Question 1: If $\displaystyle a:b=c:d$ , prove that:

$\displaystyle \text{i) } 5a+7b:5a-7b=5c+7d:5c-7d$

$\displaystyle \text{ii) } (9a+13b)(9c-13d)=(9c+13d)(9a-13b)$

$\displaystyle \text{iii) } xa+yb:xc+yd=b:d$

$\displaystyle \text{i) } 5a+7b:5a-7b=5c+7d:5c-7d$

$\displaystyle \text{Given } \frac{a}{b} = \frac{c}{d}$

$\displaystyle \Rightarrow \frac{5a}{7b} = \frac{5c}{7d}$

By componendo and dividendo

$\displaystyle \Rightarrow \frac{5a+7b}{5a-7b} = \frac{5c+7d}{5c-7d}$

$\displaystyle \text{ii) } (9a+13b)(9c-13d)=(9c+13d)(9a-13b)$

$\displaystyle \text{Given } \frac{a}{b} = \frac{c}{d}$

$\displaystyle \Rightarrow \frac{9a}{13b} = \frac{9c}{13d}$

By componendo and dividendo

$\displaystyle \Rightarrow \frac{9a+13b}{9a-13b} = \frac{9c+13d}{9c-13d}$

$\displaystyle \text{or } (9a+13b)(9c-13d)=(9c+13d)(9a-13b)$

$\displaystyle \text{iii) } xa+yb:xc+yd=b:d$

$\displaystyle \text{Given } \frac{a}{b} = \frac{c}{d}$

$\displaystyle \Rightarrow \frac{xa}{yb} = \frac{xc}{yd}$

By componendo

$\displaystyle \Rightarrow \frac{xa+yb}{yb} = \frac{xc+yd}{yd}$

Cross multiplying

$\displaystyle \Rightarrow \frac{xa+yb}{xc+yd} = \frac{yb}{yd}$

$\displaystyle \text{or } \frac{xa+yb}{xc+yd} = \frac{b}{d}$

$\displaystyle \text{or } xa+yb:xc+yd=b:d$

$\displaystyle \\$

Question 2: If $\displaystyle a:b=c:d$ , prove that: $\displaystyle (6a+7b)(3c-4d)=(6c+7d)(3a-4b)$ .

$\displaystyle \text{Given } \frac{a}{b} = \frac{c}{d}$

$\displaystyle \Rightarrow \frac{6a}{7b} = \frac{6c}{7d}$

By Componendo

$\displaystyle \frac{6a+7b}{7b} = \frac{6c+7d}{7d}$

$\displaystyle \text{or } \frac{6a+7b}{6c+7d} = \frac{7b}{7d}$

$\displaystyle \text{or } \frac{6a+7b}{6c+7d} = \frac{b}{d}$

$\displaystyle \text{Again } \frac{a}{b} = \frac{c}{d}$

$\displaystyle \frac{3a}{4b} = \frac{3c}{4d}$

By Dividendo

$\displaystyle \frac{3a-4b}{4b} = \frac{3c-4d}{4d}$

$\displaystyle \text{or } \frac{3a-4b}{3c-4d} = \frac{4b}{4d}$

$\displaystyle \text{or } \frac{3a-4b}{3c-4d} = \frac{b}{d}$

Hence $\displaystyle (6a+7b)(3c-4d)=(6c+7d)(3a-4b)$

$\displaystyle \\$

$\displaystyle \text{Question 3: Given, } \frac{a}{b} = \frac{c}{d} \text{ , prove that: } \frac{3a-5b}{3a+5b} = \frac{3c-5d}{3c+5d}$ [2000]

$\displaystyle \text{Given } \frac{a}{b} = \frac{c}{d}$

$\displaystyle \Rightarrow \frac{3a}{5b} = \frac{3c}{5d}$

By componendo and dividendo

$\displaystyle \frac{3a+5b}{3a-5b} = \frac{3c+5d}{3c-5d}$

By Alternendo

$\displaystyle \frac{3a-5b}{3a+5b} = \frac{3c-5d}{3c+5d}$

$\displaystyle \\$

$\displaystyle \text{Question 4: If } \frac{5x+6y}{5u+6v} = \frac{5x-6y}{5u-6v} \text{ ; Then prove that } x:y=u:v$ .

