Solution:

Let, \(\displaystyle \frac {2}{x} - \frac {3}{y} = 15\) ... (i)

And, \(\displaystyle \frac {8}{x} + \frac {5}{y} = 77\) ... (ii)

Substituting \(\displaystyle \frac {1}{x} = a \text{ and } \frac {1}{y} = b\),

The given equations:
 \(2a - 3b = 15\) ... (iii)
and \(8a + 5b = 77\) ... (iv)

Multiplying equation (iii) by 5,
\(10a - 15b = 75\) ... (v)

Multiplying equation (iv) by 3,
\(24a + 15b = 231\) ... (vi)

Adding (v) and (vi),

		10a	−	15b	=		75	... (v)
+		24a	+	15b	=		231	... (vi)
		34a			=		306

\(\displaystyle \therefore a = \frac {\cancelto{9}{306}}{\cancelto{1}{34}}\)

\(\displaystyle \therefore a = 9\) ... (vii)

Substituting the value of \(a\) in equation (vi),
 \(24a + 15b = 231\) ... (vi)
∴ \(24 \times 9 + 15b = 231\)
∴ \(216 + 15b = 231\)
∴ \(15b = 231 - 216\)
∴ \(15b = 15\)

\(\displaystyle \therefore b = \frac {\cancel{15}}{\cancel{15}}\)

\(\displaystyle \therefore b = 1\) ... (viii)

Substituting the value of \(a\),

 \(\displaystyle \frac {1}{x} = a\)

∴ \(\displaystyle \frac {1}{x} = 9\)

∴ \(\displaystyle x = \frac {1}{9}\) ... (ix)

Substituting the value of \(b\),

 \(\displaystyle \frac {1}{y} = b\)

∴ \(\displaystyle \frac {1}{y} = 1\)

∴ \(\displaystyle y = 1\) ... (x)

∴ Solution: \(\displaystyle \left(\frac {1}{9},\ 1\right)\)

Solution:

Let, \(\displaystyle \frac {10}{x + y} + \frac {2}{x - y} = 4\) ... (i)

And, \(\displaystyle \frac {15}{x + y} - \frac {5}{x - y} =\:-\:2\) ... (ii)

Substituting \(\displaystyle \frac {1}{x + y} = a \text{ and } \frac {1}{x - y} = b\),

The given equations:
 \(10a + 2b = 4\) ... (iii)
and \(15a - 5b =\:-\:2\) ... (iv)

Multiplying equation (iii) by 5,
\(50a + 10b = 20\) ... (v)

Multiplying equation (iv) by 2,
\(30a - 10b = -\:4\) ... (vi)

Adding (v) and (vi),

		50a	+	10b	=		20	... (v)
+		30a	−	10b	=		4	... (vi)
		80a			=		16

\(\displaystyle \therefore a = \frac {\cancelto{1}{16}}{\cancelto{5}{80}}\)

\(\displaystyle \therefore a = \frac {1}{5}\) ... (vii)

Substituting the value of \(a\) in equation (iii),
 \(10a + 2b = 4\) ... (iii)
∴ \(\displaystyle \cancelto{2}{10} \times \frac {1}{\cancelto {1}{5}} + 2b = 4\)
∴ \(2 + 2b = 4\)
∴ \(2b = 4 - 2\)
∴ \(2b = 2\)

\(\displaystyle \therefore b = \frac {\cancel{2}}{\cancel{2}}\)

\(\displaystyle \therefore b = 1\) ... (viii)

Substituting the value of \(a\),

 \(\displaystyle \frac {1}{x + y} = a\)

∴ \(\displaystyle \frac {1}{x + y} = \frac {1}{5}\)

By invertendo,
  \(\displaystyle x + y = 5\) ... (ix)

Substituting the value of \(b\),

 \(\displaystyle \frac {1}{x - y} = b\)

∴ \(\displaystyle \frac {1}{x - y} = 1\)

By invertendo,
  \(\displaystyle x - y = 1\) ... (x)

Adding (ix) and (x),

		x	+	y	=		5	... (ix)
+		x	−	y	=		1	... (x)
		2x			=		6

\(\displaystyle \therefore x = \frac {\cancelto{3}{6}}{\cancelto {1}{2}}\)

\(\displaystyle \therefore x = 3\) ... (xi)

Substituting the value of \(x\) in equation (ix),
 \(x + y = 5\) ... (ix)
∴ \(3 + y = 5\)
∴ \(y = 5 - 3\)
∴ \(y = 2\) ... (xii)

∴ Solution: \((3, 2)\)

Solution:

Let, \(\displaystyle \frac {27}{x - 2} + \frac {31}{y + 3} = 85\) ... (i)

And, \(\displaystyle \frac {31}{x - 2} + \frac {27}{y + 3} = 89\) ... (ii)

Substituting \(\displaystyle \frac {1}{x - 2} = a \text{ and } \frac {1}{y + 3} = b\),

The given equations:
 \(27a + 31b = 85\) ... (iii)
and \(31a + 27b = 89\) ... (iv)

Adding (iii) and (iv),

		27a	+	31b	=		85	... (iii)
+		31a	+	27b	=		89	... (iv)
		58a	+	58b	=		174

Dividing both sides by 58,

  \(\displaystyle \frac {\cancel{58}a}{\cancel{58}} + \frac {\bcancel{58}b}{\bcancel{58}} = \frac {\cancelto{3}{174}}{\cancelto {1}{58}}\)

