Solution:

\(\displaystyle \frac{a}{b}\) = \(\displaystyle \frac{7}{3}\) ... (Given)

Multiplying both sides by \(\displaystyle \frac{5}{3}\),

 \(\displaystyle \frac{a}{b} \times \frac{5}{3}\) = \(\displaystyle \frac{7}{3} \times \frac{5}{3}\)

∴ \(\displaystyle \frac{5a}{3b}\) = \(\displaystyle \frac{35}{9}\)

By Componendo – Dividendo,

∴ \(\displaystyle \frac{5a + 3b}{5a - 3b}\) = \(\displaystyle \frac{35 + 9}{35 - 9}\)

∴ \(\displaystyle \frac{5a + 3b}{5a - 3b}\) = \(\displaystyle \frac{44}{26}\)

∴ \(\displaystyle \frac{5a + 3b}{5a - 3b} = 22 : 13\) ... (i)

Solution:

\(\displaystyle \frac{a}{b}\) = \(\displaystyle \frac{7}{3}\) ... (Given)

Squaring both sides,

\(\displaystyle \frac{a^2}{b^2}\) = \(\displaystyle \frac{49}{9}\)

Multiplying both sides by \(\displaystyle \frac{2}{3}\),

 \(\displaystyle \frac{a^2}{b^2} \times \frac{2}{3}\) = \(\displaystyle \frac{49}{9} \times \frac{2}{3}\)

∴ \(\displaystyle \frac{2a^2}{3b^2}\) = \(\displaystyle \frac{98}{27}\)

By Componendo – Dividendo,

 \(\displaystyle \frac{2a^2 + 3b^2}{2a^2 - 3b^2}\) = \(\displaystyle \frac{98 + 27}{98 - 27}\)

∴ \(\displaystyle \frac{2a^2 + 3b^2}{2a^2 - 3b^2}\) = \(\displaystyle \frac{125}{71}\)

∴ \(\displaystyle \frac{2a^2 + 3b^2}{2a^2 - 3b^2} = 125 : 71\) ... (ii)

Solution:

\(\displaystyle \frac{a}{b}\) = \(\displaystyle \frac{7}{3}\) ... (Given)

Taking cubes of both sides,

\(\displaystyle \frac{a^3}{b^3}\) = \(\displaystyle \frac{343}{27}\)

By Dividendo,

 \(\displaystyle \frac{a^3 - b^3}{b^3} = \displaystyle \frac{343 - 27}{27}\)

∴ \(\displaystyle \frac{a^3 - b^3}{b^3} = \displaystyle \frac{316}{27}\)

∴ \(\displaystyle \frac{a^3 - b^3}{b^3} = 316 : 27\) ... (iii)

Solution:

\(\displaystyle \frac{a}{b}\) = \(\displaystyle \frac{7}{3}\) ... (Given)

Multiplying both sides by \(\displaystyle \frac{7}{9}\),

 \(\displaystyle \frac{a}{b} \times \frac{7}{9}\) = \(\displaystyle \frac{7}{3} \times \frac{7}{9}\)

∴ \(\displaystyle \frac{7a}{9b}\) = \(\displaystyle \frac{49}{27}\)

By Componendo – Dividendo,

 \(\displaystyle \frac{7a + 9b}{7a - 9b}\) = \(\displaystyle \frac{49 + 27}{49 - 27}\)

∴ \(\displaystyle \frac{7a + 9b}{7a - 9b}\) = \(\displaystyle \frac{76}{22}\)

∴ \(\displaystyle \frac{7a + 9b}{7a - 9b}\) = \(\displaystyle \frac{38}{11}\)

∴ \(\displaystyle \frac{7a + 9b}{7a - 9b} = 38 : 11\) ... (iv)

Solution:

 \(\displaystyle \frac{15a^2 + 4b^2}{15a^2 - 4b^2}\) = \(\displaystyle \frac{47}{7}\) ... (Given)

By Componendo – Dividendo,

 \(\displaystyle \frac{(15a^2 + 4b^2) + (15a^2 - 4b^2)}{(15a^2 + 4b^2) - (15a^2 - 4b^2)}\) = \(\displaystyle \frac{47 + 7}{47 - 7}\)

