help me solve these algebric equations?

Question

Problem #1
JKMO is a parallelogram
JM bisects angle OJK and angle OMK
OJ = x+5, KM = y-3
JK = 2x-4
a.Solve for x.
b.Solve for y.
c.Find the perimeter of OJKM.

Problem #2
The measure of one angle of a parallelogram is 40 more than 3 times another. Find the measure of each angle.

Answer 1

Solution #1.
As its a parallelogram , the opposite sides are equal hence
JK = OM;
OJ = KM
And as the diagnols bisect the angles it is a rhombus and has equal sides.

x +5 = y - 3 (opposite sides are equal)
2x - 4 = x + 5 (all sides are equal)
x = 9
y = 17
the perimeter is 4(OJ) or any other side as they are all equal.
thus
Perimeter = 4(x+5) = 4(9+5) = 56

Solution #2.
As the sum of all the angle is 360º so ,
let one angle be mº and the other be nº so.
m = 40 + 3n ...........eq.1
and 2m + 2n = 360º

m + n = 180º
m = 180º - n ...........eq.2
therefore equating equations 1 & 2.
40 +3n = 180 - n
4n = 140
n = 35º
so, m = 180º - 35º
m = 145º

so your angles are 145º,35º,145º,35º.

Answer 2

1. Because the line bisects the angle, all sides are equal length. Set the side equal to each other and solve.

x = 9
y = 17
P = 56

2. The sum of the two adjacent angles is 180. Solve x +3x +40 = 180. x = 35. Two angles are 35 degrees. The others are 145.

Answer 3

Edit : i am going to write it down and examine :P yet i wager the guy above me is ideal Ah nvm :P nicely bear in mind once you've (A - B)^2 or something similar you do Ax2+ -Bx2 + -Bx2xA = A^2 -B2 + -2BA