Question 884679
Solving quadratic systems. Also asked to state which conic section each equation represents. (Circle, parabolas, etc). Please explain how to solve 4x^2+9y^2=144
                                                                                             2x+3y=12 either by elimination or substitution. Also advise how would I know which conic section to chose.
<pre>
<u>{{{4x^2 + 9y^2 = 144}}}</u>
Since the coefficients (4 & 9) on the 2nd degree variables are UNEQUAL, this conic section is an ELLIPSE.

{{{2x + 3y = 12}}}
First degree variables exist, so a LINEAR EQUATION, thus a LINE graph.

{{{4x^2 + 9y^2 = 144}}} -------- eq (i)
2x + 3y = 12______2x = 12 - 3y____{{{x = (12 - 3y)/2}}} -------- eq (ii)
{{{4((12 - 3y)/2)^2 + 9y^2 = 144}}} --------- Substituting {{{(12 - 3y)/2}}} for x in eq (i)
{{{4((144 - 72y + 9y^2)/4) + 9y^2 = 144}}}
{{{cross(4)((144 - 72y + 9y^2)/cross(4)) + 9y^2 = 144}}}
{{{144 - 72y + 9y^2 + 9y^2 = 144}}}
{{{18y^2 - 72y + 144 - 144 = 0}}}
{{{18y^2 - 72y = 0}}}
{{{18y(y - 4) = 0}}}
{{{highlight_green(highlight_green(y = 4_or_y = 0))}}}

2x + 3(4) = 12 -------- Substituting 4 for y in eq (ii)
2x + 12 = 12
2x = 0
{{{x = 0/2}}}, or {{{x  = 0}}}

2x + 3(0) = 12 -------- Substituting 0 for y in eq (ii)
2x + 0 = 12
2x = 12
{{{x = 12/2}}}, or {{{x  = 6}}}
{{{highlight_green(highlight_green(x = 0_or_x = 6))}}}

Therefore, {{{highlight_green(highlight_green(x = 0_when_y = 4))}}} and {{{highlight_green(highlight_green(x = 6_when_y = 0))}}}
You can do the check!! 

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