Question

1. if (lk = mk), (lk = 7x - 10), (kn=x + 3), (mn) (diagram shows intersecting lines with points (l), (k), (m), (n), (j)). 10. if (t) is the mid - point of (overline{su}), find (x). (diagram shows a line segment (overline{su}) with mid - point (t), (st = 8x + 11), (tu = 12x - 1)). 12. if (r) is the mid - point of (overline{qs}), find (qs). (diagram shows a line segment (overline{qs}) with mid - point (r), (qr = 5x - 3), (rs = 21 - x))

Explanation:

Step1: Set up equation for problem 10

Since $T$ is mid - point of $\overline{SU}$, $ST = TU$. So, $8x + 11=12x - 1$.

Step2: Solve for $x$

Subtract $8x$ from both sides: $11 = 4x-1$. Add 1 to both sides: $12 = 4x$. Then $x = 3$.

Step3: Set up equation for problem 12

Since $R$ is mid - point of $\overline{QS}$, $QR = RS$. So, $5x - 3=21 - x$.

Step4: Solve for $x$

Add $x$ to both sides: $6x - 3=21$. Add 3 to both sides: $6x = 24$, so $x = 4$.

Step5: Find $QS$

$QS=QR + RS$. Substitute $x = 4$ into expressions for $QR$ and $RS$. $QR=5x - 3=5\times4 - 3 = 17$, $RS=21 - x=21 - 4 = 17$. So $QS=17 + 17=34$.

Answer:

For problem 10: $x = 3$
For problem 12: $QS = 34$