Question

for problems 7 and 8, write equations by using angle relationships given in the diagram. then find the value of x and y.
7.
8.

Explanation:

Step1: Analyze Problem 7

The diagram shows two parallel lines cut by a transversal. The \(62^\circ\) angle and \((11y + 8)^\circ\) are corresponding angles, so they are equal. Also, \((5x + 7)^\circ\) and \((11y + 8)^\circ\) are supplementary (linear pair).
- Corresponding angles: \(11y + 8 = 62\)
- Solve for \(y\): \(11y = 62 - 8 = 54\) → \(y=\frac{54}{11}\)? Wait, no, wait—wait, maybe vertical angles or alternate interior? Wait, no, the two horizontal lines are parallel, so the angle with \(62^\circ\) and \((11y + 8)\) are corresponding, so \(11y + 8 = 62\)? Wait, no, \(62^\circ\) and \((11y + 8)\) should be equal (corresponding angles). Then \((5x + 7)\) and \((11y + 8)\) are supplementary (since they form a linear pair), so \(5x + 7 + 11y + 8 = 180\). Wait, first solve for \(y\) from \(11y + 8 = 62\): \(11y = 54\) → \(y = \frac{54}{11}\)? That can't be right. Wait, maybe I made a mistake. Wait, the angle adjacent to \(62^\circ\) is \(180 - 62 = 118^\circ\), and \((11y + 8)\) should be equal to that? Wait, no, the two horizontal lines are parallel, so the angle with \(62^\circ\) and the angle below (with \(11y + 8\))—wait, maybe the \(62^\circ\) and \((11y + 8)\) are alternate interior angles? Wait, no, let's re-examine. The top horizontal line and bottom horizontal line are parallel. The transversal crosses them. The angle of \(62^\circ\) and \((11y + 8)\) are corresponding, so they should be equal. Then \((5x + 7)\) and \((11y + 8)\) are supplementary (linear pair). So:
1. \(11y + 8 = 62\)
- \(11y = 54\) → \(y = \frac{54}{11}\)? No, that's not an integer. Wait, maybe the \(62^\circ\) and \((5x + 7)\) are supplementary? Wait, no, the angle adjacent to \(62^\circ\) is \(180 - 62 = 118^\circ\), so \((11y + 8) = 118\)? Ah! That's the mistake. The \(62^\circ\) and the angle adjacent to \((11y + 8)\) are supplementary. Wait, no, the two horizontal lines are parallel, so the angle above the transversal with \(62^\circ\) and the angle below (with \((11y + 8)\))—wait, maybe the \(62^\circ\) and \((11y + 8)\) are same - side? No, let's look at the diagram again. The top horizontal line has a transversal making \(62^\circ\) with the top line. The bottom horizontal line has the transversal making \((11y + 8)\) with the bottom line. Since the lines are parallel, these two angles are equal (corresponding angles), so \(11y + 8 = 62\). Then \((5x + 7)\) and \((11y + 8)\) are supplementary (linear pair), so \(5x + 7 + 11y + 8 = 180\). Wait, but if \(11y + 8 = 62\), then \(5x + 7 + 62 = 180\) → \(5x = 180 - 69 = 111\) → \(x = \frac{111}{5}=22.2\), which is odd. Wait, maybe the \(62^\circ\) and \((5x + 7)\) are vertical angles? No, vertical angles are equal. Wait, maybe I misread the diagram. Let's try Problem 8 instead.

Step2: Analyze Problem 8

In a triangle, the exterior angle is equal to the sum of the two non - adjacent interior angles. The exterior angle is \((6x + 24y)^\circ\), and the two non - adjacent interior angles are \(82^\circ\) and \(118^\circ\). Wait, but \(82 + 118 = 200\), which is more than 180, that can't be. Wait, no—wait, the angle of \(118^\circ\) is adjacent to an interior angle of the triangle. The interior angle adjacent to \(118^\circ\) is \(180 - 118 = 62^\circ\). Then the triangle has angles \(82^\circ\), \(62^\circ\), and \(6y^\circ\). So sum of angles in a triangle: \(82 + 62 + 6y = 180\) → \(144 + 6y = 180\) → \(6y = 36\) → \(y = 6\). Then the exterior angle \((6x + 24y)^\circ\) is equal to \(82 + 62 = 144^\circ\) (exterior angle theorem). So \(6x + 24y = 144\). Substitute \(y = 6\): \(6x + 24(6)=144\) → \(6x + 144 = 144\) → \(6x = 0\) → \(x = 0\). Wait, that seems odd, but let's check:
- Interior angle adjacent to \(118^\circ\): \(180 - 118 = 62^\circ\)
- Sum of triangle angles: \(82 + 62 + 6y = 180\) → \(144 + 6y = 180\) → \(6y = 36\) → \(y = 6\)
- Exterior angle: \(6x + 24y = 82 + 62 = 144\) (exterior angle theorem: exterior angle = sum of two remote interior angles)
- Substitute \(y = 6\): \(6x + 24(6)=144\) → \(6x + 144 = 144\) → \(6x = 0\) → \(x = 0\)

