times x-5\times4\\ &=x^{2}+4x-5x - 20\\ &=x^{2}-x - 20 \end{align*}$$

\]
So the third blank is \(20\). The equation becomes \(x^{2}-x - 20=x^{2}-3x\).

Step4: Solve for \(x\)

Subtract \(x^{2}\) from both sides of the equation \(x^{2}-x - 20=x^{2}-3x\):
\[

$$\begin{align*} x^{2}-x - 20-x^{2}&=x^{2}-3x-x^{2}\\ -x - 20&=-3x \end{align*}$$

\]
Add \(3x\) to both sides:
\[

$$\begin{align*} -x+3x - 20&=-3x + 3x\\ 2x-20&=0 \end{align*}$$

\]
Add \(20\) to both sides:
\[

$$\begin{align*} 2x-20 + 20&=0+20\\ 2x&=20 \end{align*}$$

\]
Divide both sides by \(2\):
\[
x=\frac{20}{2}=10
\]

Answer:

The value of \(x\) is \(10\).

For the blanks:

• First blank: \(\frac{JM}{MN}\) (or \(\frac{x - 3}{x + 4}\))
• Second blank: \(x + 4\)
• Third blank: \(20\)
• Fourth blank: \(10\)