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

i) $5a+7b:5a-7b=5c+7d:5c-7d$

ii) $(9a+13b)(9c-13d)=(9c+13d)(9a-13b)$

iii) $xa+yb:xc+yd=b:d$

i) $5a+7b:5a-7b=5c+7d:5c-7d$

Given $\frac{a}{b}=\frac{c}{d}$

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

By componendo and dividendo

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

ii) $(9a+13b)(9c-13d)=(9c+13d)(9a-13b)$

Given $\frac{a}{b}=\frac{c}{d}$

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

By componendo and dividendo

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

or  $(9a+13b)(9c-13d)=(9c+13d)(9a-13b)$

iii) $xa+yb:xc+yd=b:d$

Given $\frac{a}{b}=\frac{c}{d}$

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

By componendo

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

Cross multiplying

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

or  $\frac{xa+yb}{xc+yd}=\frac{b}{d}$

or $xa+yb:xc+yd=b:d$

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

Given $\frac{a}{b}=\frac{c}{d}$

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

By Componendo

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

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

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

Again   $\frac{a}{b}=\frac{c}{d}$

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

By Dividendo

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

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

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

Hence $(6a+7b)(3c-4d)=(6c+7d)(3a-4b)$

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

Given $\frac{a}{b}=\frac{c}{d}$

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

By componendo and dividendo

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

By Alternendo

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

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

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

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

By componendo and dividendo

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

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

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

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

Given $(7a+8b)(7c-8d)=(7a-8b)(7c+8d)$

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

By componendo and dividendo

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

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

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

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Question 6:

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

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

i) Given $x=\frac{6ab}{a+b}$

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

By componendo and dividendo

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

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

Similarly

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

By componendo and dividendo

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

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

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

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

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

By componendo and dividendo

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

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

Similarly

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

By componendo and dividendo

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

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

Adding i) and ii) we get

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

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

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

Given $(a+b+c+d)(a-b-c+d)=(a+b-c-d)(a-b+c-d)$

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

By componendo and dividendo

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

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

Applying again componendo and dividendo

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

Simplifying

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

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

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

By componendo and dividendo

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

Simplifying

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

Applying again componendo and dividendo

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

Simplifying

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

$2ad=3bc$

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

Given $(a^2+b^2 )(x^2+y^2 )=(ax+by)^2$

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

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

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

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

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

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

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

Given $\frac{a}{b}=\frac{b}{c} = k$

$a = bk \ and \ b = ck$

or $a = ck^2$

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

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

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

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

$=k^2$

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

Hence LHS = RHS. Hence Proved.

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

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

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

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

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

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

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

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

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

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

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

Hence LHS = RHS. Hence Proved.

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Question 11: Using Properties of proportion, solve for $x$ :

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

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

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

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

Applying componendo and dividendo

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

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

Squaring both sides

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

$x = 20$

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

Applying componendo and dividendo

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

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

Squaring both sides

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

Applying componendo and dividendo

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

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

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

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

Applying componendo and dividendo

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

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

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

Squaring both sides

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

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

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

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

Applying componendo and dividendo

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

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

Squaring both sides

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

Applying componendo and dividendo once again

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

simplifying

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

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

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

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Question 13: Using the properties of proportion, solve for $x$ Given: $\frac{(x^4+1)}{2x^2} =\frac{17}{8}$.    [2013]

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

Applying componendo and dividendo

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

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

Taking the square root of both sides

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

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

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

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

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

Applying componendo and dividendo

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

Simplifying

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

Squaring both sides

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

Applying componendo and dividendo again

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

Simplifying

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

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

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

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

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

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

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

Applying componendo and dividendo again

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

Simplifying

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

or $nx = my$

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