Solve the following pair of linear equations:
152x – 378y = – 74
–378x + 152y = – 604

152x - 378y = -74
-378x + 152y = -604
The given pair of linear equations is
152x - 378y = -74    ...(i)
-378x + 152y = -604    ...(ii)
Adding equation (i) and equation (ii), we get
-226x - 226y = -678
⇒    x + y = 3    ...(iii)
[Dividing throughout by -226]
Subtracting equation (ii) from equation (i), we get
530x - 530y = 530
⇒    x - y = 1    ...(iv)
[Dividing throughout by 530]
Adding equation (iii) and equation (iv), we get
2x = 4


Subtracting equation (iv) from equation (iii), we get
  2y = 2

Hence, the solution of the given pair of linear equations is x = 2, y = 1.

The given system of equations may be written as

Putting the value of x in (ii), we get
(a + b)x + (a + b) y = a² + b²
⇒(a + b) (a + b) + (a + b)y = a² + b²
⇒    (a + b)² (a + b)y = a² + b²
⇒(a + b)y = (a² + b²) - (a + b)²
⇒ (a + b)y = (a² + b²) - (a² + b² + 2ab)
⇒    (a + b)y = a² + b² - a² - b² - 2ab
⇒ (a + b)y = - 2ab

Hence,

   

We know that the opposite angles of a cyclic quadrilateral are supplementary, therefore,
∠A + ∠C = 180°
⇒ 4y + 20 + 4x = 480°
⇒ 4x + 4y = 60°
⇒    x + y = 40°    ...(i)
[Dividing throughout by 4]
and ∠B + ∠D = 180°
⇒3y - 5 + 7x + 5= 180°
⇒ 7x + 3y = 180°    ...(ii)
From equation (i), we have
y = 40 - x    ...(iii)
Substituting this value of y in equation (ii), we get
7x + 3(40 - x) = 180°
⇒7x + 120 - 3x = 180°
⇒    4x = 60

Substituting x = 15 in equation (iii), we get
y = 40 - x
= 40 - 15 = 25°
Hence, required angles be
∠A = 4y + 20 = 4 × 25 + 20 = 120
∠B = 3y - 5 = 3 × 25 - 5 = 75 - 5 = 70
∠C = 4x = 4 × 15 = 60°
∠D = 7x + 5 = 7 × 15 + 5
= 105 + 5 = 110°