Solution:

(i) \(\displaystyle \frac{x}{7} = \frac{y}{3} = \frac{3x + 5y}{36} = \frac{7x - 9y}{22}\)

Solution:

\(\displaystyle \frac{a}{3} = \frac{b}{4} = \frac{c}{7}= \frac{a - 2b + 3c}{16} = \frac{2a - 2b + 2c}{6 - 8 + 14}\)

Solution:

 \(\displaystyle 5m - n = 3m + 4n\) ... (Given)

∴ \(\displaystyle 5m - 3m = 4n + n\)

∴ \(\displaystyle 2m = 5n\)

∴ \(\displaystyle \frac {m}{n} = \frac {5}{2}\) ... (i)

Squaring both sides,

 \(\displaystyle \frac {m^2}{n^2} = \frac {25}{4}\)

By Componendo – Dividendo,

 \(\displaystyle \frac {m^2 + n^2}{m^2 - n^2} = \frac {25 + 4}{25 - 4}\)

∴ \(\displaystyle \frac {m^2 + n^2}{m^2 - n^2} = \frac {29}{21}\)

i.e. \(\displaystyle \frac {m^2 + n^2}{m^2 - n^2} = 29 : 21\)

Solution:

 \(\displaystyle \frac {m}{n} = \frac {5}{2}\) ... [From (i)]

Multiplying both sides by \(\displaystyle \frac{3}{4}\),

 \(\displaystyle \frac{m}{n} \times \frac{3}{4}\) = \(\displaystyle \frac{5}{2} \times \frac{3}{4}\)

∴ \(\displaystyle \frac{3m}{4n}\) = \(\displaystyle \frac{15}{8}\)

By Componendo – Dividendo,

 \(\displaystyle \frac{3m + 4n}{3m - 4n}\) = \(\displaystyle \frac{15 + 8}{15 - 8}\)

∴ \(\displaystyle \frac{3m + 4n}{3m - 4n}\) = \(\displaystyle \frac{23}{7}\)

i.e. \(\displaystyle \frac{3m + 4n}{3m - 4n}\) = 23 : 7

Proof:

No two of a, b, c are equal. ... (Given)

∴ a − b ≠ 0, b − c ≠ 0, and c − a ≠ 0.

∴ Dividing all the terms by abc,

 \(\displaystyle \frac{\enclose{downdiagonalstrike}{a}(y + z)}{\enclose{downdiagonalstrike}{a}bc}\) = \(\displaystyle \frac{\enclose{downdiagonalstrike}{b}(z + x)}{a\enclose{downdiagonalstrike}{b}c}\) = \(\displaystyle \frac{\enclose{downdiagonalstrike}{c}(x + y)}{ab\enclose{downdiagonalstrike}{c}}\)

∴ \(\displaystyle \frac{y + z}{bc}\) = \(\displaystyle \frac{z + x}{ac}\) = \(\displaystyle \frac{x + y}{ab}\)

Using Theorem on Equal Ratios:

 Each Ratio = \(\displaystyle \frac{(x + y) - (z + x)}{ab - ac}\)

∴ Each Ratio = \(\displaystyle \frac{\enclose{downdiagonalstrike}{x} + y - z - \enclose{downdiagonalstrike}{x}}{a(b - c)}\)

∴ Each Ratio = \(\displaystyle \frac{y - z}{a(b - c)}\) ... (i)

Also,

Using Theorem on Equal Ratios:

  Each Ratio = \(\displaystyle \frac{(y + z) - (x + y)}{bc - ab}\)

∴ Each Ratio = \(\displaystyle \frac{\enclose{downdiagonalstrike}{y} + z - x - \enclose{downdiagonalstrike}{y}}{b(c - a)}\)

∴ Each Ratio = \(\displaystyle \frac{z - x}{b(c - a)}\) ... (ii)

And,

Using Theorem on Equal Ratios:

  Each Ratio = \(\displaystyle \frac{(z + x) - (y + z)}{ac - bc}\)

∴ Each Ratio = \(\displaystyle \frac{\enclose{downdiagonalstrike}{z} + x - y - \enclose{downdiagonalstrike}{z}}{c(a - b)}\)

∴ Each Ratio = \(\displaystyle \frac{x - y}{c(a - b)}\) ... (iii)

From (i), (ii) and (iii), we have:

 \(\displaystyle \frac{y - z}{a(b - c)}\) = \(\displaystyle \frac{z - x}{b(c - a)}\) = \(\displaystyle \frac{x - y}{c(a - b)}\)

Proof:

\(\displaystyle \frac{x}{3x - y - z}\) = \(\displaystyle \frac{y}{3y - z - x}\) = \(\displaystyle \frac{z}{3z - x - y}\) ... (Given)

Using Theorem on Equal Ratios,

 Each Ratio = \(\displaystyle \frac{x + y + z}{3x - y - z + 3y - z - x + 3z - x - y}\)

