| Solve : (PS)= 2 and:=-2 3-+ = 3 andry-25.x+yX- y-and-y-2and-1-10-2y.

1.	\| Solve : (PS)= 2 and:=-2 3-+ = 3 andry-25.x+yX- y-and-y-2and-1-10-2y.y1ਦੇF\|23+)
Answer» 2/x + 3/y = 13 ----(1) 5/x - 4/y = -2 ------(2) If we substitute 1/x = a and 1/y = b , we get the following pair of linear equations: 2a+3b=13 -----(3) 5a-4b = -2 -----(4) Coefficients of b are 3 and 4 and their LCM is 12 . Using the elimination method . Equation (3) × 4 8a+12b=52 Equation (4) × 3 15a-12b=-6 'b' terms have opposite sign , so we add the two equations. 23a = 46 => a = 46/23 => a = 2 Substitute the value of a in equation (3), we get => 2×2 +3b =13 => 4 + 3b = 13 => 3b = 13 - 4 => 3b = 9 => b = 9/3 => b = 3 But 1/x = a = 2 => x = 1/2 1/y = b = 3 => y = 1/3

| Solve : (PS)= 2 and:=-2 3-+ = 3 andry-25.x+yX- y-and-y-2and-1-10-2y.y1ਦੇF|23+)

Answer»

2/x + 3/y = 13 ----(1)

5/x - 4/y = -2 ------(2)

If we substitute 1/x = a and 1/y = b , we get the following pair of linear equations:

2a+3b=13 -----(3)

5a-4b = -2 -----(4)

Coefficients of b are 3 and 4 and

their LCM is 12 .

Using the elimination method .

Equation (3) × 4 8a+12b=52

Equation (4) × 3 15a-12b=-6

'b' terms have opposite sign , so we add the two equations.

23a = 46

=> a = 46/23

=> a = 2

Substitute the value of a in equation (3), we get

=> 2×2 +3b =13

=> 4 + 3b = 13

=> 3b = 13 - 4

=> 3b = 9

=> b = 9/3

=> b = 3

But 1/x = a = 2 => x = 1/2

1/y = b = 3 => y = 1/3