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