Question 58

If x+3y=-3x+y, then $$\frac{x^2}{2y^2}$$ is equal to

Solution

x + 3y = -3x + y

4x = -2y

x = $$\ -\frac{\ y}{2}$$

Therefore

$$\ \frac{\ x^2}{2y^2}\ =\ \ \frac{\ \ \frac{\ y^2}{4}}{2y^2}\ =\ \ \frac{\ y^2}{4\times\ 2y^2}=\ \frac{\ 1}{8}$$

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