Question

find the values indicated.

1. y =

lm =
no =
if lmno is a parallelogram, and lm =2y − 9 and no = y − 2, find lm. (you know you need to show all of your work!) (draw & label parallelogram first)
(no work no credit)

1. x =

rq =
qt =
rstv is a parallelogram. \\(\overline{rt}\\) and \\(\overline{sv}\\) intersect at q. rq = 5x + 1 and qt = 3x + 15. find qt. (show all of your work!) (draw and label it first)

Explanation:

Problem 6:

Step1: Recall parallelogram property

In a parallelogram, opposite sides are equal. So \( LM = NO \).
Given \( LM = 2y - 9 \) and \( NO = y - 2 \), we set up the equation:
\( 2y - 9 = y - 2 \)

Step2: Solve for \( y \)

Subtract \( y \) from both sides:
\( 2y - y - 9 = y - y - 2 \)
\( y - 9 = -2 \)

Add 9 to both sides:
\( y - 9 + 9 = -2 + 9 \)
\( y = 7 \)

Step3: Find \( LM \)

Substitute \( y = 7 \) into \( LM = 2y - 9 \):
\( LM = 2(7) - 9 = 14 - 9 = 5 \)

Step4: Find \( NO \)

Substitute \( y = 7 \) into \( NO = y - 2 \):
\( NO = 7 - 2 = 5 \)

Step1: Recall parallelogram diagonal property

In a parallelogram, the diagonals bisect each other. So \( RQ = QT \).
Given \( RQ = 5x + 1 \) and \( QT = 3x + 15 \), we set up the equation:
\( 5x + 1 = 3x + 15 \)

Step2: Solve for \( x \)

Subtract \( 3x \) from both sides:
\( 5x - 3x + 1 = 3x - 3x + 15 \)
\( 2x + 1 = 15 \)

Subtract 1 from both sides:
\( 2x + 1 - 1 = 15 - 1 \)
\( 2x = 14 \)

Divide both sides by 2:
\( x = \frac{14}{2} = 7 \)

Step3: Find \( RQ \)

Substitute \( x = 7 \) into \( RQ = 5x + 1 \):
\( RQ = 5(7) + 1 = 35 + 1 = 36 \)

Step4: Find \( QT \)

Substitute \( x = 7 \) into \( QT = 3x + 15 \):
\( QT = 3(7) + 15 = 21 + 15 = 36 \)

Answer:

\( y = 7 \)
\( LM = 5 \)
\( NO = 5 \)

Problem 7: