Answer
$(2,1),(-2,1),(2,-1),(-2,-1)$
Work Step by Step
$x^{2}-3y^{2}=1$ Equation $(1)$
$4x^{2}+5y^{2}=21$ Equation $(2)$
Multiply Equation $(1)$ by $-4$ and add with Equation $(2)$
$-4(x^{2}-3y^{2})+4x^{2}+5y^{2}=-4(1)+21$
$-4x^{2}+12y^{2}+4x^{2}+5y^{2}=-4+21$
$17y^{2}=17$
$y^{2}=1$
$y=±1$
$y=1$ or $y=-1$
Substitute $y$ values in Equation $(1)$ to get corresponding $x$ values.
Let $y=1$
$x^{2}-3y^{2}=1$
$x^{2}-3(1)^{2}=1$
$x^{2}-3=1$
$x^{2}=1+3$
$x^{2}=4$
$x=±2$
Let $y=-1$
$x^{2}-3y^{2}=1$
$x^{2}-3(-1)^{2}=1$
$x^{2}-3=1$
$x^{2}=1+3$
$x^{2}=4$
$x=±2$
$(2,1),(-2,1),(2,-1),(-2,-1)$ satisfy given equations.
Solution set: $\{(2,1),(-2,1),(2,-1),(-2,-1)\}$
