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# 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)

$$2x = 6$$ => $$x = 3$$

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)

$$8p = 2 - p = {{1} \over {4}}$$

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)

$$6x = 6 - x = 1$$
$$3 (1) + y = 4$$
$$y = 4 – 3 = 1$$