.. meta::
   :author: Shiv Shankar Dayal
   :title: Preface
   :description: Algebra
   :keywords: Algebra, ratio, proportions, variations, complex numbers,
              arithmetic progressions, geometric progressions, harmonic
              progressions, series, sequence, quadratic equations,
              permutations, combinations, lograithms, binomial theorem,
              determinant, matricesNumber System

Solutions for Proportions
*************************
1. a. Let the number is x. Then, we have,

.. math::
   \frac{3}{4}=\frac{9}{x}~
   \Rightarrow~x=12

Second and third part are left as exercise to the reader.

2. a. Let mean proportional be x so we have,

.. math::
   x^2=ab~\Rightarrow~x=\sqrt{ab}

Second part is left as an exercise.

3. a. Let third proportional be x. Then, we have,

.. math::
   9^2=7x~\Rightarrow~x=\frac{81}{7}

Second part is left as an exercise.

4. a. Let us equate the given proportion\'s ratios to k.

.. math::
   \frac{a}{b}=\frac{c}{d}~\Rightarrow~a=bk,~c=dk

Now let us evaluate L.H.S. by substituting above value for a and c.

.. math::
   \frac{a^c+ac^2}{b^d+bd^2}=\frac{b^2k^2dk+bkd^2k^2}{b^d+db^2}=k^3

Now let us evaluate R.H.S. by substituting above value for a and c.

.. math::
   \frac{(bk+dk)^3}{(b+d)^3}=k^3

.. math::
   \text{Hence,} L.H.S.= R.H.S.

5. Proceeding the same way as previous problem we have L.H.S. as

.. math::
   \frac{pb^2k^2+qb^2}{p^2k^2-qb^2}=\frac{pk^2+q}{pk^2-q}

and R.H.S. as

.. math::
   \frac{pd^2k^2+qd^2}{pd^2k^2-qd^2}=\frac{pk^2+q}{pk^2-q}

.. math::
   \text{Hence,} L.H.S.= R.H.S.

6. Processing similarly L.H.S. is

.. math::
   \frac{bk-dk}{b-d}=k

and R.H.S. is

.. math::
   \frac{\sqrt{b^2k^2+d^2k^2}}{\sqrt{b^2+d^2}} = k

.. math::
   \text{Hence,} L.H.S.= R.H.S.

7. We have computed R.H.S. to be k in last problem\'s R.H.S. So let us try to
calculate R.H.S.

.. math::
   \frac{\sqrt{bdk^2+\frac{d^3k^3}{bk}}}{\sqrt{bd+\frac{d^3}{b}}}=k

8. Since a, b, c and d are in continued proportion we can write them as

.. math::
   \frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k

.. math::
   \Rightarrow~a=dk^3, b=dk^2,~\text{and}~c=dk

First we do L.H.S.

.. math::
   \frac{a}{b+d}=\frac{dk^3}{dk^2+d}=\frac{k^3}{k^2+1}

now we do R.H.S.

.. math::
   \frac{c^3}{c^2d+d^3}=\frac{d^3k^3}{d^2k^2d+d^2}=\frac{k^3}{k^2+1}.

Problem no. 9 and 10 are left as exercises to the reader as they are similar.

11. Since b is mean proportional between a and b we can write
:math:`b^2=ac`. Now let us evaluate the expression.

.. math::
   L.H.S.=\frac{a^2-ac+c^2}{\frac{1}{a^2}-\frac{1}{ac}+\frac{1}{c^2}}

.. math::
   \Rightarrow~\frac{(a^2-ac-c^2)a^2c^2}{c^2-ac+a^2}

.. math::
   \Rightarrow~a^2c^2 = b^4

12. Equating first set of ratio to :math:`k` and second set to :math:`l` we
have following:

.. math::
   a=bk\text{ and }c=dk;~e=fl\text{ and }g=hl

Substituting these value for :math:`a, c, e \text{ and } g` we get both left
hand side and right hand side equal to

.. math::
   \frac{kl+1}{kl-1}

13. Proceeding similarly as previous problems we can say that :math:`a=bk` and
:math:`c=dk`. Now substituting for :math:`a` and :math:`c` we get L.H.S. as

.. math::
   L.H.S. = \{bk+b+dk+d\}\{bk-b-dk+d\} = \{(b+d)(k+1)\}\{(b-d)(k-1)\}

.. math::
   \Rightarrow~(b^2-d^2)(k^2-1)

Similarly for R.H.S. we have,

.. math::
   R.H.S. = \{bk-b+dk-d\}\{bk+b-dk-d\} = \{(b-d)(k+1)\}\{(b+d)(k+1)\}

.. math::
   \Rightarrow~(b^2-d^2)(k^2-1).


Remaining problems are left exercise to the reader.