∴ \(\displaystyle \frac{15a^2 + 4b^2 + 15a^2 - 4b^2}{15a^2 + 4b^2 - 15a^2 + 4b^2}\) = \(\displaystyle \frac{54}{40}\)

∴ \(\displaystyle \frac{30a^2}{8b^2}\) = \(\displaystyle \frac{27}{20}\)

∴ \(\displaystyle \frac{15a^2}{4b^2}\) = \(\displaystyle \frac{27}{20}\)

∴ \(\displaystyle \frac{a^2}{b^2}\) = \(\displaystyle \frac{27}{20} \times \frac{4}{15}\)

∴ \(\displaystyle \frac{a^2}{b^2}\) = \(\displaystyle \frac{9}{25}\)

Taking square roots of both sides,

 \(\displaystyle \frac{a}{b}\) = \(\displaystyle \pm \frac{3}{5}\)

∴\(\displaystyle \frac{a}{b}\) = \(\displaystyle + \frac{3}{5} \text{ OR } \frac{a}{b} = - \frac{3}{5}\)

(The negative value must also be considered, but it is ignored in the text book)

∴\(\displaystyle \frac{a}{b}\) = \(\displaystyle \frac{3}{5}\) ... (i)

i.e. \(\displaystyle \frac {a}{b} = 3 : 5\)

Solution:

 \(\displaystyle \frac {a}{b} = \frac {3}{5}\) ... [From (i)]

Multiplying both sides by \(\displaystyle \frac{7}{3}\),

 \(\displaystyle \frac{a}{b} \times \frac{7}{3}\) = \(\displaystyle \frac{\enclose{downdiagonalstrike}{3}}{5} \times \frac{7}{\enclose{downdiagonalstrike}{3}}\)

∴ \(\displaystyle \frac{7a}{3b}\) = \(\displaystyle \frac{7}{5}\)

By Componendo – Dividendo,

 \(\displaystyle \frac{7a + 3b}{7a - 3b}\) = \(\displaystyle \frac{7 + 5}{7 - 5}\)

∴ \(\displaystyle \frac{7a + 3b}{7a - 3b}\) = \(\displaystyle \frac{12}{2}\)

∴ \(\displaystyle \frac{7a + 3b}{7a - 3b}\) = \(\displaystyle \frac{6}{1}\)

∴ \(\displaystyle \frac{7a + 3b}{7a - 3b}\) = \(\displaystyle \frac{6}{1}\)

By Invertendo,

 \(\displaystyle \frac{7a - 3b}{7a + 3b}\) = \(\displaystyle \frac{1}{6}\) ... (iii)

Solution:

 \(\displaystyle \frac {a}{b} = \frac {3}{5}\) ... [From (i)]

By Invertendo,

 \(\displaystyle \frac {b}{a} = \frac {5}{3}\)

Squaring both sides,

 \(\displaystyle \frac {b^2}{a^2} = \frac {25}{9}\)

Multiplying both sides by \(\displaystyle \frac{1}{2}\),

 \(\displaystyle \frac {b^2}{a^2} \times \frac {1}{2} = \frac {25}{9} \times \frac {1}{2}\)

∴ \(\displaystyle \frac {b^2}{2a^2} = \frac {25}{18}\)

By Componendo – Dividendo,

 \(\displaystyle \frac {b^2 + 2a^2}{b^2 - 2a^2} = \frac {25 + 18}{25 - 18}\)

∴ \(\displaystyle \frac {b^2 + 2a^2}{b^2 - 2a^2} = \frac {43}{7}\)

By Invertendo,

 \(\displaystyle \frac {b^2 - 2a^2}{b^2 + 2a^2} = \frac {7}{43}\)

i.e. \(\displaystyle \frac {b^2 - 2a^2}{b^2 + 2a^2} = 7 : 43\) ... (iv)

Solution:

 \(\displaystyle \frac {a}{b} = \frac {3}{5}\) ... [From (i)]

By Invertendo,

 \(\displaystyle \frac {b}{a} = \frac {5}{3}\)

Taking cubes of both sides,

 \(\displaystyle \frac {b^3}{a^3} = \frac {125}{27}\)

Multiplying both sides by \(\displaystyle \frac{1}{2}\),

 \(\displaystyle \frac {b^3}{a^3} \times \frac {1}{2} = \frac {125}{27} \times \frac {1}{2}\)

