Question

directions: if (lparallel m), solve for (x) and (y).

Explanation:

Step1: Solve for \(x\) in problem 10

Since \(l\parallel m\), the corresponding - angles are equal. So, \(3x - 29=6x - 7\).
Subtract \(3x\) from both sides: \(-29 = 3x-7\).
Add 7 to both sides: \(-22 = 3x\).
Divide both sides by 3: \(x=-\frac{22}{3}\).
Also, since the sum of interior - angles on the same side of the transversal is \(180^{\circ}\) for parallel lines \(l\) and \(m\), \((3x - 29)+(8y + 17)=180\).
Substitute \(x =-\frac{22}{3}\) into the equation: \(3\times(-\frac{22}{3})-29 + 8y+17 = 180\).
Simplify: \(-22-29 + 8y+17 = 180\).
Combine like - terms: \(8y-34 = 180\).
Add 34 to both sides: \(8y=214\).
Divide by 8: \(y=\frac{107}{4}\).

Step2: Solve for \(x\) and \(y\) in problem 11

Since \(l\parallel m\), the vertical angles and corresponding angles relationships are used.
We have \(3x = 49\) (vertical angles), so \(x=\frac{49}{3}\).
Also, \(7x-23 + 11y - 1=180\) (sum of interior - angles on the same side of the transversal).
Substitute \(x=\frac{49}{3}\) into \(7x-23 + 11y - 1=180\): \(7\times\frac{49}{3}-23 + 11y-1 = 180\).
\(\frac{343}{3}-23 - 1+11y = 180\).
\(\frac{343}{3}-24 + 11y = 180\).
\(\frac{343 - 72}{3}+11y = 180\).
\(\frac{271}{3}+11y = 180\).
Subtract \(\frac{271}{3}\) from both sides: \(11y=180-\frac{271}{3}=\frac{540 - 271}{3}=\frac{269}{3}\).
\(y=\frac{269}{33}\).

Step3: Solve for \(x\) and \(y\) in problem 12

Since \(l\parallel m\), \(3x-4 + 5x - 38=180\) (sum of interior - angles on the same side of the transversal).
Combine like - terms: \(8x-42 = 180\).
Add 42 to both sides: \(8x=222\).
Divide by 8: \(x=\frac{111}{4}\).
Also, \(7y-20+3x - 4 = 90\) (right - angle relationship).
Substitute \(x=\frac{111}{4}\) into \(7y-20+3x - 4 = 90\): \(7y-20+3\times\frac{111}{4}-4 = 90\).
\(7y-24+\frac{333}{4}=90\).
\(7y=90 + 24-\frac{333}{4}\).
\(7y=\frac{360+96 - 333}{4}=\frac{123}{4}\).
\(y=\frac{123}{28}\).

Answer:

Problem 10: \(x =-\frac{22}{3},y=\frac{107}{4}\)
Problem 11: \(x=\frac{49}{3},y=\frac{269}{33}\)
Problem 12: \(x=\frac{111}{4},y=\frac{123}{28}\)