Solve the following pairs of equations by reducing them to a pair of linear equations:
(i) \({{1} \over {2x}} + {{1} \over {3y}} = 2\)
\({{1} \over {3x}} + {{1} \over {2y}} = {{13} \over {6}}\)

(ii) \({{2} \over {\sqrt{x}}} + {{3} \over {\sqrt{y}}} = 2\)
\({{4} \over {\sqrt{x}}} - {{9} \over {\sqrt{y}}} = -1\)

(iii) \({{4} \over {x}} + 3y = 14\)
\({{3} \over {x}} - 4y = 23\)

(iv) \({{5} \over {x - 1}} + {{1} \over {y - 2}} = 2\)
\({{6} \over {x - 1}} - {{3} \over {y - 2}} = 1\)

(v) \(7x - 2y = 5xy\)
\(8x + 7y = 15xy\)

(vi)\(6x + 3y - 6xy = 0\)
\(2x + 4y - 5xy = 0\)

(vii) \({{10} \over {x + y}} + {{2} \over {x - y}} = 4\)
\({{15} \over {x + y}} - {{5} \over {x - y}} = -2\)

(viii) \({{1} \over {3x + y}} + {{1} \over {3x - y}} = {{3} \over {4}}\)
\({{1} \over {2(3x + y)}} - {{1} \over {2(3x - y)}} = {{-1} \over {8}}\)

(i) \({{1} \over {2x}} + {{1} \over {3y}} = 2\)… (1)
\({{1} \over {3x}} + {{1} \over {2y}} = {{13} \over {6}}\) … (2)
Let \({{1} \over {x}}\) = p and \({{1} \over {y}}\) = q

Putting this in equation (1) and (2),
=>\({{p} \over {2}} + {{q} \over {3}} = 2 \) and \({{p} \over {3}} + {{q} \over {2}} = {{13} \over {6}}\)
=>\( 3p + 2q = 12\) and \(2p + 3q =13\)
=>\( 3p + 2q – 12 = 0 \)… (3) and \(2p + 3q – 13 = 0\) … (4)

Using cross multiplication
=>\({{p} \over {(2)(-13) - 3(-12)}} = {{q} \over {(-12)(2) - (-13)(3)}} = {{1} \over {(3)(3) - (2)(2)}}\)
=> \({{p} \over {-26 + 36}} = {{q} \over {-24 + 39}} = {{1} \over {9 - 4}}\)
=>\({{p} \over {10}} = {{q} \over {15}} = {{1} \over {5}}\)
=> p = 2 and q = 3

But \({{1} \over {x}}\) = p and \({{1} \over {y}}\) = q

Putting value of p and q in this
x = \({{1} \over {2}}\) and y = \({{1} \over {3}}\)


(ii)\({{2} \over {\sqrt{x}}} + {{3} \over {\sqrt{y}}} = 2 \)… (1)
\({{4} \over {\sqrt{x}}} - {{9} \over {\sqrt{y}}} = -1\) … (2)
Let \({{1} \over {\sqrt{x}}}\) = p and \({{1} \over {\sqrt{y}}}\)= q

Putting this in (1) and (2),
\(2p + 3q = 2\) … (3)
\(4p - 9q = -1 \)… (4)

Multiplying (3) by 2 and subtracting it from (4),
\(4p - 9q + 1 – 2 (2p + 3q - 2) = 0\)
=>\( 4p - 9q + 1 - 4p - 6q + 4 = 0\)
=>\( -15q + 5 = 0\)
=>\( q = {{5} \over {15}} = {{1} \over {3}}\)

Putting value of q in (3),
2p + 1 = 2
=> 2p = 1
=> p \( = {{1} \over {2}}\)

Putting values of p and q in (\({{1} \over {\sqrt{x}}}\)= p and \({{1} \over {\sqrt{y}}}\)= q), we get

