Question

For which values of c and d does the following pair of equations have infinitely
many solutions.
4x+7y=13
(2d-c)x + (3d-c)y= c+3d+2

Answer 1

Given:

4x+7y=13

(2d-c)x + (3d-c)y= c+3d+2

To find:

For which values of c and d does the above pair of equations have infinitely many solutions.

Solution:

The condition for the above pair of equations to have infinitely many solutions is,

a1/a2 = b1/b2 = c1/c2

where, from given we have,

a1 = 4 and a2 = 2d-c

b1 = 7 and b2 = 3d-c

c1 = -13 and c2 = -(c+3d+2)

so, we have,

4/(2d-c) = 7/(3d-c) = -13/-(c+3d+2)

now consider,

4/(2d-c) = 7/(3d-c)  

solving the above equation, we get,

c = 2d/3  .......(1)

now consider,

7/(3d-c) = -13/-(c+3d+2)

solving the above equation, we get,

c = (9d-7)/10  .......(2)

solving equations (1) and (2), we get

d = 3

substituting the value of d in any one of the equations, we get,

c = 2

Therefore, the values of c and d that makes the given pair of equations have infinitely many solutions are 2 and 3 respectively.