Question

dm - topic 2: practice #3
due september 15 at 11:59 pm
grade: 100%
finding angles in transversal problems (level 2)
transversal problems with equations (level 1)
transversal problems with equations (level 2)
answer attempt 1 out of 2
$x = square$
$y = square$

Explanation:

Step1: Use the property of corresponding - angles

Since the angles \((3x - 20)^{\circ}\) and \((2x + 3)^{\circ}\) are corresponding - angles (assuming lines \(m\) and \(n\) are parallel), they are equal. So we set up the equation \(3x-20 = 2x + 3\).
\[3x-2x=3 + 20\]
\[x=23\]

Step2: Use the property of linear - pair

The angles \((3x - 20)^{\circ}\) and \((y - 6)^{\circ}\) form a linear - pair, so their sum is \(180^{\circ}\). First, substitute \(x = 23\) into \((3x - 20)\): \(3\times23-20=69 - 20=49^{\circ}\). Then, since \(49+(y - 6)=180\), we have \(y-6=180 - 49\), \(y-6 = 131\), \(y=137\).

Answer:

\(x = 23\), \(y = 137\)