∴ \(\displaystyle \frac {b^3}{2a^3} = \frac {125}{54}\)

By Componendo – Dividendo,

 \(\displaystyle \frac {b^3 + 2a^3}{b^3 - 2a^3} = \frac {125 + 54}{125 - 54}\)

∴ \(\displaystyle \frac {b^3 + 2a^3}{b^3 - 2a^3} = \frac {179}{71}\)

By Invertendo,

 \(\displaystyle \frac {b^3 - 2a^3}{b^3 + 2a^3} = \frac {71}{179}\)

i.e. \(\displaystyle \frac {b^3 - 2a^3}{b^3 + 2a^3} = 71 : 179\) ... (v)

Solution:

\(\displaystyle \frac{3a + 7b}{3a - 7b}\) = \(\displaystyle \frac{4}{3}\) ... (Given)

By Componendo – Dividendo,

 \(\displaystyle \frac{(3a + 7b) + (3a - 7b)}{(3a + 7b) - (3a - 7b)}\) = \(\displaystyle \frac{4 + 3}{4 - 3}\)

∴ \(\displaystyle \frac{3a + 7b + 3a - 7b}{3a + 7b - 3a + 7b}\) = \(\displaystyle \frac{7}{1}\)

∴ \(\displaystyle \frac{6a}{14b}\) = \(\displaystyle \frac{7}{1}\)

∴ \(\displaystyle \frac{3a}{7b}\) = \(\displaystyle \frac{7}{1}\)

∴ \(\displaystyle \frac{a}{b}\) = \(\displaystyle \frac{7}{1} \times \frac{7}{3}\)

∴ \(\displaystyle \frac{a}{b}\) = \(\displaystyle \frac{49}{3}\)

Squaring both sides,

 \(\displaystyle \frac{a^2}{b^2}\) = \(\displaystyle \frac{2401}{9}\)

Multiplying both sides by \(\displaystyle \frac{3}{7}\),

 \(\displaystyle \frac{a^2}{b^2} \times \frac{3}{7}\) = \(\displaystyle \frac{2401}{9} \times \frac{3}{7}\)

∴ \(\displaystyle \frac{3a^2}{7b^2}\) = \(\displaystyle \frac{343}{3}\)

By Componendo – Dividendo,

 \(\displaystyle \frac{3a^2 + 7b^2}{3a^2 - 7b^2}\) = \(\displaystyle \frac{343 + 3}{343 - 3}\)

∴ \(\displaystyle \frac{3a^2 + 7b^2}{3a^2 - 7b^2}\) = \(\displaystyle \frac{346}{340}\)

∴ \(\displaystyle \frac{3a^2 + 7b^2}{3a^2 - 7b^2}\) = \(\displaystyle \frac{173}{170}\)

By Invertendo,

 \(\displaystyle \frac{3a^2 - 7b^2}{3a^2 + 7b^2}\) = \(\displaystyle \frac{170}{173}\)

i.e. \(\displaystyle \frac{3a^2 - 7b^2}{3a^2 + 7b^2}\) = 170 : 173

Solution:

\(\displaystyle \frac {x^2 + 12x - 20}{3x - 5}\) = \(\displaystyle \frac {x^2 + 8x + 12}{2x + 3}\) ... (Given)

Multiplying both sides by \(\displaystyle \frac{1}{4}\),

 \(\displaystyle \frac {x^2 + 12x - 20}{3x - 5} \times \frac {1}{4}\) = \(\displaystyle \frac {x^2 + 8x + 12}{2x + 3} \times \frac {1}{4}\)

∴ \(\displaystyle \frac {x^2 + 12x - 20}{12x - 20}\) = \(\displaystyle \frac {x^2 + 8x + 12}{8x + 12}\)

By Dividendo,

 \(\displaystyle \frac {(x^2 + 12x - 20) - (12x - 20)}{12x - 20}\) = \(\displaystyle \frac {(x^2 + 8x + 12) - (8x + 12)}{8x + 12}\)

∴ \(\displaystyle \frac {x^2 \enclose{downdiagonalstrike}{+ 12x} \enclose{downdiagonalstrike}{- 20} \enclose{downdiagonalstrike}{- 12x} \enclose{downdiagonalstrike}{+ 20}}{12x - 20}\) = \(\displaystyle \frac {x^2 \enclose{downdiagonalstrike}{+ 8x} \enclose{downdiagonalstrike}{+ 12} \enclose{downdiagonalstrike}{- 8x} \enclose{downdiagonalstrike}{- 12}}{8x + 12}\)

∴ \(\displaystyle \frac {x^2}{12x - 20}\) = \(\displaystyle \frac {x^2}{8x + 12}\)

Now, this equation is true for x = 0.