\({{1} \over {\sqrt{x}}} = {{1} \over {2}}\) and \({{1} \over {\sqrt{y}}} = {{1} \over {3}}\)
=>\({{1} \over {x}} = {{1} \over {4}}\) and \({{1} \over {y}} = {{1} \over {9}}\)
=>\( x = 4\) and \(y = 9\)


(iii) \({{4} \over {x}} + 3y = 14\) … (1)
\({{3} \over {x}} - 4y = 23\) … (2) and Let \({{1} \over {x}}= p\) … (3)

Putting (3) in (1) and (2),
\(4p + 3y = 14 \)… (4)
\(3p - 4y = 23 \)… (5)

Multiplying (4) by 3 and (5) by 4,
\(3 (4p + 3y – 14 = 0) \) and, \(4 (3p - 4y – 23 = 0)\)
=>\( 12p + 9y – 42 = 0 \)… (6) \(12p - 16y – 92 = 0\) … (7)

Subtracting (7) from (6),
\(9y - (-16y) – 42 - (-92) = 0\)
\(=> 25y + 50 = 0\)
\(=> y = \frac{-50}{25} = -2\)

Putting value of y in (4),
\(4p + 3 (-2) = 14\)
\(=> 4p – 6 = 14\)
\(=> 4p = 20\)
\(p = 5\)

Putting value of p in (3),
\({{1} \over {x}} = 5 => x = {{1} \over {5}}\)

Therefore, x = \({{1} \over {5}}\) and y = -2


(iv) \({{5} \over {x - 1}} + {{1} \over {y - 2}} = 2 \) … (1)
\({{6} \over {x - 1}} - {{3} \over {y - 2}} = 1\)… (2)
Let \({{1} \over {x - 1}} = p\) and \( {{1} \over {y - 2}} = q\)

Putting this in (1) and (2),
\(5p + q = 2\)
\(5p + q – 2 = 0\) … (3)
And, \(6p - 3q = 1\)
\(6p - 3q – 1 = 0 \)… (4)

Multiplying (3) by 3 and adding it to (4),
\(3 (5p + q - 2) + 6p - 3q – 1 = 0\)
\(15p + 3q – 6 + 6p - 3q – 1 = 0\)
\(21p – 7 = 0\)
\(p = {{1} \over {3}}\)

Putting this in (3),
\(5 ({{1} \over {3}}) + q – 2 = 0\)
\(5 + 3q = 6\)
\(3q = 6 – 5 = 1\)
\(q = {{1} \over {3}}\)

Putting values of p and q in (\({{1} \over {x - 1}} = p\) and \( {{1} \over {y - 2}} = q\)),
\({{1} \over {x - 1}} = {{1} \over {3}}\) and \( {{1} \over {y - 2}} = {{1} \over {3}}\)
\(3 = x - 1\) and \(3 = y – 2\)
\(x = 4\) and \(y = 5\)


(v) \(7x - 2y = 5xy\) … (1)
\(8x + 7y = 15xy \)… (2)

Dividing both the equations by xy,
\({{7} \over {y}} - {{2} \over {x}} = 5\)… (3)
\({{8} \over {y}} + {{7} \over {x}} = 15\)… (4)
Let \({{1} \over {x}}\) = p and \({{1} \over {y}}\) = q

Putting these in (3) and (4),
\(7q - 2p = 5 \)… (5)
\(8q + 7p = 15 \)… (6)

From equation (5),
\(2p = 7q – 5\)
\(p = {{7q – 5} \over {2}}\)

Putting value of p in (6),
\(8q + 7 (\frac{7q – 5}{2}) = 15\)
\(16q + 49q – 35 = 30\)
\(65q = 30 + 35 = 65\)
\(q = 1\)

Putting value of q in (5),
\(7 (1) - 2p = 5\)
\(- 2p = -2\)
\( p = 1\)

Putting value of p and q in (\({{1} \over {x}} = p\) and \( {{1} \over {y}} = q\)), we get
x = 1 and y = 1


