Math, asked by Anonymous, 3 months ago


\sf{Prove \:  that \:  there \:  is \:  a \:  value  \: of  \: c (\cancel= 0)  \: for \:  which  \: the \:  system}
\sf\red{6x + 3y = c - 3}
\sf\red{12x + cy = c }
\sf{has \:  infinitely \:  many \:  solutions. \:  Find \:  this \:  value}




Answers

Answered by barani7953
1

Step-by-step explanation:

6x + 3y = c - 3

12x + cy = c

has infinitely many solutions. Find this value.

SOLUTION

The given system of equation may be written as

6x + 3y - (c - 3) = 0

12 - cy - c = 0

This is of the form

`a_1x + b_1y + c_1 = 0`

`a_2x + b_2y + c_2 = 0`

Where `a_1 = 6, b_1 = 3 , c_1 = -(c - 3)`

And `a_2 = 12,b_2 = c, c_2 = -c`

For infinitely many solutions, we must have

`a_1/a_2 - b_1/b_2 = c_1/c_2`

`=> 6/12 = 13/c = (-(c - 3))/(-c)`

`=> 6/12 = 13/c = 3/c = (c- 3)/c`

`=> 6c = 12 xx 3 and 3 = (c - 3)`

`=> c = 36/6 and c - 3 = 3`

`=> c = 6 and c = 6`

Now

`a_1/a_2 = 6/12 = 1/2`

`b_1/b_2 = 3/6 = 1/2`

`c_1/c_2 = (-(6-3))/(-6) = 1/2`

`:. a_1/a_2 = b_1/b_2 = c_1/c_2`

Clearly, for this value of c, we have `a_1/a_2 = b_1/b_2 = c_1/c_2`

Hence, the given system of equations has infinitely many solutions if c = 6

Answered by Anonymous
2

Step-by-step explanation:

6x + 3y = c - 3

12x + cy = c

has infinitely many solutions. Find this value.

SOLUTION

The given system of equation may be written as

6x + 3y - (c - 3) = 0

12 - cy - c = 0

This is of the form

`a_1x + b_1y + c_1 = 0`

`a_2x + b_2y + c_2 = 0`

Where `a_1 = 6, b_1 = 3 , c_1 = -(c - 3)`

And `a_2 = 12,b_2 = c, c_2 = -c`

For infinitely many solutions, we must have

`a_1/a_2 - b_1/b_2 = c_1/c_2`

`=> 6/12 = 13/c = (-(c - 3))/(-c)`

`=> 6/12 = 13/c = 3/c = (c- 3)/c`

`=> 6c = 12 xx 3 and 3 = (c - 3)`

`=> c = 36/6 and c - 3 = 3`

`=> c = 6 and c = 6`

Now

`a_1/a_2 = 6/12 = 1/2`

`b_1/b_2 = 3/6 = 1/2`

`c_1/c_2 = (-(6-3))/(-6) = 1/2`

`:. a_1/a_2 = b_1/b_2 = c_1/c_2`

Clearly, for this value of c, we have `a_1/a_2 = b_1/b_2 = c_1/c_2`

Hence, the given system of equations has infinitely many solutions if c = 6

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