Parallel lines intersected by a common transversal form 8 angles
The required values of y and x are;
The reasons why the above values are correct are given as follows:
Given the lines are parallel lines cut by a common transversal, we have;
Angles with measure (2·y - 1)° and (9·x - 38)° are same side interior angles
Angles with measure (4·x + 23)° and (2·y - 1)° are alternate interior angles
Which gives;
(2·y - 1)°+ (9·x - 38)° = 180°...(1)
(4·x + 23)° = (2·y - 1)°...(2)
From equation (1), we get;
y = 109.5 - 4.5·x
From equation (2), we get;
y = 2·x + 12
The common solution is given by; 109.5 - 4.5·x = 2·x + 12
From 109.5 - 4.5·x = 2·x + 12, we get;
x = 15,
6) (9·x + 42)° and 15·x° are vertical angles, and therefore are equal
(9·x + 42)° = 15·x°
∴ x = 7
(4·y - 13)° and (9·x + 42)° are supplementary, therefore
(4·y - 13)° = (9·x + 42)°, which gives;
(4·y - 13)° = (9×7 + 42)° = 105°
7) (3x + 11) + (x + 21) = 180
∴ x = 37
(14·y - 4)° = 3x + 11 = 3 × 37 + 11 = 122°
8) (13·x - 33) = 10·x
∴ x = 11
10·x + 5·y = 180
10 × 11 + 5·y = 180
[tex]y = \dfrac{70^{\circ}}{5} = 14^{\circ}[/tex]
9) (27·x + 4) + (8·x + 1) = 180
∴ x = 5
y + 10 = 8·x + 1 = 8 × 5 + 1 = 41
y = 41 - 10 = 31
10) 16·x - 4 = 9·x - 6 + 58 = 9·x + 52
7·x = 56
x = 8
11) (6·x - 58)° + (x + 8)° = 90°
7·x - 50 = 90
7·x = 90 + 50 = 140
[tex]x = \dfrac{140}{7} = 20[/tex]
12) 7·x - 7 = 3·x + 45°
4·x = 52
[tex]x = \dfrac{52}{4} = 13[/tex]
x = 13°
(3·x + 45)° + 63° + (y - 6)° = 180°
(3×13 + 45)° + 63° + (y - 6)° = 180°
147 + (y - 6)° = 180°
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