∴ x = 0 is one solution of this equation. ... (i)

If x ≠ 0, then x² ≠ 0

Multiplying both sides by \(\displaystyle \frac{1}{x^2}\),

 \(\displaystyle \frac {x^2}{12x - 20} \times \frac {1}{x^2}\) = \(\displaystyle \frac {x^2}{8x + 12} \times \frac {1}{x^2}\)

∴ \(\displaystyle \frac {1}{12x - 20}\) = \(\displaystyle \frac {1}{8x + 12} \)

By Invertendo,

 \(\displaystyle 12x - 20\) = \(\displaystyle 8x + 12\)

∴ \(\displaystyle 12x - 8x\) = \(\displaystyle 12 + 20\)

∴ \(\displaystyle 4x = 32\)

∴ \(\displaystyle x = \frac{32}{4}\)

∴ \(\displaystyle x = 8\) ... (ii)

∴ 0 and 8 are the solutions of the given equation.

[Refer to Solved Example (1) on Page 68 of your Textbook which explains why there are two solutions of this equation.]

Solution:

\(\displaystyle \frac {10x^2 + 15x + 63}{5x^2 - 25x + 12}\) = \(\displaystyle \frac {2x + 3}{x - 5}\) ... (Given)

Multiplying both sides by \(\displaystyle \frac{1}{2}\),

 \(\displaystyle \frac {10x^2 + 15x + 63}{5x^2 - 25x + 12} \times \frac{1}{2}\) = \(\displaystyle \frac {2x + 3}{x - 5} \times \frac{1}{2}\)

∴ \(\displaystyle \frac {10x^2 + 15x + 63}{10x^2 - 50x + 24}\) = \(\displaystyle \frac {2x + 3}{2x - 10}\)

By Dividendo,

 \(\displaystyle \frac {(10x^2 + 15x + 63) - (10x^2 - 50x + 24)}{10x^2 - 50x + 24}\) = \(\displaystyle \frac {(2x + 3) - (2x - 10)}{2x - 10}\)

∴ \(\displaystyle \frac {\enclose{downdiagonalstrike}{10x^2} + 15x + 63 \enclose{downdiagonalstrike}{- 10x^2} + 50x - 24}{10x^2 - 50x + 24}\) = \(\displaystyle \frac {\enclose{downdiagonalstrike}{2x} + 3 \enclose{downdiagonalstrike}{- 2x} + 10}{2x - 10}\)

∴ \(\displaystyle \frac {15x + 63 + 50x - 24}{10x^2 - 50x + 24}\) = \(\displaystyle \frac {3 + 10}{2x - 10}\)

∴ \(\displaystyle \frac {65x + 39}{10x^2 - 50x + 24}\) = \(\displaystyle \frac {13}{2x - 10}\)

By cross multiplication,

 \(\displaystyle (65x + 39)(2x - 10)\) = \(\displaystyle 13(10x^2 - 50x + 24)\)

∴ \(\displaystyle \enclose{downdiagonalstrike}{130x^2} \enclose{downdiagonalstrike}{- 650x} + 78x - 390\) = \(\displaystyle \enclose{downdiagonalstrike}{130x^2} \enclose{downdiagonalstrike}{- 650x} + 312\)

∴ \(\displaystyle 78x - 390\) = \(\displaystyle 312\)

∴ \(\displaystyle 78x\) = \(\displaystyle 312 + 390\)

∴ \(\displaystyle 78x\) = \(\displaystyle 702\)

∴ \(\displaystyle x = \frac{702}{78}\)

∴ \(\displaystyle x = 9\)

∴ x = 9 is the solution of the given equation.

Solution:

 \(\displaystyle \frac {(2x + 1)^2 + (2x - 1)^2}{(2x + 1)^2 - (2x - 1)^2}\) = \(\displaystyle \frac {17}{8}\) ... (Given)

Let, \(\displaystyle (2x + 1)^2 = m\) and \(\displaystyle (2x - 1)^2 = n\).

