Math Help - Substitution & Elimination

1. Substitution & Elimination

2. Originally Posted by megs_world
From the first equation:

$5x^2 = 49 + 3y^2$

From the second equation:

$4x = 11+2y$

Hence

$(4x^2) = (11+2y)^2$

$16x^2 = 121 + 44y + 4y^2$

So we have:

$16x^2 = 121 + 44y + 4y^2$

$5x^2 = 49 + 3y^2$

Multiply the top one by 5, and the bottom by 20

$80x^2 = 5(121 + 44y + 4y^2)$

$80x^2 = 20(49 + 3y^2)$

Now eliminate.

Alternatively:

$4x = 11+2y$

$x = \frac{11+2y}{4}$

$5\bigg( \frac{11+2y}{4}\bigg)^2 = 49 + 3y^2$