∴ Each Ratio = \(\displaystyle \frac{\enclose{downdiagonalstrike}{x + y + z}}{\enclose{downdiagonalstrike}{x + y + z}}\)

∴ Each Ratio = 1 ... (∵ x + y + z ≠ 0)

Proof:

\(\displaystyle \frac{ax + by}{x + y}\) = \(\displaystyle \frac{bx + az}{x + z}\) = \(\displaystyle \frac{ay + bz}{y + z}\) ... (Given)

Using Theorem on Equal Ratios,

 Each Ratio = \(\displaystyle \frac{ax + by + bx + az + ay + bz}{x + y + x + z + y + z}\)

Rearranging the terms,

 Each Ratio = \(\displaystyle \frac{ax + bx + ay + by + az + bz}{2x + 2y + 2z}\)

∴ Each Ratio = \(\displaystyle \frac{x(a + b) + y(a + b) + z(a + b)}{2(x + y + z)}\)

∴ Each Ratio = \(\displaystyle \frac{(a + b)(\enclose{downdiagonalstrike}{x + y + z})}{2(\enclose{downdiagonalstrike}{x + y + z})}\)

∴ Each Ratio = \(\displaystyle \frac{a + b}{2}\) ... (∵ x + y + z ≠ 0)

Proof:

\(\displaystyle \frac{y + z}{a}\) = \(\displaystyle \frac{z + x}{b}\) = \(\displaystyle \frac{x + y}{c}\) ... (Given)

Using Theorem on Equal Ratios,

 Each Ratio = \(\displaystyle \frac{(z + x) + (x + y) - (y + z)}{b + c - a}\)

∴ Each Ratio = \(\displaystyle \frac{\enclose{downdiagonalstrike}{z} + x + x + \enclose{updiagonalstrike}{y} - \enclose{updiagonalstrike}{y} - \enclose{downdiagonalstrike}{z}}{b + c - a}\)

∴ Each Ratio = \(\displaystyle \frac{2x}{b + c - a}\) ... (i)

Similarly,

 Each Ratio = \(\displaystyle \frac{(x + y) + (y + z) - (z + x)}{c + a - b}\)

∴ Each Ratio = \(\displaystyle \frac{\enclose{downdiagonalstrike}{x} + y + y + \enclose{updiagonalstrike}{z} - \enclose{updiagonalstrike}{z} - \enclose{downdiagonalstrike}{x}}{c + a - b}\)

∴ Each Ratio = \(\displaystyle \frac{2y}{c + a - b}\) ... (ii)

And,

 Each Ratio = \(\displaystyle \frac{(y + z) + (z + x) - (x + y)}{a + b - c}\)

∴ Each Ratio = \(\displaystyle \frac{\enclose{downdiagonalstrike}{y} + z + z + \enclose{updiagonalstrike}{x} - \enclose{updiagonalstrike}{x} - \enclose{downdiagonalstrike}{y}}{a + b - c}\)

∴ Each Ratio = \(\displaystyle \frac{2z}{a + b - c}\) ... (iii)

From (i), (ii) and (iii),

Each Ratio = \(\displaystyle \frac{2x}{b + c - a}\) = \(\displaystyle \frac{2y}{c + a - b}\) = \(\displaystyle \frac{2z}{a + b - c}\)

Dividing each term by 2,

 \(\displaystyle \frac{x}{b + c - a}\) = \(\displaystyle \frac{y}{c + a - b}\) = \(\displaystyle \frac{z}{a + b - c}\)

Proof:

\(\displaystyle \frac{3x - 5y}{5z + 3y}\) = \(\displaystyle \frac{x + 5z}{y - 5x}\) = \(\displaystyle \frac{y - z}{x - z}\) ... (Given)

Multiplying the third ratio by \(\displaystyle \frac{5}{5},\)

\(\displaystyle \frac{3x - 5y}{5z + 3y}\) = \(\displaystyle \frac{x + 5z}{y - 5x}\) = \(\displaystyle \frac{5y - 5z}{5x - 5z}\)

Using Theorem on Equal Ratios,

 Each Ratio = \(\displaystyle \frac{(3x - 5y) + (x + 5z) + (5y - 5z)}{(5z + 3y) + (y - 5x) + (5x - 5z)}\)

∴ Each Ratio = \(\displaystyle \frac{3x - 5y + x + 5z + 5y - 5z}{5z + 3y + y - 5x + 5x - 5z}\)

∴ Each Ratio = \(\displaystyle \frac{3x - \enclose{downdiagonalstrike}{5y} + x + \enclose{updiagonalstrike}{5z} + \enclose{downdiagonalstrike}{5y} - \enclose{updiagonalstrike}{5z}}{\enclose{updiagonalstrike}{5z} + 3y + y - \enclose{downdiagonalstrike}{5x} + \enclose{downdiagonalstrike}{5x} - \enclose{updiagonalstrike}{5z}}\)

∴ Each Ratio = \(\displaystyle \frac{\enclose{downdiagonalstrike}{4}x}{\enclose{downdiagonalstrike}{4}y}\)

