Question

Solve the following system of equations:
$
  8v = 3u + 5uv \\
  6v - 5u = - 2uv \\
 $

Answer 1

Hint: First of all frame the two given equations in the form of only two variables and then either here we can use substitution method or the elimination method to get the values for the terms “u” and “v”

Complete step by step solution:
Take the given expressions:
$
  8v = 3u + 5uv \\
  6v - 5u = - 2uv \\
 $
Divide both the above equations with
\[\dfrac{{8v}}{{uv}} = \dfrac{{3u}}{{uv}} + \dfrac{{5uv}}{{uv}}\]
Common multiple from the numerator and the denominator cancel each other.
\[\dfrac{8}{u} = \dfrac{3}{v} + 5\]
Place,
 $
  \dfrac{1}{u} = a \\
  \dfrac{1}{v} = b \\
 $ …. (A)
$8a = 3b + 5$
The above equation can be re-written as –
$8a - 3b = 5$ …. (B)
Similarly divide the other equation with
$\dfrac{{6v}}{{uv}} - \dfrac{{5u}}{{uv}} = - 2\dfrac{{uv}}{{uv}}$
Common multiple from the numerator and the denominator cancel each other.
\[\dfrac{6}{u} - \dfrac{5}{v} = - 2\]
Place,
 $
  \dfrac{1}{u} = a \\
  \dfrac{1}{v} = b \\
 $
$6a - 5b = - 2$ …. (C)
Now, multiply equation (B) with and (C) with
$
  40a - 15b = 25 \\
  18a - 15b = - 6 \\
 $
Subtract above equations-
$(40a - 18a) - 15b + 15b = 25 + 6$
Simplify the above equation, like terms with the same value and the opposite sign cancels each other.
$
  22a = 31 \\
  a = \dfrac{{31}}{{22}}{\text{ }}....{\text{ (D)}} \\
 $
Place the above value in the equation (C)
$6\left( {\dfrac{{31}}{{22}}} \right) - 5b = - 2$
Make the required value the subject –
$
  \dfrac{{93}}{{11}} + 2 = 5b \\
  5b = \dfrac{{93 + 22}}{{11}} \\
  b = \dfrac{{115}}{{11(5)}} \\
  b = \dfrac{{23}}{{11}}{\text{ }}....{\text{ (E)}} \\
 $
Place the values of equation (D) and (E ) in equation (B)

$
  \dfrac{1}{u} = \dfrac{{31}}{{22}} \Rightarrow u = \dfrac{{22}}{{31}} \\
  \dfrac{1}{v} = \dfrac{{23}}{{11}} \Rightarrow v = \dfrac{{11}}{{23}} \\
 $
This is the required solution.

Note:
Be careful about the sign convention, and always remember that when you move any term from one side to the opposite side, then the sign of the terms also changes. Positive term becomes negative and the negative term becomes the positive term. Also, remember the product of two negative terms gives a positive term.