Question

directions: if ( lparallel m ), solve for ( x ) and ( y ).
9.
10.

Explanation:

Step1: Solve for x in problem 9

Set equal corresponding - angles:
$9x + 25=13x - 19$
Add 19 to both sides:
$9x+25 + 19=13x-19 + 19$
$9x + 44=13x$
Subtract 9x from both sides:
$9x-9x + 44=13x-9x$
$44 = 4x$
Divide both sides by 4:
$x=\frac{44}{4}=11$

Step2: Solve for y in problem 9

Since $x = 11$, we know that one of the angles is $9x+25=9\times11 + 25=99 + 25=124^{\circ}$
The angle $(17y + 5)^{\circ}$ is supplementary to the $124^{\circ}$ angle (linear - pair).
So, $17y+5+124 = 180$
$17y+129 = 180$
Subtract 129 from both sides:
$17y=180 - 129$
$17y = 51$
Divide both sides by 17:
$y=\frac{51}{17}=3$

Step3: Solve for x in problem 10

Set equal corresponding - angles:
$3x-29=6x - 7$
Add 29 to both sides:
$3x-29 + 29=6x-7 + 29$
$3x=6x + 22$
Subtract 6x from both sides:
$3x-6x=6x-6x + 22$
$-3x=22$
$x=-\frac{22}{3}$

Step4: Solve for y in problem 10

Since $x =-\frac{22}{3}$, one of the angles is $3x-29=3\times(-\frac{22}{3})-29=-22-29=-51^{\circ}$ (we take the absolute value for angle measure in this context).
The angle $(8y + 17)^{\circ}$ is equal to the angle $3x - 29$ (corresponding angles).
So, $8y+17=3x - 29$
Substitute $x =-\frac{22}{3}$:
$8y+17=3\times(-\frac{22}{3})-29$
$8y+17=-22-29$
$8y+17=-51$
Subtract 17 from both sides:
$8y=-51 - 17$
$8y=-68$
Divide both sides by 8:
$y=-\frac{68}{8}=-\frac{17}{2}$

Answer:

For problem 9: $x = 11,y = 3$
For problem 10: $x=-\frac{22}{3},y =-\frac{17}{2}$