Question

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Answer 1

in general a linear system is an algebraic system like this:
\[\left\{ {\begin{array}{*{20}{c}}
{ax + by = c} \\
{dx + ey = f}
\end{array}} \right.\]
where \(a,b,c,d,e,f\) are real numbers, and \(x,y\) are the unknowns.
Each equation can be graphed by a line, so the solution of that system is, if it exists, the inersection point of those lines

Answer 2

In the case of nonlinear equation, we can have the subsequent system:
\[\left\{ {\begin{array}{*{20}{c}}
{a{x^2} + bxy + c{y^2} + dx + ey + f = 0} \\
{mx + q = 0}
\end{array}} \right.\]
where, as usual, \(a,b,c,d,e,f,n,q\) are real numbers, whereas \(x,y\) are the unknows.
Now, depending on the values of the coefficients \(a,b,c,d,e,f\) the graph of first equation can be a circumference, a parabola, an ellipse or a hyperbola.
Whereas the second equation can be graphed as a line, as usual.
So, if the first equation can be graphed as a circumference, then the solution of your system are, if they exist, the intersection point of those curves:

Please, note that the system above is a second gradde system, since the grade of the first equation is 2, whereas the grade of the second equation is 1, so the product od those grades is:
\(2 \cdot 1 = 2 \)

Answer 3

grade*

Answer 4

nevertheless we can have this non-linear system:
\[\left\{ {\begin{array}{*{20}{c}}
{{a_1}{x^2} + {b_1}xy + {c_1}{y^2} + {d_1}x + {e_1}y + {f_1} = 0} \\
{{a_2}{x^2} + {b_2}xy + {c_2}{y^2} + {d_2}x + {e_2}y + {f_2} = 0}
\end{array}} \right.\]
where, as usual, \(a_1,a_2,b_1,b_2,c_1,c_2,d_1,d_2,e_1,e_2,f_1,f_2\) are real numbers, and \(x,y\) are the unknowns
Now, depending on the values of those coefficients \(a_1,a_2,b_1,b_2,c_1,c_2,d_1,d_2,e_1,e_2,f_1,f_2\)
those equation can be graphed as two circumferences, so, the solution of that system, which is a fourth grade system \(2 \cdot 2=4\), if they exist, are the intersection point of the two circumferences:

Answer 5

of course, many other examples, can be considered