SOLUTION: 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

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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.
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39630) About Me  (Show Source):
You can put this solution on YOUR website!
Ellipse, 4x%5E2%2B9y%5E2=144
. Line, 2x%2B3y=12
.

You can identify the type of conic section just by looking at those equations. The ellipse equation is almost but not yet in standard form and the line is in standard form. Identification is uncomplicated.

Solving the system of the two equations is a little more work.
Solve the linear equation for either variable and substitute into the ellipse equation and solve for the single variable. You could use the linear equation again to determine the other variable.

3y=-2x%2B12
y=-2x%2F3%2B4
-
4x%5E2%2B9%28-2x%2F3%2B4%29%5E2=144
4x%5E2%2B9%28%284%2F9%29x%5E2-16x%2F3%2B16%29=144
4x%5E2%2B4x%5E2-3%2A16x%2B9%2A16=144
8x%5E2-48x%2B151=144
8x%5E2-48x%2B7=0
Discriminant: %28-48%29%5E2-4%2A8%2A7=2080
Factoring Discriminant: .
Continuing with general solution of quadratic equation:
x=%2848%2B-+sqrt%282%5E6%2A5%2A13%29%29%2F%282%2A8%29
x=%2848%2B-+8sqrt%2865%29%29%2F%282%2A8%29
highlight%28x=%286%2B-+sqrt%2865%29%29%2F2%29, for the two x coordinates.

FINDING y COORDINATES
Found from earlier, y=-2x%2F3%2B4.
y=-%282%2F3%29x%2B4
y=-%282%2F3%29%28%286%2B-+sqrt%2865%29%29%2F2%29%2B4
y=-4%2B-+%281%2F3%29sqrt%2865%29%2B4, notice the resulting additive inverses 4 and -4.
highlight%28y=-%281%2F3%29sqrt%2865%29%29 OR highlight%28%281%2F3%29sqrt%2865%29%29.


Careful which coordinate goes with which other coordiante.
-
x,y: 6-sqrt%2865%29%2F2, sqrt%2865%29%2F3
-
x,y: 6%2Bsqrt%2865%29%2F2, -sqrt%2865%29%2F3

Answer by MathTherapy(10557) About Me  (Show Source):
You can put this solution on YOUR website!
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.


4x%5E2+%2B+9y%5E2+=+144

Since the coefficients (4 & 9) on the 2nd degree variables are UNEQUAL, this conic section is an ELLIPSE.



2x+%2B+3y+=+12

First degree variables exist, so a LINEAR EQUATION, thus a LINE graph.



4x%5E2+%2B+9y%5E2+=+144 -------- eq (i)

2x + 3y = 12______2x = 12 - 3y____x+=+%2812+-+3y%29%2F2 -------- eq (ii)

4%28%2812+-+3y%29%2F2%29%5E2+%2B+9y%5E2+=+144 --------- Substituting %2812+-+3y%29%2F2 for x in eq (i)

4%28%28144+-+72y+%2B+9y%5E2%29%2F4%29+%2B+9y%5E2+=+144

cross%284%29%28%28144+-+72y+%2B+9y%5E2%29%2Fcross%284%29%29+%2B+9y%5E2+=+144

144+-+72y+%2B+9y%5E2+%2B+9y%5E2+=+144

18y%5E2+-+72y+%2B+144+-+144+=+0

18y%5E2+-+72y+=+0

18y%28y+-+4%29+=+0

highlight_green%28highlight_green%28y+=+4_or_y+=+0%29%29



2x + 3(4) = 12 -------- Substituting 4 for y in eq (ii)

2x + 12 = 12

2x = 0

x+=+0%2F2, or x++=+0



2x + 3(0) = 12 -------- Substituting 0 for y in eq (ii)

2x + 0 = 12

2x = 12

x+=+12%2F2, or x++=+6

highlight_green%28highlight_green%28x+=+0_or_x+=+6%29%29



Therefore, highlight_green%28highlight_green%28x+=+0_when_y+=+4%29%29 and highlight_green%28highlight_green%28x+=+6_when_y+=+0%29%29

You can do the check!! 



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