Solve $13 x^{2} - 12 y^{2} = 40; 6 x^{2} - 8 y^{2} = 16$ and hence find the value of $x^{2} + y^{2}$

Open in App

Solution

afte putting x=2 and y= 1 both the equations are satisy therefor
the correct ans is $x^{2} + y^{2}$ = 5

Suggest Corrections

Similar questions

13 x^{2} - 12 y^{2} a n d 6 x^{2} - 8 y^{2}

y^{2} - \frac{1}{8 y^{2}}

y \frac{(y - 3)}{12 y^{2}} \times \frac{72 y^{2}}{y^{2} - 3^{2}}

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

\frac{x^{2}}{16} + \frac{9}{b^{2}} = 1

\frac{1}{x + y} - \frac{1}{2 x} = \frac{1}{30}, \frac{5}{x + y} + \frac{1}{x} = \frac{4}{3}

2 x^{2} - y^{2}

Join BYJU'S Learning Program