Answer
$(5,1),(5,-1),(-5,1),(-5,-1)$
Work Step by Step
$x^{2}+y^{2}=26$ Equation $(1)$
$x^{2}-2y^{2}=23$ Equation $(2)$
Multiply Equation $(1)$ by $2$ then add with Equation $(2)$
$2(x^{2}+y^{2})+x^{2}-2y^{2}=2(26)+23$
$2x^{2}+2y^{2}+x^{2}-2y^{2}=52+23$
$3x^{2}=75$
$x^{2}=25$
$x=±5$
$x=5$ or $x=-5$
Substitute $x$ values in Equation $(1)$ to get corresponding $y$ values.
Let $x=5$
$x^{2}+y^{2}=26$
$5^{2}+y^{2}=26$
$25+y^{2}=26$
$y^{2}=1$
$y=±1$
Let $x=-5$
$x^{2}+y^{2}=26$
$(-5)^{2}+y^{2}=26$
$25+y^{2}=26$
$y^{2}=1$
$y=±1$
$(5,1),(5,-1),(-5,1),(-5,-1)$ satisfy the given equations.
Solutions are $(5,1),(5,-1),(-5,1),(-5,-1)$
