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