∴ The given equation:

 \(\displaystyle \frac {m + n}{m - n} = \frac {17}{8}\)

By Componendo – Dividendo,

 \(\displaystyle \frac {(m + n) + (m - n)}{(m + n) - (m - n)}\) = \(\displaystyle \frac {17 + 8}{17 - 8}\)

∴ \(\displaystyle \frac {m \enclose{downdiagonalstrike}{+ n} + m \enclose{downdiagonalstrike}{- n}}{\enclose{updiagonalstrike}{m} + n \enclose{updiagonalstrike}{- m} + n}\) = \(\displaystyle \frac {25}{9}\)

∴ \(\displaystyle \frac {\enclose{downdiagonalstrike}{2}m}{\enclose{downdiagonalstrike}{2}n}\) = \(\displaystyle \frac {25}{9}\)

∴ \(\displaystyle \frac {m}{n}\) = \(\displaystyle \frac {25}{9}\)

Substituting the values of m and n,

  \(\displaystyle \frac {(2x + 1)^2}{(2x - 1)^2}\) = \(\displaystyle \frac {25}{9}\)

Taking square roots of both sides,

  \(\displaystyle \frac {2x + 1}{2x - 1}\) = \(\displaystyle \frac {5}{3}\)

By cross multiplication,

  \(\displaystyle 3 (2x + 1)\) = \(\displaystyle 5(2x - 1)\)

∴ \(\displaystyle 6x + 3\) = \(\displaystyle 10x - 5\)

∴ \(\displaystyle 6x - 10x\) = \(\displaystyle - 5 - 3\)

∴ \(\displaystyle - 4x = - 8\)

i.e. \(\displaystyle 4x = 8\)

∴ \(\displaystyle x = \frac {4}{2}\)

∴ \(\displaystyle x = 2\)

∴ x = 2 is the solution of the given equation.

Solution:

 \(\displaystyle \frac {\sqrt {4x + 1} + \sqrt {x + 3}}{\sqrt {4x + 1} - {\sqrt{x + 3}}}\) = \(\displaystyle \frac {4}{1}\) ... (Given)

Let, \(\displaystyle \sqrt {4x + 1} = m\) and \(\displaystyle \sqrt {x + 3} = n\).

∴ The given equation:

 \(\displaystyle \frac {m + n}{m - n} = \frac {4}{1}\)

By Componendo – Dividendo,

 \(\displaystyle \frac {(m + n) + (m - n)}{(m + n) - (m - n)}\) = \(\displaystyle \frac {4 + 1}{4 - 1}\)

∴ \(\displaystyle \frac {m \enclose{downdiagonalstrike}{+ n} + m \enclose{downdiagonalstrike}{- n}}{\enclose{updiagonalstrike}{m} + n \enclose{updiagonalstrike}{- m} + n}\) = \(\displaystyle \frac {5}{3}\)

∴ \(\displaystyle \frac {\enclose{downdiagonalstrike}{2}m}{\enclose{downdiagonalstrike}{2}n}\) = \(\displaystyle \frac {5}{3}\)

∴ \(\displaystyle \frac {m}{n}\) = \(\displaystyle \frac {5}{3}\)

Substituting the values of m and n,

  \(\displaystyle \frac {\sqrt {4x + 1}}{\sqrt {x + 3}}\) = \(\displaystyle \frac {5}{3}\)

Squaring both sides,

  \(\displaystyle \frac {4x + 1}{x + 3}\) = \(\displaystyle \frac {25}{9}\)

By cross multiplication,

  \(\displaystyle 9(4x + 1)\) = \(\displaystyle 25(x + 3)\)

∴ \(\displaystyle 36x + 9\) = \(\displaystyle 25x + 75\)

∴ \(\displaystyle 36x - 25x\) = \(\displaystyle 75 - 9\)

∴ \(\displaystyle 11x\) = \(\displaystyle 66\)

∴ \(\displaystyle x = \frac {66}{11}\)

∴ \(\displaystyle x = 6\)

∴ x = 6 is the solution of the given equation.

Solution:

Here, we will use a different technique.

