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\({7 \over 2x+1}+{13 \over y+2}=27\) ;

\({13 \over 2x+1} + {7 \over y+2 } =33\)

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\(7({1 \over 2x+1}) + 13({1 \over y+ 2})=27\) ...(I)

\(13({1 \over 2x +1})+ {7({1 \over y+2})}=33\) ...(II)

Replace \(1\over 2x+1\) by a and \(1 \over y+2\) by b in equations (I) and (II)

we get,

7a + 13b = 27 ...(III)

13a + 7b = 33 ...(IV)

[In the two equations above, the coefficients of x and y are interchanged so use ASA]

Adding eq. (III) and (IV)

7a + 13 b = 27

__+ 13a +7 b = 33 __

20a +20 b = 60

a + b = 3 ...(V)

Subtracting eq. (III) and (IV)

7a +13b = 27

13a +7 b = 33

__(–) (–) (–)__

–6a +6 b = –6

–a + b = –1 ...(VI)

Adding eq. (V) and (VI)

a + b =3

__+ –a + b = –1__

2b = 2

b =1

Place b = 1 in eq. (V)

a + 1 = 3

∴ a = 3 – 1

∴ a =2

Resubstituting the values of a and b

\({1 \over 2x+1 } =2\) and \({1 \over y+2} = 1\)

2x + 1 = \(1 \over 2\) and y + 2 = 1

\(2x={1\over 2}-1\) and y =1 – 2

∴ \(2x ={-1\over 2}\) and ∴ y = –1

∴\(x={-1\over 4}\)

∴ (x, y) = \(({-1 \over 4}, -1)\) is the required solutions.