Question

2. Find the values of a and b for which the following system of equations has infinite number of solutions:

$5x + 3y = 15$
,

$(a + b)x + (4a + b)y = 1$

Standard:- 10

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Answer 1

Heya !!

5X + 3Y - 15 = 0------(1)

And,

( a + b )X + (4a + b )Y = 1

( a + b )X + (4a + b )Y - 1 = 0 ---------(2)

These equations are of the form of A1X + B1Y + C1 = 0 and A2X + B2Y + C2 = 0

Where,

A1 = 5 , B1 = 3 and C1 = -15

And,

A2 = (a + b ) , B2 = (4a + b ) and C2 = -1.

Therefore,

A1/A2 = 5/a + b , B1/B2 = 3/(4a + b ) and C1/C2 = -15/-1 = 15

The given equations have infinitely many solutions.

Then , A1/A2 = B1/B2 = C1/C2

=> 5/a + b = 3/(4a + b ) = 15/1

=> 5/a + b = 3/4a + b and 3/4a + b = 15

=> 5(4a + b ) = 3 ( a + b) and 15 ( 4a +b ) = 3

=> 20a +5b = 3a + 3b and 60a + 15b = 3

=> 17a + 2b = 0 -----------(1)

And,

60a + 15b - 3 = 0 ----------(2)

From equation (1) we get,

17a + 2b = 0

17a = -2b

a = -2b/17 ----(3)

Putting the value of A in equation (2)

60a + 15b = 3

60 × -2b/17 + 15b = 3

-120b /17 + 15b = 3

-120b + 255b = 51

135b = 51

b = 51/135

b = 17/45

Putting the value of B in equation (3)

A = -2b / 17 = -2 × 17/45 / 17

A = -34/45/17 = -34/45 × 1/17

A = -2/45


Hence,



A = -2/45 and B = 17/45