(5) In question 5, the angles marked 3x + 10 and 5x + 90 on the two parallel lines are not congruent but they are SUPPLEMENTARY angles. This is because angle 3x + 10 and the angle next to it both lie on a straight line and that means they both sum up to 180 degrees. Observe carefully that the angle that lies next to angle 3x + 10 is corresponding to angle 5x + 90. Let the unlabelled angle (beside 3x + 10) be angle A, then this means;
[tex]\begin{gathered} \angle3x+10+\angle A=180 \\ \angle A\cong\angle5x+90 \\ \text{If therefore (3x+10)}+A=180,\text{ then} \\ 3x+10+5x+90=180 \\ (\text{This is because A}\cong5x+90) \\ 3x+10+5x+90=180 \\ 8x+100=180 \\ \text{Subtract 100 from both sides of the equation} \\ 8x=80 \\ \text{Divide both sides of the equation by 8} \\ x=10 \end{gathered}[/tex]
When x = 10, then,
3x + 10 = 3(10) + 10
3x + 10 = 30 + 10
3x + 10 = 40
Also
5x + 90 = 5(10) + 90
5x + 90 = 50 + 90
5x + 90 = 140
(6) In question number 6, the two parallel lines are cut by a transversal and that forms a right angle as indicated. Observe that the angle indicated as a right angle is CONGRUENT to angle 3p - 6. This means;
[tex]\begin{gathered} 3p-6=90 \\ \text{Add 6 to both sides of the equation;} \\ 3p=96 \\ \text{Divide both sides of the equation by 3} \\ p=32 \end{gathered}[/tex]
If p = 32, then
3p - 6 = 3(32) - 6
3p - 6 = 96 - 6
3p - 6 = 90
