How to find c and y so that each quadrilateral is a parallelogram

Question

1st problem asks find x and y so that each quadrilateral is a parallelogram
and gives you these numbers (5x+29)° (5y-9)° (3y+15)° (7x-11)° i am assuming you have to find x and y?
2 is the same thing just with these numbers -4x-2 2y+8 3y-5 -3x+4
Im not quite sure how to figure how to solve these correctly, so if someone can explain or show me how i would be very grateful :)
So http://germantownhs.scsk12.org/~b1edwards/Site/Geometry_Summer_School_files/8%20WS%20answers.pdf has the picture on page three problems 7 an 8 Sadly they have the answers on it too.

Answer 1

1. The opposite internal angles of a parallelogram are equal, and adjacent angles are supplementary, but which of the given angles are opposite and which are adjacent? We know that all angles must be positive, so 7x-11>0 and 5y-9>0 so x>11/7 or 1.57 and y>1.8. We also know that all the angles of a parallelogram can be determined if just one is known, because of the relationships. This means that if we take any pair of angles we know that they are (a) equal or (b) supplementary. Take the first pair: 5x+29 and 5y-9: if (a), 5x+29=5y-9, so 5(y-x)=38 and y=(38+5x)/5; or if (b), 5(y+x)=160, or y+x=32 and y=32-x. Also, the remaining pair 3y+15 and 7x-11: if (a), 7x-3y=26, y=(7x-26)/3; or if (b), 7x+3y=176 and y=(176-7x)/3. We can see that if (a) is applied to the first pair at least one of x or y will contain a fraction. If (a) is applied to the second pair, which can be written 2x-8+(x-2)/3, x must be x=5, 8, 11, ..., 3n+2 for y to be an integer (where n is a positive integer) and y is 3, 10, 17, ..., and if (b), which can be written 58-2x-(x-2)/3, x must be 3n+2 for y to be an integer (where n is an integer 0<n<8, implying x cannot exceed 23 and y is between 5 and 54 (5, 12, 19, ...). So we have to apply (b) to the first pair, that is, the first pair must be supplementary, so (b) must also apply to the second pair. That establishes a relationship between x and y: y=32-x and 7x+3y=176=7x+96-3x (substituting for y). Therefore, 4x=176-96=80 and x=20, so y=12. Therefore the four angles are 5x+29=129, 5y-9=51, 3y+15=51, 7x-11=129.
2. We apply the same logic as in (1). Apply (a) to the first pair: -4x-2=2y+8, 2y+4x=-10, y+2x=-5 and (b): 2y-4x=174, y=87+2x. Apply (a) the second pair: 3y-5=-3x+4, 3(y+x)=9, y+x=3 and (b): 3(y-x)=181. The latter equation would create a fraction in x or y or both, so we conclude that (a) must be applied to both pairs. Therefore, y=-5-2x and y+x=3, and -5-2x=3-x, x=-8 and y=11. So the angles appear to be: -4x-2=30, 2y+8=30, 3y-5=28, -3x+4=28. Clearly, this is wrong! Apply (b) to each pair: y=87+2x, 3(y-x)=3(87+x)=181. 261+3x=181 and x=-80/3 and y=87-160/3=(261-160)/3=101/3. The angles are: -4x-2=320/3-2=314/3, 2y+8=202/3+8=226/3, 3y-5=101-5=96, -3x+4=84. Still wrong! At this point I'm suspicious, so I checked out the link given in the question, and discovered that part (2) was not a set of angles, as implied in the text of the question above, but the lengths of semi-diagonals. The diagonals of a parallelogram bisect one another so there will be two pairs of semi-diagonals with the same length. That's similar to the (a) situation of opposite equal angles and the first solution identified two values 28 and 30, identifying x as -8 and y as 11. These values are the lengths of the semi-diagonals, so the diagonals have lengths 56 and 60. However, the textbook shows in question 8 exactly which measurement applies to each semi-diagonal, whereas the solution above does not assume any connection between the equal expressions, hence the alternative answer. If it had been known which two expressions were used for each diagonal, the logic would have been much simpler: -4x-2=-3x+4, x=-6; 2y+8=3y-5, y=13.