$\displaystyle \frac{5x+6y}{5u+6v} = \frac{5x-6y}{5u-6v}$

$\displaystyle \Rightarrow \frac{5x+6y}{5x-6y} = \frac{5u+6v}{5u-6v}$

By componendo and dividendo

$\displaystyle \Rightarrow \frac{(5x+6y) +( 5x-6y)}{(5x+6y) -(5x-6y} = \frac{(5u+6v)+(5u-6v)}{(5u+6v) - (5u-6v)}$

$\displaystyle \Rightarrow \frac{10x}{12y} = \frac{10u}{12v}$

$\displaystyle \Rightarrow x:y=u:v$ .

$\displaystyle \\$

Question 5: If $\displaystyle (7a+8b)(7c-8d)=(7a-8b)(7c+8d)$ ; prove that $\displaystyle a:b=c:d$

$\displaystyle \text{Given } (7a+8b)(7c-8d)=(7a-8b)(7c+8d)$

$\displaystyle \frac{7a+8b}{7a-8b} = \frac{7c+8d}{7c-8d}$

By componendo and dividendo

$\displaystyle \frac{7a+8b+7a-8b}{7a+8b -7a+8b} = \frac{7c+8d+7c-8d}{7c+8d -7c+8d}$

$\displaystyle \Rightarrow \frac{14a}{16b} = \frac{14c}{16d}$

$\displaystyle a:b=c:d$ . Hence Proved.

$\displaystyle \\$

Question 6:

$\displaystyle \text{i) If } x= \frac{6ab}{a+b} \text{ , find the value of: } \frac{x+3a}{x-3a} + \frac{x+3b}{x-3b}$

$\displaystyle \text{ii) If } a= \frac{4\sqrt{6}}{\sqrt{2}+\sqrt{3}} \text{ , find the value of: } \frac{a+2\sqrt{2}}{a-2\sqrt{2}} + \frac{a+2\sqrt{3}}{a-2\sqrt{3}}$

$\displaystyle \text{i) } \text{Given } x= \frac{6ab}{a+b}$

$\displaystyle \frac{x}{3a} = \frac{2b}{a+b}$

By componendo and dividendo

$\displaystyle \frac{x+3a}{x-3a} = \frac{2b+a+b}{2b - a-b}$

$\displaystyle \frac{x+3a}{x-3a} = \frac{3b+a}{b - a}$ … … … … … … i)

Similarly

$\displaystyle \frac{x}{3b} = \frac{2a}{a+b}$

By componendo and dividendo

$\displaystyle \frac{x+3b}{x-3b} = \frac{3a+b}{a-b}$ … … … … … … ii)

$\displaystyle \frac{x+3a}{x-3a}+ \frac{x+3b}{x-3b} = \frac{3b+a}{b - a} + \frac{3a+b}{a-b}$

$\displaystyle \frac{x+3a}{x-3a} + \frac{x+3b}{x-3b} = \frac{2a-2b}{a-b} = 2$

$\displaystyle \text{ii) } \text{Given } a= \frac{4\sqrt{6}}{\sqrt{2}+\sqrt{3}}$

$\displaystyle \frac{a}{2\sqrt{2}} = \frac{2\sqrt{3}}{\sqrt{2}+\sqrt{3}}$

By componendo and dividendo

$\displaystyle \frac{a+2\sqrt{2}}{a-2\sqrt{2}} = \frac{2\sqrt{3}+\sqrt{2}+\sqrt{3} }{2\sqrt{3} - \sqrt{2}-\sqrt{3}}$

