I don't know how to substitute or eliminate with squares; please help me.
From the first equation:
$\displaystyle 5x^2 = 49 + 3y^2$
From the second equation:
$\displaystyle 4x = 11+2y $
Hence
$\displaystyle (4x^2) = (11+2y)^2 $
$\displaystyle 16x^2 = 121 + 44y + 4y^2 $
So we have:
$\displaystyle 16x^2 = 121 + 44y + 4y^2 $
$\displaystyle 5x^2 = 49 + 3y^2$
Multiply the top one by 5, and the bottom by 20
$\displaystyle 80x^2 = 5(121 + 44y + 4y^2) $
$\displaystyle 80x^2 = 20(49 + 3y^2)$
Now eliminate.
Alternatively:
$\displaystyle 4x = 11+2y $
$\displaystyle x = \frac{11+2y}{4} $
$\displaystyle 5\bigg( \frac{11+2y}{4}\bigg)^2 = 49 + 3y^2$