Question

How do you solve for $ x $ and $ y $ if the equations are
 $ \left( {\dfrac{5}{{x + y}}} \right) - \left( {\dfrac{2}{{x - y}}} \right) + 1 = 0 $ and $ \left( {\dfrac{{15}}{{x + y}}} \right) + \left( {\dfrac{7}{{x - y}}} \right) - 10 = 0 $

Answer 1

Hint: In order to solve the given system of equations, use the Gabriel Cramer’s rule by assuming $ x + y = p $ and $ x - y = q $ .Rewrite the equations as per the assumptions. Compare both the equations with $ \dfrac{a}{p} + \dfrac{b}{q} = c $ to get the values of the variable $ a,b,c $ . Now calculate $ D,{D_x},{D_y} $ which are equal to $ D = \left| {\begin{array}{*{20}{c}}
  {{a_1}}&{{b_1}} \\
  {{a_2}}&{{b_2}}
\end{array}} \right| $ , $ {D_x} = \left| {\begin{array}{*{20}{c}}
  {{c_1}}&{{b_1}} \\
  {{c_2}}&{{b_2}}
\end{array}} \right| $ and $ {D_y} = \left| {\begin{array}{*{20}{c}}
  {{a_1}}&{{c_1}} \\
  {{a_2}}&{{c_2}}
\end{array}} \right| $ .According to Cramer’s rule $ \dfrac{1}{p} = \dfrac{{{D_x}}}{D} $ and $ \dfrac{1}{q} = \dfrac{{{D_y}}}{D} $ simplifying these, you will get two equations in variable $ x $ and $ y $ . Solve them using elimination of variable methods to get the required value of variables $ x $ and $ y $ .

Complete step by step solution:
We are given a system of equation in variables $ x $ and $ y $ as
$ \left( {\dfrac{5}{{x + y}}} \right) - \left( {\dfrac{2}{{x - y}}} \right) + 1 = 0 $ and $ \left( {\dfrac{{15}}{{x + y}}} \right) + \left( {\dfrac{7}{{x - y}}} \right) - 10 = 0 $
In order to solve the given system of equations, we will be using Gabriel Cramer’s method .
To use this method , let us assume $ x + y = p $ and $ x - y = q $ , try to rewrite the given system of equations according to this assumption.
So , we get the first equation equal to
$ \left( {\dfrac{5}{p}} \right) - \left( {\dfrac{2}{q}} \right) + 1 = 0 \Rightarrow \dfrac{5}{p} - \dfrac{2}{q} = - 1 $
Let $ {a_1} = 5,{b_1} = - 2,{c_1} = - 1 $
Now Similarly rewriting the second equation , we get
$ \Rightarrow \left( {\dfrac{{15}}{p}} \right) + \left( {\dfrac{7}{q}} \right) - 10 = 0 \Rightarrow \dfrac{{15}}{p} + \dfrac{7}{q} = 10 $
Let $ {a_2} = 15,{b_2} = 7,{c_2} = 10 $
Now we are going to calculate $ D,{D_x},{D_y} $ which are equal to $ D = \left| {\begin{array}{*{20}{c}}
  {{a_1}}&{{b_1}} \\
  {{a_2}}&{{b_2}}
\end{array}} \right| $ , $ {D_x} = \left| {\begin{array}{*{20}{c}}
  {{c_1}}&{{b_1}} \\
  {{c_2}}&{{b_2}}
\end{array}} \right| $ and $ {D_y} = \left| {\begin{array}{*{20}{c}}
  {{a_1}}&{{c_1}} \\
  {{a_2}}&{{c_2}}
\end{array}} \right| $ respectively by putting the above assumed values ,
We get
$
  D = \left| {\begin{array}{*{20}{c}}
  {{a_1}}&{{b_1}} \\
  {{a_2}}&{{b_2}}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
  5&{ - 2} \\
  {15}&7
\end{array}} \right| = 7 \times 5 - \left( { - 2} \right) \times 15 \\
  D = 35 + 30 \\
  D = 65 \\
$
Also,
\[
  {D_x} = \left| {\begin{array}{*{20}{c}}
  {{c_1}}&{{b_1}} \\
  {{c_2}}&{{b_2}}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
  { - 1}&{ - 2} \\
  {10}&7
\end{array}} \right| = - 1 \times 7 - \left( { - 2} \right) \times 10 \\
  {D_x} = 20 - 7 \\
  {D_x} = 13 \\
 \]
And at the last
\[
  {D_y} = \left| {\begin{array}{*{20}{c}}
  {{a_1}}&{{c_1}} \\
  {{a_2}}&{{c_2}}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
  5&{ - 1} \\
  {15}&{10}
\end{array}} \right| = 10 \times 50 - \left( { - 1} \right) \times 15 \\
  {D_y} = 50 + 15 \\
  {D_y} = 65 \\
 \]
Hence, we have obtained the value for $ D,{D_x},{D_y} $ as $ 65,13,65 $ respectively
Now as we can clearly see that $ D \ne 0; $ then by Cramer’s Rule we can write
 $ \dfrac{1}{p} = \dfrac{{{D_x}}}{D} $ or
 $ p = \dfrac{D}{{{D_x}}} = \dfrac{{65}}{{13}} = 5 $
And similarly , according to Cramer’s rule we can also write
$ \dfrac{1}{q} = \dfrac{{{D_y}}}{D} $ or
$ q = \dfrac{D}{{{D_y}}} = \dfrac{{65}}{{65}} = 1 $

Now putting the $ p = x + y\,and\,q = x - y $ as we have assumed earlier we have
$ x + y = 5 $ ---(1)
$ x - y = 1 $ ----(2)
To solve both the equation, we are going to add them i.e. LHS of equation 1 with LHS of equation 2 and same with the RHS side also in order to eliminate the variable $ y $
$
  x + y + x - y = 5 + 1 \\
  2x = 6 \\
  x = 3 \\
$
Putting this value of $ x $ in equation (1) and solving it for variable $ y $ , we get the value of variable $ y $ as
$
  3 + y = 5 \\
  y = 5 - 3 \\
  y = 2 \\
  $
 $ \therefore x = 3,y = 2 $
Therefore, the solution to the given system of equations is $ x = 3\,and\,y = 2 $

Additional Information:
Linear Equation in two variables: A linear equation is an equation which can be represented in the form of $ ax + by + c $ where $ x $ and $ y $ are the unknown variables and c is the number known where $ a \ne 0,b \ne 0 $ .
The degree of the variable in the linear equation is of the order 1.

Note: 1.You can also use substitution to solve the pair of linear equations in the end to solve the equations
2.If the value of determinant $ D $ is equal to zero, then the Cramer’s Rule does not apply and then the given system of equations has either no solution or infinite solutions.
3.Calculate the determinant properly with proper positioning of signs.
4.Cramer’s Rule is named for the Mathematician Gabriel Cramer.