5)
For the figure, lines AB and CD are crossed by a transversal line FE.
If lines AB || CD, then the angles, ∠BHK, and ∠HKD are supplementary angles, which means they add up to 180º.
So we can say that:
[tex]\angle\text{BHK+}\angle\text{HKD}=180º[/tex]We know that
∠BHK= (2x+50)º
∠HKD= (3x)º
Replace both measures in the expression above and you can determine an equation with "x" as the only unknown
[tex](2x+50)+(3x)=180[/tex]Order the like terms together and simplify
[tex]\begin{gathered} 2x+3x+20=180 \\ 5x+20=180 \end{gathered}[/tex]Now you can determine the value of x. First, pass 50 to the other side of the equal sign by applying the opposite operation to both sides of it:
[tex]\begin{gathered} 5x+50-20=180-50 \\ 5x=130 \end{gathered}[/tex]Second, divide both sides of the expression by 5 to reach the value of x:
[tex]\begin{gathered} \frac{5x}{5}=\frac{130}{5} \\ x=26 \end{gathered}[/tex]Once determined the value of x, you can calculate the measures of the angles ∠BHK and ∠HKD
6)
[tex]\begin{gathered} \angle BHK=2x+50 \\ \angle BHK=2\cdot26+50 \\ \angle BHK=52+50 \\ \angle BHK=102º \end{gathered}[/tex]7)
[tex]\begin{gathered} \angle HKD=3x \\ \angle HKD=3\cdot26 \\ \angle HKD=78º \end{gathered}[/tex]