Question

1. parking in the parking lot shown, the lines that mark the width of each space are parallel. m∠1=(2x - 3y)° m∠2=(x + 3y)° find x and y.

Explanation:

Step1: Use the property of parallel lines (corresponding angles)

Since the lines are parallel, \(\angle1\) and \(\angle2\) are supplementary (they form a linear - pair of angles adjacent to a transversal cutting parallel lines). So, \(m\angle1 + m\angle2=180^{\circ}\).
Substitute the given expressions: \((2x - 3y)+(x + 3y)=180\).
Simplify the left - hand side: \(2x-3y+x + 3y=180\), which gives \(3x=180\).
Solve for \(x\): \(x = 60\).

Step2: Use the property of parallel lines (alternate - interior angles or other relevant properties, assume another relationship if needed, but since we have found \(x\) and if we assume a standard parking - lot angle relationship where \(\angle1=\angle2\) (if it's a case of congruent angles due to parallel lines and a transversal in a symmetric parking - lot design, for example)

Set \(2x-3y=x + 3y\) (if we assume \(\angle1=\angle2\) based on parallel - line symmetry in the parking - lot).
Substitute \(x = 60\) into \(2x-3y=x + 3y\).
We get \(2\times60-3y=60 + 3y\).
\(120-3y=60 + 3y\).
Add \(3y\) to both sides: \(120=60 + 6y\).
Subtract \(60\) from both sides: \(6y=60\).
Solve for \(y\): \(y = 10\).

Answer:

\(x = 60,y = 10\)