Question

lines m and n with a transversal, angles: (6x-16)°, (3x+17)°, (2y+14)°. answer attempt 2 out of 2. x = box, y = box

Explanation:

Step1: Solve for \( x \)

Since lines \( m \) and \( n \) are parallel (implied by the corresponding angles being equal), the angle \( (6x - 16)^\circ \) and \( (3x + 17)^\circ \) are equal (corresponding angles). So we set up the equation:
\( 6x - 16 = 3x + 17 \)
Subtract \( 3x \) from both sides:
\( 6x - 3x - 16 = 3x - 3x + 17 \)
\( 3x - 16 = 17 \)
Add 16 to both sides:
\( 3x - 16 + 16 = 17 + 16 \)
\( 3x = 33 \)
Divide both sides by 3:
\( x = \frac{33}{3} = 11 \)

Step2: Solve for \( y \)

First, find the measure of the angle \( (3x + 17)^\circ \) by substituting \( x = 11 \):
\( 3(11) + 17 = 33 + 17 = 50^\circ \)
Now, the angle \( (3x + 17)^\circ \) and \( (2y + 14)^\circ \) are supplementary (they form a linear pair), so their sum is \( 180^\circ \). Wait, no, actually, looking at the diagram, \( (3x + 17)^\circ \) and \( (2y + 14)^\circ \) are adjacent and form a linear pair? Wait, no, actually, since \( (6x - 16)^\circ = (3x + 17)^\circ = 50^\circ \), and \( (2y + 14)^\circ \) and \( (3x + 17)^\circ \) are supplementary? Wait, no, maybe \( (2y + 14)^\circ \) is equal to \( (6x - 16)^\circ \)? Wait, no, let's re-examine. Wait, actually, \( (3x + 17)^\circ \) and \( (2y + 14)^\circ \) are supplementary? Wait, no, the angle \( (3x + 17)^\circ \) and \( (2y + 14)^\circ \) are adjacent and form a linear pair, so they should add up to \( 180^\circ \)? Wait, no, that can't be. Wait, no, actually, \( (3x + 17)^\circ \) and \( (2y + 14)^\circ \) are vertical angles? No, wait, the correct approach: since \( (6x - 16)^\circ \) and \( (3x + 17)^\circ \) are equal (corresponding angles), so \( x = 11 \), so \( (6x - 16) = 50^\circ \). Then, \( (2y + 14)^\circ \) and \( (6x - 16)^\circ \) are equal? Wait, no, maybe \( (2y + 14)^\circ \) is equal to \( (6x - 16)^\circ \) because they are corresponding angles? Wait, no, let's look at the diagram again. The two angles \( (6x - 16)^\circ \) and \( (2y + 14)^\circ \) are actually equal because they are corresponding angles? Wait, no, maybe \( (3x + 17)^\circ \) and \( (2y + 14)^\circ \) are supplementary. Wait, let's calculate \( (3x + 17)^\circ \) when \( x = 11 \): \( 3*11 + 17 = 50^\circ \). Then, \( (2y + 14)^\circ \) should be equal to \( 180 - 50 = 130^\circ \)? No, that doesn't make sense. Wait, no, maybe I made a mistake. Wait, actually, \( (3x + 17)^\circ \) and \( (2y + 14)^\circ \) are adjacent and form a linear pair, so they add up to \( 180^\circ \). Wait, no, that would mean \( 50 + (2y + 14) = 180 \), so \( 2y + 64 = 180 \), \( 2y = 116 \), \( y = 58 \). But that seems off. Wait, no, maybe \( (6x - 16)^\circ \) and \( (2y + 14)^\circ \) are equal. Let's check: \( 6x - 16 = 2y + 14 \). We know \( x = 11 \), so \( 6*11 - 16 = 66 - 16 = 50 \). So \( 50 = 2y + 14 \), so \( 2y = 36 \), \( y = 18 \). Ah, that makes more sense. So the correct approach is: \( (6x - 16)^\circ \) and \( (2y + 14)^\circ \) are equal (corresponding angles), and \( (6x - 16)^\circ = (3x + 17)^\circ \) (corresponding angles). So first, solve \( 6x - 16 = 3x + 17 \) to get \( x = 11 \). Then, substitute \( x = 11 \) into \( 6x - 16 \) to get \( 50 \), then set \( 2y + 14 = 50 \) to solve for \( y \).

So, solving \( 2y + 14 = 50 \):
Subtract 14 from both sides:
\( 2y + 14 - 14 = 50 - 14 \)
\( 2y = 36 \)
Divide by 2:
\( y = \frac{36}{2} = 18 \)

Answer:

\( x = 11 \), \( y = 18 \)