Answer
$(2\sqrt 2,1),(-2\sqrt 2,1),(2\sqrt 2,-1),(-2\sqrt 2,-1)$
Work Step by Step
$4x^{2}+3y^{2} = 35$ Equation $(1)$
$5x^{2}+2y^{2} = 42$ Equation $(2)$
Multiply Equation $(1)$ by $2$ and Equation $(2)$ by $-3$
$8x^{2}+6y^{2} = 70$ Equation $(3)$
$-15x^{2}-6y^{2} = -126$ Equation $(4)$
Add Equation $(3)$ and Equation $(4)$
$8x^{2}+6y^{2} -15x^{2}-6y^{2} = 70-126$
$-7x^{2} = -56$
$x^{2} = \frac{-56}{-7}$
$x^{2} = 8$
$x = \sqrt 8$
$x = \sqrt (4 \times 2)$
$x = ±2\sqrt 2$
$x = 2\sqrt 2$ or $x = -2\sqrt 2$
Substitute $ x$ values in Equation $(1)$
Let $x = 2\sqrt 2$
$4x^{2}+3y^{2} = 35$
$4(2\sqrt 2)^{2}+3y^{2} = 35$
$4(8)+3y^{2} = 35$
$32+3y^{2} = 35$
$3y^{2} = 3$
$y^{2} = 1$
$y = ±\sqrt 1$
$y = ±1$
Let $x = -2\sqrt 2$
$4x^{2}+3y^{2} = 35$
$4(-2\sqrt 2)^{2}+3y^{2} = 35$
$4(8)+3y^{2} = 35$
$32+3y^{2} = 35$
$3y^{2} = 3$
$y^{2} = 1$
$y = ±1$
$(2\sqrt 2,1),(-2\sqrt 2,1),(2\sqrt 2,-1),(-2\sqrt 2,-1)$ satisfy the given equations.
So, the solutions set is $\{(2\sqrt 2,1),(-2\sqrt 2,1),(2\sqrt 2,-1),(-2\sqrt 2,-1)\}$
