Question

1. solve for x.
2. solve for x.

Explanation:

Step1: Use mid - segment theorem for problem 5

In the first triangle, since $DE$ is a mid - segment (by the equal - length markings), we know that $DE=\frac{1}{2}JL$. So $2x - 11=\frac{1}{2}(2x - 2)$.
\[

$$\begin{align*} 2(2x - 11)&=2x - 2\\ 4x-22&=2x - 2 \end{align*}$$

\]

Step2: Solve the equation for $x$

Subtract $2x$ from both sides: $4x-2x-22=2x-2x - 2$, which gives $2x-22=-2$. Then add 22 to both sides: $2x-22 + 22=-2+22$, so $2x = 20$. Divide both sides by 2: $x = 10$.

Step3: Use mid - segment theorem for problem 6

In the second triangle, since $RS$ is a mid - segment (by the equal - length markings), we know that $RS=\frac{1}{2}JH$. So $x + 1=\frac{1}{2}(x + 10)$.
\[

$$\begin{align*} 2(x + 1)&=x + 10\\ 2x+2&=x + 10 \end{align*}$$

\]

Step4: Solve the equation for $x$

Subtract $x$ from both sides: $2x-x+2=x-x + 10$, which gives $x+2 = 10$. Then subtract 2 from both sides: $x=8$.

Answer:

For problem 5: $x = 10$
For problem 6: $x = 8$