Question

Lines a and b are cut by transversal f. At the intersection of lines f and a, the top left angle is 96 degrees. At the intersection of lines b and f, the bottom right angle is (6 x minus 36) degrees.
What must be the value of x so that lines a and b are parallel lines cut by transversal f?

Answer 1

To determine the value of \( x \) for which lines \( a \) and \( b \) are parallel when cut by transversal \( f \), we can use the property of corresponding angles. When two lines are parallel, corresponding angles formed by a transversal are equal.

Given:
- The angle at the intersection of lines \( f \) and \( a \) (top left angle) is \( 96^\circ \).
- The angle at the intersection of lines \( b \) and \( f \) (bottom right angle) is \( 6x - 36 \) degrees.

Since these angles are corresponding angles, we can set them equal to each other:

\[
96 = 6x - 36
\]

To solve for \( x \):
1. Add \( 36 \) to both sides:

\[
96 + 36 = 6x
\]

\[
132 = 6x
\]

2. Divide both sides by \( 6 \):

\[
x = \frac{132}{6} = 22
\]

Thus, the value of \( x \) that makes the lines \( a \) and \( b \) parallel is

\[
\boxed{22}
\]