Solving for x and y

**Stuck Man** · Nov 28th 2011, 12:00 AM

solve [(x^2)/25 + (y^2)/4 -25=0, (x^2)/16 + (y^2)/9 -29=0, x, y] - Wolfram|Alpha

I am not certain what the method is. I have made and equation y^2=... from the first equation and put it into the other but I did not get +or-20.

**Quacky** · Nov 28th 2011, 01:22 AM

$\frac{x^2}{25}+\frac{y^2}{4}=25$

Multiply through by $4$ :

$\frac{4x^2}{25}+y^2=100$

$y^2=100-\frac{4x^2}{25}$

What did you get when you substituted this in and tried to solve?

**Stuck Man** · Nov 28th 2011, 03:12 AM

X^2=1299600/2601 X=+/- 22.35 x is +/-20.

**Quacky** · Nov 28th 2011, 03:22 AM

I'm fairly sure that might be overcomplicating things.

$y^2=100-\frac{4x^2}{25}$

Substituting this into $\frac{x^2}{16}+\frac{y^2}{9}-29=0$ gives:

$\frac{x^2}{16}+\frac{100-\frac{4x^2}{25}}{9}-29=0$

Multiplying by $9:$

$\frac{9}{16}x^2+100-\frac{4}{25}x^2-261=0$

Which leads to:

$\frac{161}{400}x^2=161$

Does that clarify?

**Stuck Man** · Nov 28th 2011, 03:36 AM

I have typed the hyperbola correctly by accident. The book wrote 21 not 29 and I found that error. Also it had a negative sign instead of positive.

**Quacky** · Nov 28th 2011, 03:44 AM

The joy of serendipity.

**Wilmer** · Nov 28th 2011, 08:41 AM

Another way:

x^2/25 + y^2/4 = 25 [1]
x^2/16 + y^2/9 = 29 [2]

[1] *100: 4x^2 + 25y^2 = 2500 [1]
[2] * 144: 9x^2 + 16y^2 = 4176 [1]

36x^2 + 225y^2 = 22500 : [1]*9
-36x^2 - 64y^2 = -16704 : [2]*-9

Add above: 161y^2 = 5796
Now solve for y, then substitute to solve for x

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