(vi)\( 6x + 3y - 6xy = 0 \)… (1)
\(2x + 4y - 5xy = 0\) … (2)

Dividing both the equations by xy,
\({{6} \over {y}} + {{3} \over {x}} - 6 = 0\)… (3)
\({{2} \over {y}} + {{4} \over {x}} - 5 = 0\)… (4)

Let \({{1} \over {x}} = p\) and \( {{1} \over {y}} = q\)

Putting these in (3) and (4),
\(6q + 3p – 6 = 0\) … (5)
\(2q + 4p – 5 = 0\) … (6)

From (5),
\(3p = 6 - 6q\)
\(p = 2 - 2q\)

Putting this in (6),
\(2q + 4 (2 - 2q) – 5 = 0\)
\(2q + 8 - 8q – 5 = 0\)
\(-6q = -3\)
\(q = {{1} \over {2}}\)

Putting value of q in (p = 2 – 2q), we get
\(p = 2 – 2 ({{1} \over {2}}) = 2 – 1 = 1\)

Putting values of p and q in (\({{1} \over {x}} = p\) and \( {{1} \over {y}} = q\)), we get
x = 1 and y = 2


(vii) \({{10} \over {x + y}} + {{2} \over {x - y}} = 4\)… (1)
\(\frac{15}{(x + y)} - \frac{5}{(x - y)} = -2\)…(2)
Let \({{1} \over {(x + y)}} = p\) and \({{1} \over {(x - y)}} = q\)

Putting this in (1) and (2),
\(10p + 2q = 4\) … (3)
\(15p - 5q = -2\) … (4)

From equation (3),
\(2q = 4 - 10p\)
\(q = 2 - 5p\) … (5)

Putting this in (4),
\(15p – 5 (2 - 5p) = -2\)
\(15p – 10 + 25p = -2\)
\(40p = 8\) => \(p = {{1} \over {5}}\)

Putting value of p in (5),
\(q = 2 – 5 ({{1} \over {5}}) = 2 – 1 = 1\)

Putting values of p and q in (\({{1} \over {x + y}} = p\) and \( {{1} \over {x - y}} = q\)), we get
\(x + y = 5\) … (6) and \(x – y = 1\) … (7)

Adding (6) and (7),
\(2x = 6\) => \(x = 3\)

Putting x = 3 in (7), we get
\(3 – y = 1\)
\(y = 3 – 1 = 2\)

Therefore, x = 3 and y = 2


(viii)\({{1} \over {3x + y}} + {{1} \over {3x - y}} = {{3} \over {4}}\) … (1)
\({{1} \over {2(3x + y)}} - {{1} \over {2(3x - y)}} = {{-1} \over {8}}\) … (2)
Let \({{1} \over {(3x + y)}} = p\) and \( {{1} \over {(3x - y)}} = q \) … (2)

Putting this in (1) and (2),
\(p + q = {{3} \over {4}}\) and \({{p} \over {2}} - {{q} \over {2}} = {{1} \over {8}}\)
\(4p + 4q = 3\) … (3) and \(4p - 4q = -1\) … (4)

Adding (3) and (4),
\(8p = 2 - p = {{1} \over {4}}\)

Putting value of p in (3),
\(4 ({{1} \over {4}}) + 4q = 3\)
\(1 + 4q = 3\)
\(4q = 3 – 1 = 2\)
\(q = {{1} \over {2}}\)

Putting value of p and q in \({{1} \over {(3x + y)}} = p\) and \( {{1} \over {(3x - y)}} = q \), we get
\({{1} \over {(3x + y)}} = {{1} \over {4}}\) and \( {{1} \over {(3x - y)}} = {{1} \over {2}} \)
\(3x + y = 4 \)… (5) and \(3x – y = 2 \)… (6)

Adding (5) and (6),
\(6x = 6 - x = 1\)

Putting x = 1 in (5) ,
\(3 (1) + y = 4\)
\( y = 4 – 3 = 1\)

Therefore, x = 1 and y = 1