Question

1. determine the values of x and y.
2. ∠1 and ∠2 are complementary. m∠2 is six more than three times the m∠1. find the measure of both angles.

Explanation:

Step1: Set up equation for x

The angles $(5x + 6)^{\circ}$ and $(17x-2)^{\circ}$ are vertical - angles, and vertical angles are equal. So we set up the equation $5x + 6=17x - 2$.
$5x+6 = 17x - 2$
$6 + 2=17x - 5x$
$8 = 12x$
$x=\frac{8}{12}=\frac{2}{3}$

Step2: Find the value of y

The angle $(5x + 6)^{\circ}$ and $y^{\circ}$ are supplementary (a linear - pair), so $(5x + 6)+y = 180$.
Substitute $x = \frac{2}{3}$ into $5x+6$:
$5\times\frac{2}{3}+6=\frac{10}{3}+6=\frac{10 + 18}{3}=\frac{28}{3}$
Then $\frac{28}{3}+y = 180$
$y=180-\frac{28}{3}=\frac{540 - 28}{3}=\frac{512}{3}=170\frac{2}{3}$

Step3: Solve for angles in problem 11

Let $m\angle1=x$ and $m\angle2 = y$.
Since $\angle1$ and $\angle2$ are complementary, $x + y=90$.
Also, $y = 3x+6$.
Substitute $y = 3x + 6$ into $x + y=90$:
$x+(3x + 6)=90$
$4x+6 = 90$
$4x=90 - 6=84$
$x = 21$
$y=3x + 6=3\times21+6=63 + 6=69$

Answer:

$x=\frac{2}{3}$, $y = \frac{512}{3}$; $m\angle1 = 21^{\circ}$, $m\angle2=69^{\circ}$