directions: if ( l parallel m ), solve for ( x ) and ( y ).
9.
(9x + 25)° (13x − 19)° (17y + 5)° with lines ( l ) and ( m ).
10.
(3x − 29)° (8y + 17)° (6x − 7)° with lines ( l ) and ( m ).
11.
49° (7x − 23)° (3x)° (11y − 1)° with lines ( l ) and ( m ).
12.
(7y − 20)° (5x − 38)° (3x − 4)° with a right angle and lines ( l ) and ( m ).

Question

Accepted Answer

Let's solve problem 9 first. Since $ l \parallel m $, the corresponding angles are equal. So we set $ 9x + 25 = 13x - 19 $ (assuming they are corresponding angles, or alternate interior angles, depending on the diagram).

### Step 1: Solve for $ x $
We have the equation $ 9x + 25 = 13x - 19 $.
Subtract $ 9x $ from both sides: $ 25 = 4x - 19 $.
Add 19 to both sides: $ 44 = 4x $.
Divide both sides by 4: $ x = 11 $.

### Step 2: Solve for $ y $
Now that we know $ x = 11 $, we can find the measure of the angle $ 13x - 19 $ (or $ 9x + 25 $). Let's calculate $ 13(11) - 19 = 143 - 19 = 124^\circ $? Wait, no, wait: $ 9(11) + 25 = 99 + 25 = 124^\circ $? Wait, no, maybe the angles are supplementary? Wait, maybe I made a mistake. Wait, if $ l \parallel m $, and the transversal cuts them, maybe the angles $ (9x + 25)^\circ $ and $ (13x - 19)^\circ $ are alternate interior angles, so they should be equal. Wait, but let's check the calculation again.

Wait, $ 9x + 25 = 13x - 19 $
$ 25 + 19 = 13x - 9x $
$ 44 = 4x $
$ x = 11 $. Then the angle is $ 9(11) + 25 = 99 + 25 = 124^\circ $. Now, the angle $ (17y + 5)^\circ $ and the angle we just found (124°) are supplementary? Wait, no, maybe they are same - side interior angles? Wait, the diagram shows that $ (17y + 5)^\circ $ and the angle $ (13x - 19)^\circ $ are adjacent? Wait, maybe I misread the diagram. Wait, the problem says "If $ l \parallel m $, solve for $ x $ and $ y $". Let's assume that $ (9x + 25)^\circ $ and $ (13x - 19)^\circ $ are equal (alternate interior angles), so we found $ x = 11 $. Then, the angle $ (17y + 5)^\circ $ and $ (9x + 25)^\circ $ are supplementary (since they are same - side interior angles or linear pair? Wait, maybe the angle $ (17y + 5)^\circ $ and the angle $ (9x + 25)^\circ $ are supplementary. So $ (9x + 25)+(17y + 5)=180 $. We know $ x = 11 $, so $ 9(11)+25 + 17y+5 = 180 $
$ 99 + 25+5+17y = 180 $
$ 129+17y = 180 $
$ 17y = 180 - 129 = 51 $
$ y = 3 $

### Answer for problem 9:
$ x = 11 $, $ y = 3 $

Now let's solve problem 10. Since $ l \parallel m $, the alternate interior angles are equal. So $ 3x - 29 = 6x - 7 $ (assuming they are alternate interior angles).

### Step 1: Solve for $ x $
We have the equation $ 3x - 29 = 6x - 7 $
Subtract $ 3x $ from both sides: $ - 29 = 3x - 7 $
Add 7 to both sides: $ - 22 = 3x $? Wait, that can't be right. Maybe the angles $ (3x - 29)^\circ $ and $ (6x - 7)^\circ $ are supplementary? Wait, if $ l \parallel m $, and the transversal cuts them, maybe $ (3x - 29)+(8y + 17)+(6x - 7)=180 $? No, maybe the angle $ (8y + 17)^\circ $ and $ (6x - 7)^\circ $ are equal? Wait, let's re - examine the diagram. The lines $ l $ and $ m $ are parallel, and there are two transversals? Wait, maybe the angle $ (3x - 29)^\circ $ and $ (6x - 7)^\circ $ are alternate interior angles, but if that gives a negative value, maybe they are same - side interior angles. So $ (3x - 29)+(6x - 7)=180 $
$ 9x-36 = 180 $
$ 9x=180 + 36=216 $
$ x = 24 $

### Step 2: Solve for $ y $
Now, the angle $ (8y + 17)^\circ $ and $ (3x - 29)^\circ $ are supplementary? Wait, no, if $ l \parallel m $, and the transversal forms a triangle? Wait, maybe the angle $ (8y + 17)^\circ $ and $ (6x - 7)^\circ $ are equal. Wait, we found $ x = 24 $, so $ 6x - 7=6\times24 - 7 = 144 - 7 = 137^\circ $. Then $ 8y+17 = 180 - 137=43 $? No, that doesn't make sense. Wait, maybe I made a mistake in identifying the angles. Let's try again.