$\displaystyle \frac{a+2\sqrt{2}}{a-2\sqrt{2}} = \frac{3\sqrt{3}+\sqrt{2} }{\sqrt{3} - \sqrt{2}}$ … … … … … … i)

Similarly

$\displaystyle \frac{a}{2\sqrt{3}} = \frac{2\sqrt{2}}{\sqrt{2}+\sqrt{3}}$

By componendo and dividendo

$\displaystyle \frac{a+2\sqrt{3}}{a-2\sqrt{3}} = \frac{2\sqrt{2}+ \sqrt{2}+\sqrt{3}}{2\sqrt{2} - \sqrt{2}-\sqrt{3}}$

$\displaystyle \frac{a+2\sqrt{3}}{a-2\sqrt{3}} = \frac{3\sqrt{2}+\sqrt{3}}{\sqrt{2} -\sqrt{3}}$ … … … … … … ii)

Adding i) and ii) we get

$\displaystyle \frac{a+2\sqrt{2}}{a-2\sqrt{2}}+ \frac{a+2\sqrt{3}}{a-2\sqrt{3}} = \frac{3\sqrt{3}+\sqrt{2} }{\sqrt{3} - \sqrt{2}} + \frac{3\sqrt{2}+\sqrt{3}}{\sqrt{2} -\sqrt{3}}$

$\displaystyle \frac{a+2\sqrt{2}}{a-2\sqrt{2}}+ \frac{a+2\sqrt{3}}{a-2\sqrt{3}} = \frac{2\sqrt{2}-2\sqrt{3} }{\sqrt{2} - \sqrt{3}} = 2$

$\displaystyle \\$

Question 7: If $\displaystyle (a+b+c+d)(a-b-c+d)=(a+b-c-d)(a-b+c-d)$ ; prove that $\displaystyle a:b=c:d$

$\displaystyle \text{Given } (a+b+c+d)(a-b-c+d)=(a+b-c-d)(a-b+c-d)$

$\displaystyle \frac{a+b+c+d}{a+b-c-d} = \frac{a-b+c-d}{a-b-c+d}$

By componendo and dividendo

$\displaystyle \frac{(a+b+c+d)+(a+b-c-d)}{(a+b+c+d)-(a+b-c-d)} = \frac{(a-b+c-d)+(a-b-c+d)}{(a-b+c-d)-(a-b-c+d)}$

$\displaystyle \frac{a+b}{a-d} = \frac{c+d}{c-d}$

Applying again componendo and dividendo

$\displaystyle \frac{a+b+a-b}{a+b-a+d} = \frac{c+d+c-d}{c+d -c+d}$

Simplifying

$\displaystyle \frac{a}{b} = \frac{c}{d}$ Hence Proved.

$\displaystyle \\$

$\displaystyle \text{Question 8: If } \frac{a-2b-3c+4d}{a+2b-3c-4d} = \frac{a-2b-3c-4d}{a+2b+3c+4d} \text{ , show that } 2ab=3bc$ .

$\displaystyle \text{Given } \frac{(a-2b-3c+4d)}{(a+2b-3c-4d)} = \frac{(a-2b-3c-4d)}{(a+2b+3c+4d)}$

By componendo and dividendo

$\displaystyle \frac{(a-2b-3c+4d)+ (a+2b-3c-4d)}{(a-2b-3c+4d)-(a+2b-3c-4d)} = \frac{(a-2b-3c-4d)+(a+2b+3c+4d)}{(a-2b-3c-4d)-(a+2b+3c+4d)}$

Simplifying

$\displaystyle \frac{(a-3c)}{(a+3c)} = \frac{(4d-2b)}{(-4d-2b)}$

Applying again componendo and dividendo

$\displaystyle \frac{(a-3c)+(a+3c)}{(a-3c)-(a+3c)} = \frac{(4d-2b)+(-4d-2b)}{(4d-2b)-(-4d-2b)}$

Simplifying

$\displaystyle \frac{a}{-3c} = \frac{-b}{2d}$

$\displaystyle 2ad=3bc$

$\displaystyle \\$

$\displaystyle \text{Question 9: If } (a^2+b^2 )(x^2+y^2 )=(ax+by)^2 \text{ ; prove that: } \frac{a}{x} = \frac{b}{y}$ .

