Respuesta :

The answer to this question is 2
this is because it is the highest before it starts to turn into fractions and square roots.

I hope this helps!!!!!

Answer:

The greatest possible number of solutions is 2.

Step-by-step explanation:

Given system of equations,

[tex]x^2+y^2=9-----(1)[/tex]

[tex]9x+2y=16[/tex]

[tex]9x=16-2y[/tex]

[tex]\implies x=\frac{16-2y}{9}-----(2)[/tex]

From equation (1),

[tex](\frac{16-2y}{9})^2+y^2=9[/tex]

[tex]\frac{(16-2y)^2}{81}+y^2=9[/tex]

[tex]\frac{256-64y+4y^2}{81}+y^2=9[/tex]

[tex]\frac{256-64y+4y^2+81y^2}{81}=9[/tex]

[tex]85y^2-64y+256=729[/tex]

[tex]85y^2-64y-473=0[/tex]

By quadratic formula we get,

[tex]\implies y = \frac{32+9\sqrt{509}}{85}\text{ or }y=\frac{32-9\sqrt{509}}{85}[/tex]

By substituting these values in equation (2),

We get,

[tex]x=\frac{144+2\sqrt{509}}{85}\text{ or }x=\frac{144-2\sqrt{509}}{85}[/tex]

Hence, all possible solutions of the given system of equations are,

[tex](\frac{144+2\sqrt{509}}{85},\frac{32+9\sqrt{509}}{85})\text{ and }(\frac{144-2\sqrt{509}}{85},\frac{32-9\sqrt{509}}{85})[/tex]

Option second is correct.

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