Question

Solve the following system of linear equations: \[{x^2} - xy + {y^2} = 21,\;{y^2} - 2xy + 15 = 0\]

Answer 1

Hint: Here, in the question, a system of two equations is given. And we are asked to solve the given system of equations. Solving a system of equations means to find the values of the variables present in the equations. We will use a substitution method first to get the value of one variable in terms of another variable. This will get us a final value of one variable which we will put back in the given equation to find the value of another variable.

Complete step-by-step solution:
Given set of equations:
\[{x^2} - xy + {y^2} = 21\;\;\; \ldots \left( 1 \right) \\
  {y^2} - 2xy + 15 = 0\;\;\; \ldots \left( 2 \right) \]
Simplifying equation \[\left( 2 \right)\], we get,
\[ {y^2} + 15 = 2xy \\
   \Rightarrow x = \dfrac{{{y^2} + 15}}{{2y}}\;\;\;\, \ldots \left( 3 \right) \]
Substituting the value of \[x\] in equation \[\left( 1 \right)\], we get
\[{\left( {\dfrac{{{y^2} + 15}}{{2y}}} \right)^2} - \left( {\dfrac{{{y^2} + 15}}{{2y}}} \right)y + {y^2} - 21 = 0\]
Simplifying this equation, we obtain,
\[\dfrac{{{y^4} + 225 + 30{y^2}}}{{4{y^2}}} - \left( {\dfrac{{{y^2} + 15}}{2}} \right) + {y^2} - 21 = 0\]
Adding all the terms by LCM method, we get,
\[\dfrac{{{y^4} + 225 + 30{y^2} - 2{y^4} - 30{y^2} + 4{y^4} - 84{y^2}}}{{4{y^2}}} = 0\]
Multiplying\[4{y^2}\]both sides, we get,
\[ {y^4} + 225 + 30{y^2} - 2{y^4} - 30{y^2} + 4{y^4} - 84{y^2} = 0 \\
   \Rightarrow 3{y^4} - 84{y^2} + 225 = 0 \]
Dividing by \[3\] both sides, we get,
\[{y^4} - 28{y^2} + 75 = 0\]
We can also write the above equation as,
\[{y^4} - 3{y^2} - 25{y^2} + 75 = 0\]
Taking \[{y^2}\] common from the first two terms and \[ - 25\] common from last two terms, we get,
\[{y^2}\left( {{y^2} - 3} \right) - 25\left( {{y^2} - 3} \right) = 0\]
Now, taking \[\left( {{y^2} - 3} \right)\] common again, we get,
\[\left( {{y^2} - 3} \right)\left( {{y^2} - 25} \right) = 0\]
Now, product of two numbers can be zero only and only if at-least one of the two numbers is zero
If\[\left( {{y^2} - 3} \right) = 0\]
\[
   \Rightarrow {y^2} = 3 \\
   \Rightarrow y = \pm \sqrt 3 \]
And, if\[\left( {{y^2} - 25} \right) = 0\]
\[ \Rightarrow {y^2} = 25 \\
   \Rightarrow y = \pm 5 \]
Putting back the value of \[y\] in equation \[\left( 3 \right)\], we get,
If\[y = \sqrt 3 \]
\[x = 3\sqrt 3 \]
If\[y = - \sqrt 3 \]
\[x = - 3\sqrt 3 \]
If\[y = 5\]
\[x = 4\]
If\[y = - 5\]
\[x = - 4\]
Hence, the solutions for given system of linear equations are: \[\left( {5,4} \right),\left( { - 5, - 4} \right),\left( {\sqrt 3 ,3\sqrt 3 } \right),\left( { - \sqrt 3 , - 3\sqrt 3 } \right)\]

Note: Whenever we face such types of questions, we should use a substitution method only to get the solution. One thing to keep in mind here is if in the end we get a value of the square of the variable, then, it is necessary to take both positive and negative values separately. Both the cases will give different solutions.