[SOLVED] solving a system

**satis** · March 6th 2010, 12:02 PM

Can someone point me in the right direction to solve this system of equations?

$x^2 + y^2 = 25$
$3x^2 - 16y = 0$

I'm trying to use substitution to get the answer. However, regardless of which direction I go, the answers get hopelessly convoluted. Can someone point me at how to start?

**skeeter** · March 6th 2010, 12:38 PM

Originally Posted by satis

Can someone point me in the right direction to solve this system of equations?

$x^2 + y^2 = 25$
$3x^2 - 16y = 0$

I'm trying to use substitution to get the answer. However, regardless of which direction I go, the answers get hopelessly convoluted. Can someone point me at how to start?

$x^2 = 25 - y^2$

$3(25-y^2) - 16y = 0$

$75 - 3y^2 - 16y = 0$

$3y^2 + 16y - 75 = 0$

$(3y+25)(y-3) = 0$

$y = -\frac{25}{3}$

$y = 3$

$x^2 = 25 - \left(-\frac{25}{3}\right)^2 = -\frac{400}{9}<br />$
no solution for x in this equation. this y-value is invalid.

$x^2 = 25 - 9 = 16$

$x = \pm 4$

so, two solutions ... (4,3) and (-4,3)

**satis** · March 6th 2010, 12:46 PM

excellent, thank you, that was precisely what I needed.

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