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Consider the systems of equations below.
\begin{tabular}{|l|l|l|}
\hline \multicolumn{1}{|c|}{ System A } & \multicolumn{1}{c|}{ System B } & \multicolumn{1}{c|}{ System C } \\
\hline[tex]$x^2+y^2=17$[/tex] & [tex]$y=x^2-7x+10$[/tex] & [tex]$y=-2x^2+9$[/tex] \\
[tex]$y=-\frac{1}{2}x$[/tex] & [tex]$y=-6x+5$[/tex] & [tex]$8x-y=-17$[/tex] \\
\hline
\end{tabular}

Determine the number of real solutions for each system of equations.

System A has [tex]$\square$[/tex] real solutions.

System B has [tex]$\square$[/tex] real solutions.

System C has [tex]$\square$[/tex] real solutions.

Sagot :

GinnyAnswer
Sure, let's consider each system of equations individually and determine the number of real solutions for each.

System A:
The system consists of:
[tex]\[ x^2 + y^2 = 17 \][/tex]
[tex]\[ y = -\frac{1}{2}x \][/tex]

Through analysis, we find that this system has 2 real solutions.

System B:
The system consists of:
[tex]\[ y = x^2 - 7x + 10 \][/tex]
[tex]\[ y = -6x + 5 \][/tex]

Through analysis, we observe that this system has 2 real solutions.

System C:
The system consists of:
[tex]\[ y = -2x^2 + 9 \][/tex]
[tex]\[ 8x - y = -17 \][/tex]

Through analysis, it is determined that this system has 1 real solution.

So, the final number of solutions for each system is:

System A has \( \boxed{2} \) real solutions.

System B has \( \boxed{2} \) real solutions.

System C has [tex]\( \boxed{1} \)[/tex] real solution.
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