Question

Solve the following simultaneous equations: `7/(2X+1)+13/(Y+2)=27,13/(2X+1)+7/(Y+2)=33`

 

 

Answer 1

`7/(2x+1)+13/(y+2)=27`                             .........(1)

`13/(2x+1)+7/(y+2)=33`                           ...........(2)

Substituting`1/(2x+1)=m` and`1/(y+2)=n` in equations (1) and (2), we get

7m+13n=27    .............(3)

and 13m+7n=33  ..........(4)

Adding equations (3) and (4), we get

20m+20n=60

∴m+n=3      ...............(5)

Subtracting equation (3) from equation (4), we get

6m-6n=6

∴m-n= 1        ...........(6)

Adding equations (5) and (6), we get

2m= 4

∴ m= 2

Substituting m = 2 in equation (5), we get

2 - n= 1

∴ n = 1

Resubstituting the values of m and n, we get

`1/(2x+1)=m= 2`

⇒ 2x + 1=`1/2`

⇒ x = `(-1)/4`

and `1/(y+2)=n =1`

⇒ y +2 =1

⇒ y = -1

∴ x=`(-1)/4`  and y= -1.