Question

**#1.) ∠def and ∠feg are linear pairs. the m∠def=(x² + 3x + 69)° and the m∠feg=(4x + 103)°. find the value of x and the measure of each angle.
#2.) (sat question) in the figure below, what is the value of x?
(x + 40)° x°
**#3.) use the diagram below to find the sum of x and y.
75° (2x - 5)°
(4y + 25)°

Explanation:

Step1: Use linear - pair property for #1

Since $\angle DEF$ and $\angle FEG$ are linear pairs, $m\angle DEF + m\angle FEG=180^{\circ}$. So, $(x^{2}+3x + 69)+(4x + 103)=180$.
Simplify the left - hand side: $x^{2}+3x+4x + 69 + 103=180$, which gives $x^{2}+7x+172 = 180$.
Rearrange to get a quadratic equation: $x^{2}+7x - 8=0$.
Factor the quadratic equation: $(x + 8)(x - 1)=0$.
Set each factor equal to zero: $x+8 = 0$ or $x - 1=0$. So, $x=-8$ or $x = 1$.
When $x = 1$: $m\angle DEF=1^{2}+3\times1 + 69=73^{\circ}$, $m\angle FEG=4\times1+103 = 107^{\circ}$.
When $x=-8$: $m\angle DEF=(-8)^{2}+3\times(-8)+69=64-24 + 69=109^{\circ}$, $m\angle FEG=4\times(-8)+103=103 - 32 = 71^{\circ}$.

Step2: Use supplementary - angle property for #2

The two angles $(x + 40)^{\circ}$ and $x^{\circ}$ are supplementary, so $(x + 40)+x=180$.
Combine like terms: $2x+40 = 180$.
Subtract 40 from both sides: $2x=180 - 40=140$.
Divide both sides by 2: $x = 70$.

Step3: Use vertical - angle and supplementary - angle properties for #3

First, since $75^{\circ}$ and $(2x - 5)^{\circ}$ are vertical angles, $2x-5=75$.
Add 5 to both sides: $2x=75 + 5=80$.
Divide both sides by 2: $x = 40$.
Second, since $(4y + 25)^{\circ}$ and $75^{\circ}$ are supplementary, $(4y + 25)+75=180$.
Combine like terms: $4y+100 = 180$.
Subtract 100 from both sides: $4y=180 - 100=80$.
Divide both sides by 4: $y = 20$.
The sum of $x$ and $y$ is $x + y=40+20=60$.

Answer:

For #1: When $x = 1$, $m\angle DEF = 73^{\circ}$, $m\angle FEG = 107^{\circ}$; when $x=-8$, $m\angle DEF = 109^{\circ}$, $m\angle FEG = 71^{\circ}$.
For #2: $x = 70$.
For #3: $x + y=60$.