If $ l \parallel m $, then the angle $ (3x - 29)^\circ $ and $ (6x - 7)^\circ $ are alternate interior angles, so $ 3x - 29=6x - 7$ gives $ - 3x=22$, $ x =-\frac{22}{3}$, which is not possible. So they must be same - side interior angles, so $ (3x - 29)+(6x - 7)=180$
$ 9x-36 = 180$
$ 9x = 216$
$ x = 24$

Now, the angle $ (8y + 17)^\circ $ and $ (3x - 29)^\circ $ are vertical angles or something? Wait, $ 3x - 29=3\times24 - 29 = 72 - 29 = 43^\circ $. Then, if $ (8y + 17)^\circ $ is equal to $ (3x - 29)^\circ $ (alternate interior angles with another transversal), then $ 8y+17 = 43$
$ 8y=43 - 17 = 26$
$ y=\frac{26}{8}=\frac{13}{4}=3.25$

### Answer for problem 10:
$ x = 24 $, $ y = 3.25 $ (or $ \frac{13}{4} $)

For problem 11: We have a 49° angle, $ (3x)^\circ $, $ (7x - 23)^\circ $ and $ (11y - 1)^\circ $. Since vertical angles are equal, and linear pairs sum to 180°. Let's assume that $ 3x+49 + 7x - 23=180 $ (since they are on a straight line).

### Step 1: Solve for $ x $
$ 10x + 26 = 180 $
$ 10x=180 - 26 = 154 $
$ x = 15.4 $? Wait, no, maybe $ 3x = 49 $? No, that doesn't fit. Wait, maybe $ 7x - 23 = 49 $ (vertical angles or corresponding angles). So $ 7x=49 + 23 = 72 $
$ x=\frac{72}{7}\approx10.29 $. No, this is getting confusing. Wait, the lines $ l $ and $ m $ are parallel, so $ (11y - 1)^\circ $ and $ 49^\circ $ are equal (corresponding angles). So $ 11y-1 = 49 $
$ 11y=50 $
$ y=\frac{50}{11}\approx4.55 $

And for $ x $, maybe $ 3x+7x - 23 = 180 - 49 $ (since the sum of angles around a point is 360, but maybe a straight line). $ 10x-23 = 131 $
$ 10x=154 $
$ x = 15.4 $

### Answer for problem 11:
$ x = 15.4 $, $ y=\frac{50}{11}$ (or approximately $ x = 15.4 $, $ y\approx4.55 $)

For problem 12: Since $ l \parallel m $, and there is a right angle, the sum of the angles in the triangle is 180°. So $ (7y - 20)+(5x - 38)+(3x - 4)+90 = 180 $? Wait, no, the right angle is between the two non - parallel sides. So the angles $ (7y - 20)^\circ $, $ (5x - 38)^\circ $ and $ (3x - 4)^\circ $ form a triangle with a right angle. So $ (7y - 20)+(5x - 38)+(3x - 4)=90 $ (since the right angle is 90°, and the sum of angles in a triangle is 180, so the other two angles plus the right angle? Wait, no, the lines $ l $ and $ m $ are parallel, so the angle $ (5x - 38)^\circ $ and $ (3x - 4)^\circ $ are related. Wait, maybe $ 5x - 38+3x - 4 = 90 $ (since they are same - side interior angles with the right angle). So $ 8x-42 = 90 $
$ 8x=132 $
$ x = 16.5 $

Then, $ 7y - 20=90-(5x - 38 + 3x - 4) $? No, maybe $ 7y - 20 $ and $ (3x - 4)^\circ $ are equal (alternate interior angles). So $ 7y - 20=3x - 4 $. We know $ x = 16.5 $, so $ 7y-20 = 3\times16.5-4=49.5 - 4 = 45.5 $
$ 7y=45.5 + 20 = 65.5 $
$ y=\frac{65.5}{7}\approx9.36 $

### Answer for problem 12:
$ x = 16.5 $, $ y\approx9.36 $ (or $ x=\frac{33}{2}$, $ y=\frac{131}{14}$)

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Explanation:

Step 1: Solve for \( x \)

Step 2: Solve for \( y \)

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Step 1: Solve for \( x \)

Step 2: Solve for \( y \)