Math  /  Geometry

QuestionIn Exercises 3 and 4 , solve for yy in terms of xx.
3. y29+x24=1\frac{y^{2}}{9}+\frac{x^{2}}{4}=1
4. x236+y225=1\frac{x^{2}}{36}+\frac{y^{2}}{25}=1

Studdy Solution
Take the square root of both sides to solve for y y :
y=±2525x236 y = \pm \sqrt{25 - \frac{25x^{2}}{36}}
The solutions are:
For equation 3: y=±99x24 y = \pm \sqrt{9 - \frac{9x^{2}}{4}}
For equation 4: y=±2525x236 y = \pm \sqrt{25 - \frac{25x^{2}}{36}}

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