We know that,
\(\displaystyle (a + b)^2 = a^2 + 2ab + b^2\)

Thus, \(\displaystyle (2x + 3)^2\)

\(\displaystyle = (2x)^2 + 2 \times 2x \times 3 + (3)^2\)

\(\displaystyle = 4x^2 + 12x + 9\) ... (i)

If we substitute this value in the given equation, it becomes,

 \(\displaystyle \frac {(4x + 1)^2 + (2x + 3)^2}{(2x + 3)^2}\) = \(\displaystyle \frac {61}{36}\)

Now, let \(\displaystyle (4x + 1)^2 = m\) and \(\displaystyle (2x + 3)^2 = n\)

∴ The given equation:

 \(\displaystyle \frac {m + n}{n} = \frac {61}{36}\)

By Dividendo,

 \(\displaystyle \frac {m + n - n}{n}\) = \(\displaystyle \frac {61 - 36}{36}\)

∴ \(\displaystyle \frac {m \enclose{downdiagonalstrike}{+ n} \enclose{downdiagonalstrike}{- n}}{n}\) = \(\displaystyle \frac {25}{36}\)

∴ \(\displaystyle \frac {m}{n}\) = \(\displaystyle \frac {25}{36}\)

Substituting the values of m and n,

  \(\displaystyle \frac {(4x + 1)^2}{(2x + 3)^2}\) = \(\displaystyle \frac {25}{36}\)

Taking square roots of both sides,

  \(\displaystyle \frac {4x + 1}{2x + 3}\) = \(\displaystyle \frac {5}{6}\)

By cross multiplication,

  \(\displaystyle 6 (4x + 1)\) = \(\displaystyle 5(2x + 3)\)

∴ \(\displaystyle 24x + 6\) = \(\displaystyle 10x + 15\)

∴ \(\displaystyle 24x - 10x\) = \(\displaystyle 15 - 6\)

∴ \(\displaystyle 14x = 9\)

∴ \(\displaystyle x = \frac {9}{14}\)

∴ \(\displaystyle x = \frac {9}{14}\) is the solution of the given equation.

Solution:

 \(\displaystyle \frac {(3x - 4)^3 - (x + 1)^3}{(3x - 4)^3 + (x + 1)^3}\) = \(\displaystyle \frac {61}{189}\) ... (Given)

Let, \(\displaystyle (3x - 4)^3 = m\) and \(\displaystyle (x + 1)^3 = n\).

∴ The given equation:

 \(\displaystyle \frac {m - n}{m + n} = \frac {61}{189}\)

By Componendo – Dividendo,

 \(\displaystyle \frac {(m - n) + (m + n)}{(m - n) - (m + n)}\) = \(\displaystyle \frac {61 + 189}{61 - 189}\)

∴ \(\displaystyle \frac {m \enclose{downdiagonalstrike}{- n} + m \enclose{downdiagonalstrike}{+ n}}{\enclose{updiagonalstrike}{m} - n \enclose{updiagonalstrike}{- m} - n}\) = \(\displaystyle \frac {250}{-128}\)

∴ \(\displaystyle \frac {\enclose{downdiagonalstrike}{2}m}{-\enclose{downdiagonalstrike}{2}n}\) = \(\displaystyle \frac {125}{-64}\)

∴ \(\displaystyle \frac {m}{-n}\) = \(\displaystyle \frac {125}{-64}\)

i.e. \(\displaystyle \frac {m}{n}\) = \(\displaystyle \frac {125}{64}\)

Substituting the values of m and n,

  \(\displaystyle \frac {(3x - 4)^3}{(x + 1)^3}\) = \(\displaystyle \frac {125}{64}\)

Taking cube roots of both sides,

  \(\displaystyle \frac {3x - 4}{x + 1}\) = \(\displaystyle \frac {5}{4}\)

By cross multiplication,

  \(\displaystyle 4 (3x - 4)\) = \(\displaystyle 5(x + 1)\)

∴ \(\displaystyle 12x - 16\) = \(\displaystyle 5x + 5\)

∴ \(\displaystyle 12x - 5x\) = \(\displaystyle 5 + 16\)

∴ \(\displaystyle 7x = 21\)

∴ \(\displaystyle x = \frac {21}{7}\)

∴ \(\displaystyle x = 3\)

∴ x = 3 is the solution of the given equation.