Question

1. if m∠deg=(5x - 4)°, m∠gef=(7x - 8)°, m∠deh=(9y + 5)°, find the values of x and y.
2. ∠r and ∠s are complementary angles. if m∠r=(12x - 3)° and m∠s=(7x - 2)°, find m∠r.
3. ∠p and ∠q are supplementary angles. if m∠p=(4x + 1)° and m∠q=(9x - 3)°, find m∠q.
4. ∠1 and ∠2 form a linear pair. the measure of ∠2 is six more than twice the measure of ∠1. find m∠2.
5. ∠j and ∠k are complementary angles. the measure of ∠j is 18 less than the measure of ∠k. find the measure of each angle.
6. if (overline{uw}) bisects ∠tuv, m∠tuw=(13x - 5)° and m∠wuv=(7x + 31)°, find the value of x.
7. if (overline{mo}) bisects ∠pmn, m∠pmn = 74° and m∠omn=(2x + 7)°, find the value of x.
8. if (overline{ef}) bisects ∠ceb, m∠cef=(7x + 21)° and m∠feb=(10x - 3)°, find the measure of ∠deb.

Explanation:

Step1: Use angle - relationship equations

1. Since $\angle R$ and $\angle S$ are complementary angles, $m\angle R + m\angle S=90^{\circ}$. So, $(12x - 3)+(7x - 2)=90$.

Combine like - terms: $12x+7x-3 - 2 = 90$, which simplifies to $19x-5 = 90$.
Add 5 to both sides: $19x=90 + 5=95$.
Divide both sides by 19: $x = 5$.
Then $m\angle R=(12x - 3)^{\circ}=(12\times5-3)^{\circ}=(60 - 3)^{\circ}=57^{\circ}$.

1. Since $\angle P$ and $\angle Q$ are supplementary angles, $m\angle P + m\angle Q = 180^{\circ}$. So, $(4x + 1)+(9x - 3)=180$.

Combine like - terms: $4x+9x+1 - 3 = 180$, which simplifies to $13x-2 = 180$.
Add 2 to both sides: $13x=180 + 2=182$.
Divide both sides by 13: $x = 14$.
Then $m\angle Q=(9x - 3)^{\circ}=(9\times14-3)^{\circ}=(126 - 3)^{\circ}=123^{\circ}$.

1. Since $\angle1$ and $\angle2$ form a linear pair, $m\angle1 + m\angle2=180^{\circ}$. Let $m\angle1=x$, then $m\angle2=2x + 6$.

So, $x+(2x + 6)=180$.
Combine like - terms: $3x+6 = 180$.
Subtract 6 from both sides: $3x=180 - 6=174$.
Divide both sides by 3: $x = 58$.
Then $m\angle2=(2x + 6)^{\circ}=(2\times58+6)^{\circ}=(116 + 6)^{\circ}=122^{\circ}$.

1. Since $\angle J$ and $\angle K$ are complementary angles, $m\angle J + m\angle K=90^{\circ}$. Let $m\angle K=x$, then $m\angle J=x - 18$.

So, $(x - 18)+x=90$.
Combine like - terms: $2x-18 = 90$.
Add 18 to both sides: $2x=90 + 18=108$.
Divide both sides by 2: $x = 54$.
So, $m\angle K = 54^{\circ}$ and $m\angle J=(54 - 18)^{\circ}=36^{\circ}$.

1. Since $\overline{UW}$ bisects $\angle TUV$, $m\angle TUW=m\angle WUV$. So, $13x-5=7x + 31$.

Subtract $7x$ from both sides: $13x-7x-5=31$, which simplifies to $6x-5 = 31$.
Add 5 to both sides: $6x=31 + 5=36$.
Divide both sides by 6: $x = 6$.

1. Since $\overline{MO}$ bisects $\angle PMN$, $m\angle PMN = 2m\angle OMN$. Given $m\angle PMN = 74^{\circ}$ and $m\angle OMN=(2x + 7)^{\circ}$, then $74 = 2(2x + 7)$.

First, divide both sides by 2: $37=2x + 7$.
Subtract 7 from both sides: $2x=37 - 7=30$.
Divide both sides by 2: $x = 15$.

1. Since $\overline{EF}$ bisects $\angle CEB$, $m\angle CEF=m\angle FEB$. So, $7x + 21=10x-3$.

Subtract $7x$ from both sides: $21=10x-7x-3$, which simplifies to $21 = 3x-3$.
Add 3 to both sides: $3x=21 + 3=24$.
Divide both sides by 3: $x = 8$.
$m\angle CEF=(7x + 21)^{\circ}=(7\times8+21)^{\circ}=(56 + 21)^{\circ}=77^{\circ}$.
$m\angle CEB=2m\angle CEF = 154^{\circ}$.
Assuming $\angle DEB$ and $\angle CEB$ are supplementary (if they form a linear pair), $m\angle DEB=180^{\circ}-m\angle CEB=180 - 154=26^{\circ}$.

Answer:

1. $m\angle R = 57^{\circ}$
2. $m\angle Q = 123^{\circ}$
3. $m\angle2 = 122^{\circ}$
4. $m\angle J = 36^{\circ}$, $m\angle K = 54^{\circ}$
5. $x = 6$
6. $x = 15$
7. $m\angle DEB = 26^{\circ}$