Respuesta :

pkmnmushy
One of the things you can do here is rearrange the equations like this: 
x2 +y2 -13 = 0 
y = 2x + 4 
After that what you can do is substitute: 
x2 + (2x + 4)2 -13 = 0 
x2 + (4x2 +16x + 16) -13 = 0 
5x2 +16x + 3 = 0 
x = (-16 + sqrt(16^2 - 4*5*3))/(2*5) and (-16 - sqrt(16^2 - 4*5*3))/(2*5) 
x = (-16 + 14)/10 and (-16 - 14)/10 
x = -0.2 and -3 

y = 2x + 4 
y = 2*(-0.2) + 4 = 3.6 
y = 2*(-3) + 4 = -2 
The answer will be 
(-0.2,3.6) and (-3,-2)
Hope this helps you greatly

Answer: The value of (x,y)=(3,2) and

[tex](x,y)=(\frac{1}{5},\frac{-18}{5})[/tex]

Step-by-step explanation:

Since we have given that

[tex]x^2+y^2=13\\\\2x-y=4[/tex]

From second equation i.e.

[tex]2x-y=4[/tex]

[tex]\implies 2x-4=y[/tex]

Put it in the first equation:

[tex]x^2+y^2=13\\\\x^2+(2x-4)^2=13\\\\x^2+4x^2+16-16x=13\\\\5x^2-16x+16=13\\\\5x^2-16x=13-16\\\\5x^2-16x=-3\\\\5x^2-16x+3=0[/tex]

[tex]\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]

So, we will apply this to get the roots:

[tex]\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\=\frac{16+\sqrt{\left(-16\right)^2-4\cdot \:5\cdot \:3}}{2\cdot \:5}=3\\\\\frac{16-\sqrt{\left(-16\right)^2-4\cdot \:5\cdot \:3}}{2\cdot \:5}=\frac{1}{5}[/tex]

Hence the value of x are

[tex]\frac{1}{5},2[/tex]

Therefore, the value of y will be

[tex]y=2x-4[/tex]

If we put [tex]x=\frac{1}{5}[/tex]

[tex]y=2\times \frac{1}{5}-4\\\\y=\frac{2}{5}-4\\\\y=\frac{-18}{4}[/tex]

If we put x=3

then

[tex]y=2x-4\\\\y=2\times 3-4\\\\y=6-4\\\\y=2[/tex]


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