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A system of linear equations is given by the tables.

\begin{tabular}{|c|c|}
\hline
[tex]$x$[/tex] & [tex]$y$[/tex] \\
\hline
-1 & 1 \\
\hline
0 & 3 \\
\hline
1 & 5 \\
\hline
2 & 7 \\
\hline
\end{tabular}
\begin{tabular}{|c|c|}
\hline
[tex]$x$[/tex] & [tex]$y$[/tex] \\
\hline
-2 & -7 \\
\hline
0 & -1 \\
\hline
2 & 5 \\
\hline
4 & 11 \\
\hline
\end{tabular}

The first equation of this system is [tex]$y = \square x + 3$[/tex].

The second equation of this system is [tex]$y = 3x - \square 1$[/tex].

The solution of the system is ([tex]$\square, \square$[/tex]).

Sagot :

GinnyAnswer
Let's analyze the given tables to determine the equations of the lines and the solution to the system.

For the first table:

[tex]\(\begin{tabular}{|c|c|} \hline x & y \\ \hline -1 & 1 \\ \hline 0 & 3 \\ \hline 1 & 5 \\ \hline 2 & 7 \\ \hline \end{tabular}\)[/tex]

To find the equation of the line, let's determine the slope ([tex]\(m\)[/tex]) and the y-intercept ([tex]\(b\)[/tex]) of the line.

1. Calculate the slope (m):
- Using points (0, 3) and (1, 5):
[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - 3}{1 - 0} = 2 \][/tex]

2. Find the y-intercept (b):
- Using the point (0, 3):
[tex]\[ b = 3 \][/tex]

Thus, the first equation is:
[tex]\[ y = 2x + 3 \][/tex]

For the second table:

[tex]\(\begin{tabular}{|c|c|} \hline x & y \\ \hline -2 & -7 \\ \hline 0 & -1 \\ \hline 2 & 5 \\ \hline 4 & 11 \\ \hline \end{tabular}\)[/tex]

Again, we need to find the slope ([tex]\(m\)[/tex]) and the y-intercept ([tex]\(b\)[/tex]).

1. Calculate the slope (m):
- Using points (0, -1) and (2, 5):
[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - (-1)}{2 - 0} = \frac{5 + 1}{2} = 3 \][/tex]

2. Find the y-intercept (b):
- Using the point (0, -1):
[tex]\[ b = -1 \][/tex]

Thus, the second equation is:
[tex]\[ y = 3x - 1 \][/tex]

Now, let's find the solution to the system of equations by solving for the intersection point of the two lines:
[tex]\[ y = 2x + 3 \][/tex]
[tex]\[ y = 3x - 1 \][/tex]

To find the intersection:
1. Set the equations equal to each other:
[tex]\[ 2x + 3 = 3x - 1 \][/tex]
2. Solve for [tex]\(x\)[/tex]:
[tex]\[ 3 + 1 = 3x - 2x \implies 4 = x \implies x = 4 \][/tex]

3. Substitute [tex]\(x = 4\)[/tex] back into one of the original equations to find [tex]\(y\)[/tex]:
[tex]\[ y = 2(4) + 3 = 8 + 3 = 11 \][/tex]

Therefore, the solution to the system of equations is:
[tex]\[ (4, 11) \][/tex]

So, filling in the blanks:

The first equation of this system is [tex]\(y= \)[/tex] [tex]\(\boxed{2}\)[/tex] [tex]\(x+3\)[/tex].

The second equation of this system is [tex]\(y=3 x-\)[/tex] [tex]\(\boxed{-1}\)[/tex].

The solution of the system is [tex]\( ( \boxed{4} , \boxed{11} )\)[/tex].
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