Question

camelot lane, \overleftrightarrow{tf}, intersects arthur street, \overleftrightarrow{mw}, and tales road, \overleftrightarrow{cr}, as shown where $m\angle mgh = (6x - 34)\degree$ and $m\angle chg = 52\degree$.
diagram: camelot lane (tf) intersects arthur street (mw) at g (with $\angle mgh = (6x - 34)\degree$) and tales road (cr) at h (with $\angle chg = 52\degree$), with mw and cr horizontal, tf as a transversal.
determine the value of $x$ that shows that $\overleftrightarrow{mw} \parallel \overleftrightarrow{cr}$.
$x = \square$

Explanation:

Step1: Identify Alternate Interior Angles

If \( \overleftrightarrow{MW} \parallel \overleftrightarrow{CR} \), then \( \angle MGH \) and \( \angle CHG \) are alternate interior angles, so they are equal. Thus, \( 6x - 34 = 52 \).

Step2: Solve for \( x \)

Add 34 to both sides: \( 6x = 52 + 34 \)
\( 6x = 86 \)? Wait, no, 52 + 34 is 86? Wait, 52 + 34: 50 + 30 = 80, 2 + 4 = 6, so 86? Wait, no, wait, 6x - 34 = 52. Add 34: 6x = 52 + 34 = 86? Wait, no, 52 + 34 is 86? Wait, 52 + 30 is 82, plus 4 is 86. Then divide by 6: \( x = \frac{86}{6} \)? Wait, no, that can't be. Wait, maybe I made a mistake. Wait, alternate interior angles: \( \angle MGH \) and \( \angle CHG \) – wait, \( \angle MGH \) is (6x - 34)°, and \( \angle CHG \) is 52°. Wait, maybe they are same - side interior? No, no, if MW || CR, and TF is the transversal, then \( \angle MGH \) and \( \angle CHG \) are alternate interior angles, so they should be equal. Wait, 6x - 34 = 52. Then 6x = 52 + 34 = 86? Wait, 86 divided by 6 is 14.333... That doesn't seem right. Wait, maybe I misidentified the angles. Wait, \( \angle MGH \) and \( \angle GHR \) would be same - side, but \( \angle CHG \) and \( \angle MGH \): wait, let's look at the diagram. MW is horizontal, CR is horizontal, TF is the transversal. So at G, angle MGH is (6x - 34)°, at H, angle CHG is 52°. So if MW || CR, then alternate interior angles: \( \angle MGH \) and \( \angle GHR \) – no, wait, \( \angle CHG \) and \( \angle MGH \): maybe they are corresponding? Wait, no, let's think again. If two parallel lines are cut by a transversal, alternate interior angles are equal. So \( \angle MGH \) and \( \angle CHG \) – wait, maybe \( \angle MGH \) and \( \angle GHR \) are same - side, but \( \angle CHG \) is equal to \( \angle MGH \) as alternate interior. Wait, maybe I made a mistake in the equation. Wait, 6x - 34 = 52. Let's solve again:

\( 6x - 34 = 52 \)

Add 34 to both sides:

\( 6x = 52 + 34 \)

\( 6x = 86 \)? Wait, 52 + 34 is 86? Wait, 52 + 30 is 82, plus 4 is 86. Then \( x=\frac{86}{6}=\frac{43}{3}\approx14.33 \). But that seems odd. Wait, maybe the angle is supplementary? Wait, if they are same - side interior angles, they would be supplementary. Let's check: if MW || CR, then same - side interior angles are supplementary. So \( (6x - 34)+52 = 180 \). Then \( 6x + 18 = 180 \). \( 6x = 162 \). \( x = 27 \). Ah! That makes more sense. I misidentified the angles. \( \angle MGH \) and \( \angle CHG \) are same - side interior angles, so they should be supplementary. So the correct equation is \( (6x - 34)+52 = 180 \).

Step1: Set up Supplementary Angles Equation

Since \( \overleftrightarrow{MW} \parallel \overleftrightarrow{CR} \), same - side interior angles \( \angle MGH \) and \( \angle CHG \) are supplementary. So:
\( (6x - 34)+52 = 180 \)

Step2: Simplify the Equation

Combine like terms: \( 6x + 18 = 180 \)

Step3: Solve for \( x \)

Subtract 18 from both sides: \( 6x = 180 - 18 \)
\( 6x = 162 \)
Divide both sides by 6: \( x=\frac{162}{6}=27 \)

Answer:

\( x = \boxed{27} \)