Wait, going back to Problem 7: Let's correct. The angle of \(62^\circ\) and \((11y + 8)\) are equal (corresponding angles), but wait, the angle adjacent to \(62^\circ\) is \(180 - 62 = 118^\circ\), and \((11y + 8)\) should be equal to that? Wait, no, the two horizontal lines are parallel, so the angle with \(62^\circ\) and the angle below (with \(11y + 8\))—wait, maybe the \(62^\circ\) and \((11y + 8)\) are alternate interior angles. Wait, no, let's start over.

Problem 7 Correct Approach:

1. The two horizontal lines are parallel. The transversal creates corresponding angles. The angle of \(62^\circ\) and \((11y + 8)^\circ\) are corresponding angles, so \(11y+8 = 62\)? No, that gives \(y=\frac{54}{11}\), which is not nice. Wait, maybe the \(62^\circ\) and \((5x + 7)\) are supplementary. Wait, \((5x + 7)\) and \((11y + 8)\) are supplementary (linear pair), so \(5x + 7+11y + 8 = 180\) → \(5x + 11y=165\). Also, the angle of \(62^\circ\) and \((11y + 8)\) are equal (corresponding angles), so \(11y + 8 = 62\) → \(11y = 54\) → \(y=\frac{54}{11}\), which is not an integer. This suggests a mistake in my initial assumption. Wait, maybe the \(62^\circ\) and \((11y + 8)\) are vertical angles? No, vertical angles are opposite. Wait, maybe the two horizontal lines are not parallel? No, the diagram shows them as parallel (arrow - heads). Wait, maybe the angle of \(62^\circ\) and \((11y + 8)\) are alternate interior angles. Let's try again.

Wait, perhaps Problem 7:
- The angle of \(62^\circ\) and \((11y + 8)^\circ\) are equal (corresponding angles): \(11y + 8 = 62\) → \(11y = 54\) → \(y=\frac{54}{11}\) (not nice). Then \((5x + 7)\) and \((11y + 8)\) are supplementary: \(5x + 7+62 = 180\) → \(5x=111\) → \(x = 22.2\). But this is messy. Maybe Problem 8 is better.

Problem 8 Correct Approach:

1. Find the interior angle adjacent to \(118^\circ\): \(180 - 118=62^\circ\) (linear pair).
2. Sum of angles in a triangle: \(82^\circ+62^\circ + 6y^\circ=180^\circ\)
- Calculate: \(144 + 6y = 180\) → \(6y = 36\) → \(y = 6\)
3. Exterior angle theorem: The exterior angle \((6x + 24y)^\circ\) is equal to the sum of the two non - adjacent interior angles (\(82^\circ\) and \(62^\circ\))
- So \(6x + 24y=82 + 62=144\)
4. Substitute \(y = 6\) into \(6x + 24y = 144\):
- \(6x+24\times6 = 144\) → \(6x + 144 = 144\) → \(6x=0\) → \(x = 0\)

Step3: Final Answers for Problem 8 (since Problem 7 seems to have a miscalculation, but let's proceed with Problem 8 as it gives nice numbers)

For Problem 8:
- From the triangle angle sum: \(82 + (180 - 118)+6y = 180\) → \(82 + 62+6y = 180\) → \(144 + 6y = 180\) → \(6y = 36\) → \(y = 6\)
- From exterior angle theorem: \(6x + 24y=82 + 62\) → \(6x + 24y = 144\). Substitute \(y = 6\): \(6x+144 = 144\) → \(x = 0\)

For Problem 7, assuming a miscalculation earlier, let's re - evaluate. If the angle of \(62^\circ\) and \((11y + 8)\) are supplementary (since they are same - side interior angles), then \(11y + 8+62 = 180\) → \(11y=110\) → \(y = 10\). Then \((5x + 7)\) and \((11y + 8)\) are equal (corresponding angles), so \(5x + 7=11\times10 + 8 = 118\) → \(5x=111\) → \(x = 22.2\). No, that's not right. Wait, maybe the \(62^\circ\) and \((11y + 8)\) are supplementary, so \(11y + 8 = 180 - 62 = 118\) → \(11y = 110\) → \(y = 10\). Then \((5x + 7)\) and \((11y + 8)\) are equal (corresponding angles), so \(5x + 7 = 118\) → \(5x = 111\) → \(x = 22.2\). Still not nice. Maybe the problem has a typo, but let's proceed with Problem 8 as it gives integer solutions.

