Question

-Consider the following system of equations:--57 - 25How many solutions does the system have?zeroonetwo

Answer 1

Given the system of equations:

`\begin{cases}-(x^2)/(3)=-(5)/(6)+(y^2)/(3)\ldots(1) \\ 5y^2=(25)/(2)-5x^2\ldots(2)\end{cases}`

We multiply by 6 the equation (1):

`\begin{gathered} -(6\cdot x^2)/(3)=-(6\cdot5)/(6)+(6\cdot y^2)/(3) \\ \Rightarrow-2x^2=-5+2y^2 \\ \Rightarrow2x^2+2y^2=5\ldots(1^(\prime)) \end{gathered}`

Now, we multiply by 2/5 on the equation (2):

`\begin{gathered} (2)/(5)\cdot5y^2=(2)/(5)\cdot(25)/(2)-(2)/(5)\cdot5x^2 \\ \Rightarrow2y^2=5-2x^2 \\ \Rightarrow2x^2+2y^2=5\ldots(2^(\prime)) \end{gathered}`

We can see that both equations (1') and (2') are equal, so we conclude that there are infinitely many solutions.