Question

find the value of 'a' and 'b' for which the following system of linear equations has infinite number of solutions 2x+3y=7, (a+b+1)x + (a+2b+2)y=4(a+b)+1

Answer 1

Hey Mate !

Here is your solution :

=> 2x + 3y = 7

=> 2x + 3y - 7 = 0 ----------- ( 1 )

Here,

=> a1 = 2

=> b1 = 3

=> c1 = -7

And,

=> ( a + b + 1 )x + ( a + 2b + 2 )y = 4( a + b ) + 1

=> ( a + b + 1 )x + ( a + 2b + 2 )y - 4( a + b ) - 1=0 ------------ ( 2 )

Here,

=> a2 = ( a + b + 1 )

=> b2 = ( a + 2b + 2 )

=> c2 = -4( a + b ) - 1 = - [ 4( a + b ) + 1 ]

For any two linear equations in two equations to have infinte solutions,

=> a1/a2 = b1/b2 = c1/c2

=> 2/( a + b + 1 ) = 3/( a + 2b + 2 ) = -7 /-[ 4( a + b ) + 1 ]

Now,

=> 2/( a + b + 1 ) = 3/( a + 2b + 2 )

=> 2( a + 2b + 2 ) = 3( a + b + 1 )

=> 2a + 4b + 4 = 3a + 3b + 3

=> 4 - 3 = 3a - 2a + 3b - 4b

=> 1 = a - b

=> a - b = 1 -------------------- ( 3 )

And,

=> 2/( a + b + 1 ) = -7/-[ 4( a + b ) + 1 ]

=> 2/( a + b + 1 ) = 7/[ 4( a + b ) + 1 ]

=> 2[ 4( a + b ) + 1 ] = 7( a + b + 1 )

=> 8( a + b ) + 2 = 7a + 7b + 7

=> 8a + 8b + 2 = 7a + 7b + 7

=> 8a - 7a + 8b - 7b = 7 - 2

=> a + b = 5 ---------------- ( 4 )

Adding ( 3 ) and ( 4 ),

=> a - b + a + b = 1 + 5

=> 2a = 6

=> a = 6 ÷ 2

=> a = 3

By substituting the value of a in ( 1 ),

=> a - b = 1

=> ( 3 ) - b = 1

=> 3 -b = 1

=> 3 -1 = b

=> 2 = b

Hence, a = 3 and b = 2.

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Hope it helps !! ^_^

Answer 2

Step-by-step explanation:

I hope the pic may help you