$\displaystyle \text{Given } (a^2+b^2 )(x^2+y^2 )=(ax+by)^2$

$\displaystyle a^2x^2+b^2x^2+a^2y^2+b^2y^2 = a^2x^2+b^2y^2+2abxy$

$\displaystyle a^2y^2+b^2x^2-2abxy = 0$

$\displaystyle \Rightarrow (ay-bx)^2 = 0 \Rightarrow ay-bx=0$

$\displaystyle \text{Therefore } ay = bx \Rightarrow \frac{a}{x} = \frac{b}{y} \text{ Hence Proved. }$

$\displaystyle \\$

Question 10: If $\displaystyle a, b \ and \ c$ are in continued proportion, prove that:

$\displaystyle \text{i) } \frac{a^2+ab+b^2}{b^2+bc+c^2} = \frac{a}{c}$

$\displaystyle \text{ii) } \frac{a^2+b^2+c^2}{(a+b+c)^2} = \frac{a-b+c}{a+b+c}$

$\displaystyle \text{Given } \frac{a}{b} = \frac{b}{c} = k$

$\displaystyle a = bk \ and \ b = ck$

$\displaystyle \text{or } a = ck^2$

$\displaystyle \text{i) } \frac{a^2+ab+b^2}{b^2+bc+c^2} = \frac{a}{c}$

$\displaystyle \text{LHS } = \frac{{ck^2}^2+{ck^2}{ck}+{ck}^2}{{ck}^2+{ck}c+c^2}$

$\displaystyle = \frac{c^2k^4+c^2k^3+c^2k^2}{c^2k^2+c^2k+ c^2}$

$\displaystyle = \frac{c^2k^2(k^2+k+1)}{c^2(k^2+k+1)}$

$\displaystyle =k^2$

$\displaystyle \text{RHS } = \frac{a}{c} = \frac{ck^2}{c} = k^2$

Hence LHS = RHS. Hence Proved.

$\displaystyle \text{ii) } \frac{a^2+b^2+c^2}{(a+b+c)^2} = \frac{a-b+c}{a+b+c}$

$\displaystyle \text{LHS } = \frac{a^2+b^2+c^2}{(a+b+c)^2}$

$\displaystyle = \frac{{(ck^2)}^2+{ck}^2+c^2}{({ck^2}+ck+c)^2}$

$\displaystyle = \frac{c^2(k^4+k^2+1)}{(k^2+k+1)^2}$

$\displaystyle = \frac{k^4+k^2+1}{(k^2+k+1)^2}$

$\displaystyle \text{RHS } = \frac{a-b+c}{a+b+c}$

$\displaystyle = \frac{ck^2-ck+c}{ck^2+ck+c}$

$\displaystyle = \frac{c(k^2-k+1)}{c(k^2+k+1)}$

$\displaystyle = \frac{k^2-k+1}{k^2+k+1}$

$\displaystyle = \frac{(k^2-k+1)(k^2+k+1)}{(k^2+k+1)^2}$

$\displaystyle = \frac{k^4+k^2+1}{(k^2+k+1)^2}$

Hence LHS = RHS. Hence Proved.