∴ Each Ratio = \(\displaystyle \frac{x}{y}\)

Solution:

\(\displaystyle \frac {16x^2 - 20x + 9}{8x^2 + 12x + 21}\) = \(\displaystyle \frac {4x - 5}{2x + 3}\) ... (Given)

Multiplying the numerator and denominator of the second ratio by − 4x,

 \(\displaystyle \frac {16x^2 - 20x + 9}{8x^2 + 12x + 21}\) = \(\displaystyle \frac {4x - 5}{2x + 3} \times \frac{-4x}{-4x}\)

∴ \(\displaystyle \frac {16x^2 - 20x + 9}{8x^2 + 12x + 21}\) = \(\displaystyle \frac {-16x^2 + 20x}{-8x^2 - 12x}\)

Using Theorem on Equal Ratios,

  Each Ratio = \(\displaystyle \frac {(16x^2 - 20x + 9) + (- 16x^2 + 20x)}{(8x^2 + 12x + 21) + (- 8x^2 - 12x)}\)

∴ Each Ratio = \(\displaystyle \frac {16x^2 - 20x + 9 -16x^2 + 20x}{8x^2 + 12x + 21 -8x^2 - 12x}\)

∴ Each Ratio = \(\displaystyle \frac {\enclose{downdiagonalstrike}{16x^2} - \enclose{updiagonalstrike}{20x} + 9 - \enclose{downdiagonalstrike}{16x^2} + \enclose{updiagonalstrike}{20x}}{\enclose{downdiagonalstrike}{8x^2} + \enclose{updiagonalstrike}{12x} + 21 - \enclose{downdiagonalstrike}{8x^2} - \enclose{updiagonalstrike}{12x}}\)

∴ Each Ratio = \(\displaystyle \frac {9}{21}\)

∴ Each Ratio = \(\displaystyle \frac {3}{7}\)

∴ \(\displaystyle \frac {16x^2 - 20x + 9}{8x^2 + 12x + 21}\) = \(\displaystyle \frac {4x - 5}{2x + 3}\) = \(\displaystyle \frac {3}{7}\)

Now, we will solve this considering the 2^nd and the 3^rd ratio.

 \(\displaystyle \frac {4x - 5}{2x + 3}\) = \(\displaystyle \frac {3}{7}\)

By Cross Multiplication,

 \(\displaystyle 7(4x - 5)\) = \(\displaystyle 3(2x + 3)\)

∴ \(\displaystyle 28x - 35\) = \(\displaystyle 6x + 9\)

∴ \(\displaystyle 28x - 6x\) = \(\displaystyle 9 + 35\)

∴ \(\displaystyle 22x\) = \(\displaystyle 44\)

∴ \(\displaystyle x\) = \(\displaystyle \frac{44}{22}\)

∴ \(\displaystyle x\) = 2

∴ \(\displaystyle x\) = 2 is the solution of the given equation.

Solution:

\(\displaystyle \frac {5y^2 + 40y - 12}{5y + 10y^2 - 4}\) = \(\displaystyle \frac {y + 8}{1 + 2y}\) ... (Given)

Multiplying the numerator and denominator of the second ratio by − 5y,

 \(\displaystyle \frac {5y^2 + 40y - 12}{5y + 10y^2 - 4}\) = \(\displaystyle \frac {y + 8}{1 + 2y} \times \frac{-5y}{-5y}\)

∴ \(\displaystyle \frac {5y^2 + 40y - 12}{5y + 10y^2 - 4}\) = \(\displaystyle \frac {-5y^2 - 40y}{-5y - 10y^2}\)

Using Theorem on Equal Ratios,

 Each Ratio = \(\displaystyle \frac {(5y^2 + 40y - 12) + (- 5y^2 - 40y)}{(5y + 10y^2 - 4) + (- 5y - 10y^2)}\)

∴ Each Ratio = \(\displaystyle \frac {5y^2 + 40y - 12 - 5y^2 - 40y}{5y + 10y^2 - 4 - 5y - 10y^2}\)

∴ Each Ratio = \(\displaystyle \frac {\enclose{downdiagonalstrike}{5y^2} + \enclose{updiagonalstrike}{40y} - 12 - \enclose{downdiagonalstrike}{5y^2} - \enclose{updiagonalstrike}{40y}}{\enclose{updiagonalstrike}{5y} + \enclose{downdiagonalstrike}{10y^2} - 4 - \enclose{updiagonalstrike}{5y} - \enclose{downdiagonalstrike}{10y^2}}\)

∴ Each Ratio = \(\displaystyle \frac {-12}{-4}\)

∴ Each Ratio = \(\displaystyle \frac {3}{1}\)

∴ \(\displaystyle \frac {5y^2 + 40y - 12}{5y + 10y^2 - 4}\) = \(\displaystyle \frac {y + 8}{1 + 2y}\) = \(\displaystyle \frac {3}{1}\)