∴ \(a + b = 3\) ... (v)

Subtracting (iii) from (iv),

		31a	+	27b	=		89	... (iv)
−	⊕	27a	⊕	31b	=	⊕	85	... (iii)
	−		−			−
		4a	−	4b	=		4

Dividing both sides by 4,

  \(\displaystyle \frac {\cancel{4}a}{\cancel{4}} - \frac {\bcancel{4}b}{\bcancel{4}} = \frac {\cancel{4}}{\cancel{4}}\)

∴ \(a - b = 1\) ... (vi)

Adding (v) and (vi),

		a	+	b	=		3	... (v)
+		a	−	b	=		1	... (vi)
		2a			=		4

\(\displaystyle a = \frac {\cancelto {2}{4}}{\cancelto {1}{2}}\)

\(\displaystyle a = 2\) ... (vii)

Substituting the value of a in equation (v),
 \(a + b = 3\) ... (v)
∴ \(2 + b = 3\)
∴ \(b = 3 - 2\)
∴ \(b = 1\) ... (viii)

Substituting the value of \(a\),

 \(\displaystyle \frac {1}{x - 2} = a\)

∴ \(\displaystyle \frac {1}{x - 2} = 2\)

∴ \( 1 = 2(x - 2)\)
∴ \( 1 = 2x - 4\)
∴ \( 1 + 4 = 2x\)
∴ \( 5 = 2x\)
i.e. \( 2x = 5\)

\(\displaystyle \therefore x = \frac {5}{2}\) ... (ix)

Substituting the value of \(b\),

 \(\displaystyle \frac {1}{y + 3} = b\)

∴ \(\displaystyle \frac {1}{y + 3} = 1\)

∴ \(1 = 1(y + 3)\)
∴ \(1 = y + 3\)
∴ \(1 - 3 = y\)
∴ \(-\:2 = y\)
i.e. \(y = -\:2\) ... (x)

∴ Solution: \(\displaystyle \left(\frac {5}{2},\ -2\right)\)

Solution:

 \(\displaystyle \frac {1}{3x + y} + \frac {1}{3x - y} = \frac {3}{4}\)

Multiplying both sides by 4,

 \(\displaystyle \frac {1}{3x + y} \times 4 + \frac {1}{3x - y} \times 4 = \frac {3}{\cancel {4}} \times \cancel {4}\)

∴ \(\displaystyle \frac {4}{3x + y} + \frac {4}{3x - y} = 3\) ... (i)

and

 \(\displaystyle \frac {1}{2(3x + y)} - \frac {1}{2(3x - y)} = - \frac {1}{8}\)

Multiplying both sides by 8,

 \(\displaystyle \frac {1}{\cancelto{1}{2}(3x + y)} \times \cancelto {4}{8} - \frac {1}{\cancelto {1}{2}(3x - y)} \times \cancelto {4}{8} = - \frac {1}{\cancel {8}} \times \cancel {8}\)

∴ \(\displaystyle \frac {4}{3x + y} - \frac {4}{3x - y} = -\:1\) ... (ii)

Substituting \(\displaystyle \frac {1}{3x + y} = a\) and \(\displaystyle \frac {1}{3x - y} = b\),

The given equations:
 \(4a + 4b = 3\) ... (iii)
 \(4a - 4b = - 1\) ... (iv)

Adding (iii) and (iv),

		4a	+	4b	=		3	... (iii)
+		4a	−	4b	=	−	1	... (iv)
		8a			=		2

\(\displaystyle a = \frac {\cancelto {1}{2}}{\cancelto {4}{8}}\)

\(\displaystyle a = \frac {1}{4}\) ... (v)

Substituting the value of \(a\) in equation (iii),
 \(4a + 4b = 3\) ... (iii)

∴ \(\displaystyle \cancel {4} \times {\frac {1}{\cancel {4}}} + 4b = 3\)

∴ \(1 + 4b = 3\)
∴ \(4b = 3 - 1\)
∴ \(4b = 2\)

∴ \(\displaystyle b = \frac {\cancelto {1}{2}}{\cancelto {2}{4}}\)

∴ \(\displaystyle b = \frac {1}{2}\) ... (vi)

Resubstituting the value of \(a\),

 \(\displaystyle \frac {1}{3x + y} = a\)

∴ \(\displaystyle \frac {1}{3x + y} = \frac {1}{4}\)

By invertendo,
 \(3x + y = 4\) ... (vii)

Resubstituting the value of \(b\),

 \(\displaystyle \frac {1}{3x - y} = b\)

∴ \(\displaystyle \frac {1}{3x - y} = \frac {1}{2}\)

By invertendo,
 \(3x - y = 2\) ... (viii)

Adding (vii) and (viii),

		3x	+	y	=		4	... (vii)
+		3x	−	y	=		2	... (viii)
		6x			=		6

\(\displaystyle x = \frac {\cancelto {1}{6}}{\cancelto {1}{6}}\)

\(\displaystyle x = 1\) ... (ix)

Substituting the value of \(x\) in equation (vii),
 \(3x + y = 4\) ... (vii)
∴ \(3(1) + y = 4\)
∴ \(3 + y = 4\)
∴ \(y = 4 - 3\)
∴ \(y = 1\) ... (x)

∴ Solution: \((1, 1)\)