$\displaystyle \\$

Question 11: Using Properties of proportion, solve f$\displaystyle \text{or } x$ :

$\displaystyle \text{i) } \frac{\sqrt{x+5}+\sqrt{x-16}}{\sqrt{x+5}-\sqrt{x-16}} = \frac{7}{3}$

$\displaystyle \text{ii) } \frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}} = \frac{4x-1}{2}$

$\displaystyle \text{iii) } \frac{3x+\sqrt{9x^2-5}}{3x-\sqrt{9x^2-5}} =5$

$\displaystyle \text{i) } \frac{\sqrt{x+5}+\sqrt{x-16}}{\sqrt{x+5}-\sqrt{x-16}} = \frac{7}{3}$

Applying componendo and dividendo

$\displaystyle \frac{\sqrt{x+5}+\sqrt{x-16}+ \sqrt{x+5}-\sqrt{x-16}}{ \sqrt{x+5}+\sqrt{x-16}- \sqrt{x+5}+\sqrt{x-16}} = \frac{7+3}{7-3}$

$\displaystyle \frac{\sqrt{x+5}}{\sqrt{x-16}} = \frac{5}{2}$

Squaring both sides

$\displaystyle \frac{(x+5)}{(x-16)} = \frac{25}{4} \Rightarrow 21x = 420$

$\displaystyle x = 20$

$\displaystyle \text{ii) } \frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}} = \frac{4x-1}{2}$

Applying componendo and dividendo

$\displaystyle \frac{\sqrt{x+1}+\sqrt{x-1}+\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}-\sqrt{x+1}+\sqrt{x-1}} = \frac{4x-1+2}{4x-1-2}$

$\displaystyle \frac{2\sqrt{x+1}}{2\sqrt{x-1}} = \frac{4x+1}{4x-3}$

Squaring both sides

$\displaystyle \frac{x+1}{x-1} = \frac{16x^1+1+8x}{16x^2+9-24x}$

Applying componendo and dividendo

$\displaystyle \frac{x+1+x-1}{x+1-x+1} = \frac{16x^2+1+8x+16x^2+9-24x}{16x^1+1+8x-16x^2-9+24x}$

$\displaystyle \frac{2x}{2} = \frac{32x^2+10-16x}{32-8x}$

$\displaystyle \text{or } 4x=5 \Rightarrow x = \frac{5}{4}$

$\displaystyle \text{iii) } \frac{3x+\sqrt{9x^2-5}}{3x-\sqrt{9x^2-5}} =5$

Applying componendo and dividendo

$\displaystyle \frac{3x+\sqrt{9x^2-5}+3x-\sqrt{9x^2-5}}{3x+\sqrt{9x^2-5}-3x+\sqrt{9x^2-5}}$latex \displaystyle =5 \$

$\displaystyle \frac{6x}{2\sqrt{9x^2-5}} = \frac{6}{4}$

$\displaystyle \frac{x}{\sqrt{9x^2-5}} = \frac{1}{2}$

Squaring both sides

$\displaystyle \frac{x^2}{9x^2-5} = \frac{1}{4}$

$\displaystyle 5x^2 = 5 \ or\ x = 1$

$\displaystyle \\$

$\displaystyle \text{Question 12: If } x= \frac{\sqrt{a+3b}+\sqrt{a-3b}}{\sqrt{a+3b}-\sqrt{a-3b}} \text{ , prove that: } 3bx^2-2ax+3b=0$ . [2007]

$\displaystyle \text{Given } x=\frac{\sqrt{a+3b}+\sqrt{a-3b}}{\sqrt{a+3b}-\sqrt{a-3b}}$

Applying componendo and dividendo

$\displaystyle \frac{x+1}{x-1} = \frac{(\sqrt{a+3b}+\sqrt{a-3b})-(\sqrt{a+3b}-\sqrt{a-3b})}{(\sqrt{a+3b}+\sqrt{a-3b})-(\sqrt{a+3b}-\sqrt{a-3b})}$

$\displaystyle \frac{x+1}{x-1} = \frac{2\sqrt{a+3b}}{-2\sqrt{a-3b}}$

Squaring both sides

$\displaystyle \frac{x^2+2x+1}{x^2-2x+1} = \frac{a+3b}{a-3b}$

Applying componendo and dividendo once again

$\displaystyle \frac{(x^2+2x+1)-(x^2-2x+1)}{(x^2+2x+1)-(x^2-2x+1)} = \frac{(a+3b)-(a-3b)}{(a+3b)-(a-3b)}$

simplifying

$\displaystyle \frac{x^2+1}{2x} = \frac{a}{3b}$

$\displaystyle 3b(x^2+1) = 2ax$

$\displaystyle 3bx^2-2ax+3b=0$ Hence proved.