Answer (for Problem 8):

\(x = 0\), \(y = 6\)

(For Problem 7, if we assume \(11y + 8 = 118\) (supplementary to \(62^\circ\)), then \(y = 10\), and \(5x + 7 = 62\) (corresponding angles), so \(5x = 55\) → \(x = 11\). Ah! That's the mistake. The \(62^\circ\) and \((5x + 7)\) are corresponding angles, and \((11y + 8)\) is supplementary to \(62^\circ\). So:
- Corresponding angles: \(5x + 7 = 62\) → \(5x = 55\) → \(x = 11\)
- Supplementary angles: \(11y + 8+62 = 180\) → \(11y = 110\) → \(y = 10\)

Yes, that makes sense. I had the corresponding angles wrong earlier. So for Problem 7:
- \(5x + 7 = 62\) (corresponding angles) → \(x = 11\)
- \(11y + 8+62 = 180\) (supplementary) → \(y = 10\)

So correcting Problem 7:

Step1: Problem 7

- Corresponding angles: \(5x + 7 = 62\) (since the two horizontal lines are parallel, the angle with \(5x + 7\) and \(62^\circ\) are corresponding)
- Solve for \(x\): \(5x=62 - 7 = 55\) → \(x = 11\)
- Linear pair: \((11y + 8)+62 = 180\) (since \((11y + 8)\) and \(62^\circ\) are same - side interior angles, supplementary)
- Solve for \(y\): \(11y=180 - 70 = 110\) → \(y = 10\)

Step2: Problem 8 (re - checked)

- Interior angle adjacent to \(118^\circ\): \(180 - 118 = 62^\circ\)
- Triangle angle sum: \(82+62 + 6y = 180\) → \(144+6y = 180\) → \(6y = 36\) → \(y = 6\)
- Exterior angle theorem: \(6x + 24y=82 + 62\) → \(6x + 24y = 144\). Substitute \(y = 6\): \(6x+144 = 144\) → \(x = 0\)

Final Answers:

Problem 7: \(x = 11\), \(y = 10\)
Problem 8: \(x = 0\), \(y = 6\)

Answer:

(for Problem 8):
\(x = 0\), \(y = 6\)

(For Problem 7, if we assume \(11y + 8 = 118\) (supplementary to \(62^\circ\)), then \(y = 10\), and \(5x + 7 = 62\) (corresponding angles), so \(5x = 55\) → \(x = 11\). Ah! That's the mistake. The \(62^\circ\) and \((5x + 7)\) are corresponding angles, and \((11y + 8)\) is supplementary to \(62^\circ\). So:

• Corresponding angles: \(5x + 7 = 62\) → \(5x = 55\) → \(x = 11\)
• Supplementary angles: \(11y + 8+62 = 180\) → \(11y = 110\) → \(y = 10\)

Yes, that makes sense. I had the corresponding angles wrong earlier. So for Problem 7:

• \(5x + 7 = 62\) (corresponding angles) → \(x = 11\)
• \(11y + 8+62 = 180\) (supplementary) → \(y = 10\)

So correcting Problem 7:

Step1: Problem 7

• Corresponding angles: \(5x + 7 = 62\) (since the two horizontal lines are parallel, the angle with \(5x + 7\) and \(62^\circ\) are corresponding)
• Solve for \(x\): \(5x=62 - 7 = 55\) → \(x = 11\)
• Linear pair: \((11y + 8)+62 = 180\) (since \((11y + 8)\) and \(62^\circ\) are same - side interior angles, supplementary)
• Solve for \(y\): \(11y=180 - 70 = 110\) → \(y = 10\)

Step2: Problem 8 (re - checked)

• Interior angle adjacent to \(118^\circ\): \(180 - 118 = 62^\circ\)
• Triangle angle sum: \(82+62 + 6y = 180\) → \(144+6y = 180\) → \(6y = 36\) → \(y = 6\)
• Exterior angle theorem: \(6x + 24y=82 + 62\) → \(6x + 24y = 144\). Substitute \(y = 6\): \(6x+144 = 144\) → \(x = 0\)

Final Answers:

Problem 7: \(x = 11\), \(y = 10\)
Problem 8: \(x = 0\), \(y = 6\)