.. 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

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Solutions for Variations Problems
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1. Since x varies as y we can write x=yk where k is proportionality constant
which means :math:`k=\frac{8}{15}.` Now when y=10 x=10*8/15
i.e. :math:`x=\frac{16}{3}.`

2. Since x varies inversely as y we have :math:`xy=k` i.e. k=21 since x=3 and
y=7. Clearly, when x=7 y will become 3.

3. Given :math:`x^2=ky^3` and when x=3, y=4 which implies k=9/64. So our
problem is:

.. math::
   y = \frac{1*64}{3*9} = \frac{4}{3}.

4. Since x varies as y and z jointly we have x=yzk. From given values of x, y
and z let us compute k.

.. math::
   k = \frac{x}{yz} = 2*\frac{5}{3}*\frac{27}{10} = 9

Now we need to compute z when x=54 and y=3

.. math::
   z = \frac{x}{ky} = \frac{54}{3*9} = 2.

5. Given a=ck1 and b=ck2 where k1 and k2 are some constants. :math:`a\pm b`
implies :math:`c(k1\pm k2)` therefore it varies as c. Similarly,
:math:`\sqrt{ab}` implies :math:`c\sqrt{k1k2}` therefore it also varies as c.

6. Since a varies as bc can write a=bck. This means b = (a/ck) = ak1/c where
k1=1/k and since k is constant k1 will also be constant. The above we can also
write as b=k1/(c/a) so we can say b varies inversely as c/a.

7. From problem statement

.. math::
   a=\frac{bk}{c}~\Rightarrow~k=\frac{ca}{b}

   \Rightarrow~k=\frac{9}{14}*\frac{2}{3}*\frac{7}{3} = 1

   \therefore~b=ca=\sqrt{48*75} = 60

8. Since x varies as y x=yk. Clearly :math:`x^2+y^2=y^2(k^2+1)` and
:math:`x^2-y^2=y^2(k^2-1)` now :math:`k^2+1` and :math:`k^2-1` are constants
since k is a constant. Therefore, :math:`x^2+y^2` varies as :math:`x^2-y^2`.

9. Let y=z1+z2 now given that z1=xk1 and z2=x/k2 where k1 and k2 are some
constants.

Given when y=6 the x=4. This implies:

.. math::
   6=4k1+\frac{4}{k2}

Also, when y=10/3 then x=3. This implies

.. math::
   \frac{10}{3} = 3k1+\frac{3}{k2}

Now these two equations involving k1 and k2 can be solved to get their values
and then the relation between x and y can be established.

Rest of the problems are left as exercise to the reader.