To calculate: The solution for the system of equations 5x^{2}+y^{2}

Calculation:
Consider the provided system of equations:
${\displaystyle 5x^2+y^2=14}$.......(1)
And,
${\displaystyle x^2-2y^2=-17}$.....(2)
Multiply by 2 in equation (1) $10x^2+2y^2=28$......(3)
Add equations (3) and (2):
${\displaystyle 11x^2=11}$
${\displaystyle x^2=1}$
${\displaystyle x=\pm 1}$
Now, substitute 1 for xin equation (2) and obtain the value of y:
$(1{)}^2-2y^2=-17$
${\displaystyle 1-2y^2=-17}$
${\displaystyle -2y^2=-18}$
${\displaystyle y^2=9}$
${\displaystyle y=\pm 3}$
Again, put ${\displaystyle x=}$ and ${\displaystyle y=-3}$ in equation ${\displaystyle 5x^2+y^2=14}$:
${\displaystyle 5{\left(1\right)}^2+{(-3)}^2\stackrel{?}{=}14}$
${\displaystyle 5+9\stackrel{?}{=}14}$
${\displaystyle 14\stackrel{?}{=}14}$
Put ${\displaystyle x=1}$ and ${\displaystyle y=—3}$ in equation ${\displaystyle x^2-2y^2=-17}$
${\displaystyle {\left(1\right)}^2-2{(-3)}^2\stackrel{?}{=}-17}$
${\displaystyle 1-18\stackrel{?}{=}-17}$
${\displaystyle -17\stackrel{?}{=}-17}$
Again, put ${\displaystyle x=-1}$ and ${\displaystyle y=3}$ in equation ${\displaystyle 5x^2+y^2=14}$
${\displaystyle 5{(-1)}^2+{\left(3\right)}^2\stackrel{?}{=}14}$
${\displaystyle 5+9\stackrel{?}{=}14}$
${\displaystyle 14\stackrel{?}{=}14}$
Put ${\displaystyle x=-1}$ and ${\displaystyle y=3}$ in equation ${\displaystyle x^2-2y^2=-17}$
${\displaystyle {(-1)}^2-2{\left(3\right)}^2\stackrel{?}{=}-17}$
${\displaystyle 1-18\stackrel{?}{=}-17}$
${\displaystyle -17\stackrel{?}{=}-17}$
Again, put ${\displaystyle x=-1}$ and ${\displaystyle y=-3}$ in equation ${\displaystyle 5x^2+y^2=14}$
${\displaystyle 5{(-1)}^2+{(-3)}^2\stackrel{?}{=}14}$
${\displaystyle 5+9\stackrel{?}{=}14}$

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