Question

Solve
${y^2} + yz + {z^2} = 49$ ……..(1),
${z^2} + zx + {x^2} = 19$ ……….(2),
${x^2} + xy + {y^2} = 39$ ……….(3).

Answer 1

Hint: The number of equations given are 3 and the number of variables that are unknown are 3. So we can find the solution as the number of equations and variables are equal. Solve the three equations by subtracting one from the other to make it simple.

Complete step by step answer:
 Given that
${y^2} + yz + {z^2} = 49$ ……..(1),
${z^2} + zx + {x^2} = 19$ ……….(2),
${x^2} + xy + {y^2} = 39$ ……….(3).
Subtracting equation (2) from equation (1), we get
$ \Rightarrow $ $\left( {{y^2} + yz + {z^2}} \right) - ({z^2} + zx + {x^2})$= 49-19
$ \Rightarrow $ $\left( {{y^2} + yz + {z^2}} \right) - ({z^2} + zx + {x^2})$ = 30
$ \Rightarrow $ ${y^2}$ + $yz$ + ${z^2}$ $ - $ ${z^2}$ $ - $ $zx$ $ - {x^2}$ = 30
$ \Rightarrow $ ${y^2}$ + $yz$ $ - $ $zx - $ ${x^2}$ = 30
$ \Rightarrow {y^2}$ $ - {x^2}$ + $z(y - x)$ = 30
   [ we know that the formula of ${a^2}$ $ - $ ${b^2}$ = $\left( {a + b} \right)\left( {a - b} \right)$ ]
  Applying the above formulae in the above equation we get
  $\left( {y + x} \right)\left( {y - x} \right)$ + $z(y - x)$ = 30
$ \Rightarrow $ $(y - x)\left( {x + y + z} \right)$ = 30 ……….. ( 4 ) .
Similarly by subtracting equation (3) from equation (1) we get
$ \Rightarrow $ ( ${y^2}$ + $yz$ + ${z^2}$ ) $ - $ ( ${x^2}$ + $xy$ + ${y^2}$ ) = 49 $ - $ 39
$ \Rightarrow {y^2}$ + $yz$ + ${z^2}$ $ - $ ${x^2}$ $ - $ $xy$ $ - $ ${y^2}$ = 10
$ \Rightarrow $ ${z^2}$ $ - $ ${x^2}$ + $yz$ $ - $ $xy$ = 10
   [ we know that the formula of ${a^2}$ $ - $ ${b^2}$ = $\left( {a + b} \right)\left( {a - b} \right)$ ]
  Applying the above formulae in the above equation we get
        $\left( {z + x} \right)\left( {z - x} \right)$ + $y\left( {z - x} \right)$ = 10
$ \Rightarrow $ $\left( {z - x} \right)\left( {x + y + z} \right)$ = 10 ………. ( 5 ) .
  Dividing equation ( 4 ) and equation ( 5 ) , we get
            $\dfrac{{\left( {y - x} \right)\left( {x + y + z} \right)}}{{\left( {z - x} \right)\left( {x + y + z} \right)}}$ = $\dfrac{{30}}{{10}}$
               $\dfrac{{\left( {y - x} \right)}}{{\left( {z - x} \right)}}$ = 3
By multiplying $\left( {z - x} \right)$ on both sides we get
           $\left( {y - x} \right)$ = $3\left( {z - x} \right)$
             $\left( {y - x} \right)$ = 3$z$ $ - $ 3$x$
               $y = 3z - 2x$
Substituting $y = 3z - 2x$ in the equation ( 3 ) , we get
${x^2}$ + $x\left( {3z - 2x} \right)$ + ${\left( {3z - 2x} \right)^2}$ = 39
$ \Rightarrow {x^2}$ + $3xz$ $ - $ $2{x^2}$ + $9{z^2}$ + $4{x^2}$ $ - $ 12$xz$ = 39
$ \Rightarrow $3${x^2}$ $ - $ 9$xz$ + 9${z^2}$ = 39
By dividing the above equation with 3 on both sides we get
${x^2} - 3xz + 3{z^2}$ = 13 ………( 6 )
We have equation ( 2 ) and equation ( 6 ) as homogenous in $x$ and $z$ , then by substituting $z$ = $mx$ in them , we get
${x^2}\left( {{m^2} + m + 1} \right)$ = 19 and ${x^2}\left( {3{m^2} - 3m + 1} \right)$ = 13
By dividing both the equation we get
  $\dfrac{{{m^2} + m + 1}}{{3{m^2} - 3m + 1}}$ = $\dfrac{{19}}{{13}}$
\[13{m^2} + 13m + 13\] = \[57{m^2} - 57m + 19\]
By solving the above equation, we get $m$ = 11 and $m$ = $\dfrac{2}{3}$
Substituting the value of $m$ in the above equations and solving for $x,y,z$ we get
$x = \pm 2,y = \pm 5,z = \pm 3$ and $x = \pm \dfrac{{11}}{{\sqrt 7 }},y = \pm \dfrac{{19}}{{\sqrt 7 }},z = \pm \dfrac{1}{{\sqrt 7 }}$ .

Note:
Do not overlook the solutions and find all the possible answers because there can multiple answer type questions in these types of problems. Remember the formula for the roots of a second degree polynomial equation i.e. roots of a${x^2}$+bx+c=0. We can also find solutions for these types of questions in many different ways i.e. matrix method, elimination method etc…