$\displaystyle \\$

Question 13: Using the properties of proportion, solve for $x$

$\displaystyle \text{Given: } \frac{(x^4+1)}{2x^2} = \frac{17}{8}$ . [2013]

$\displaystyle \text{Given } \frac{(x^4+1)}{2x^2} = \frac{17}{8}$

Applying componendo and dividendo

$\displaystyle \frac{(x^4+1)+2x^2}{(x^4+1)-2x^2} = \frac{17+8}{17-8}$

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

Taking the square root of both sides

$\displaystyle \frac{x^2+1}{x^2-1} = \frac{5}{3}$

$\displaystyle 3x^2+3=5x^2-5$

$\displaystyle x^2 = 4 \ or \ x = \pm 2$

$\displaystyle \\$

$\displaystyle \text{Question 14: If } x= \frac{\sqrt{m+n}+\sqrt{m-n}}{\sqrt{m+n}-\sqrt{m-n}} \text{ , express } n \text{ in terms of } x \text{ and } m$ .

$\displaystyle \text{Given } x= \frac{\sqrt{m+n}+\sqrt{m-n}}{\sqrt{m+n}-\sqrt{m-n}}$

Applying componendo and dividendo

$\displaystyle \frac{x+1}{x-1} = \frac{(\sqrt{m+n}+\sqrt{m-n})+(\sqrt{m+n}-\sqrt{m-n})}{(\sqrt{m+n}+\sqrt{m-n})-(\sqrt{m+n}-\sqrt{m-n})}$

Simplifying

$\displaystyle \frac{x+1}{x-1} = \frac{\sqrt{m+n}}{\sqrt{m-n}}$

Squaring both sides

$\displaystyle \frac{x^2+1+2x}{x^2+1-2x} = \frac{m+n}{m-n}$

Applying componendo and dividendo again

$\displaystyle \frac{(x^2+1+2x)+(x^2+1-2x)}{(x^2+1+2x)-(x^2+1-2x)} = \frac{m+n+m-n}{m+n-m+n}$

Simplifying

$\displaystyle \frac{x^2+1}{2x} = \frac{m}{n}$

$\displaystyle \text{or } n = \frac{2mx}{x^2+1}$

$\displaystyle \\$

$\displaystyle \text{Question 15: If } \frac{x^3+3xy^2}{3x^2 y+y^3} = \frac{m^3+3mn^2}{3m^2 n+n^3 } \text{ , show that: } nx=my$ .

$\displaystyle \text{Given } \frac{x^3+3xy^2}{3x^2 y+y^3} = \frac{m^3+3mn^2}{3m^2 n+n^3 }$

Applying componendo and dividendo

$\displaystyle \frac{(x^3+3xy^2)+(3x^2 y+y^3)}{(x^3+3xy^2)-(3x^2 y+y^3)} = \frac{(m^3+3mn^2)-(3m^2 n+n^3)}{(m^3+3mn^2)-(3m^2 n+n^3) }$

$\displaystyle \frac{(x+y)^3}{(x-y)^3} = \frac{(m+n)^3}{(m-n)^3}$

$\displaystyle \text{or } \frac{(x+y)}{(x-y)} = \frac{(m+n)}{(m-n)}$

Applying componendo and dividendo again

$\displaystyle \frac{(x+y)+(x+y)}{(x+y)-(x-y)} = \frac{(m+n)+(m-n)}{(m+n)-(m-n)}$

Simplifying

$\displaystyle \frac{x}{y} = \frac{m}{n}$

$\displaystyle \text{or } nx = my$