Let's identify the equations for each table and then find the solution to the system of equations.
1. For the first table of values:
[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
-1 & 1 \\
\hline
0 & 3 \\
\hline
1 & 5 \\
\hline
2 & 7 \\
\hline
\end{array}
\][/tex]
We observe that when [tex]\( x = -1 \)[/tex], [tex]\( y = 1 \)[/tex]; when [tex]\( x = 0 \)[/tex], [tex]\( y = 3 \)[/tex]; when [tex]\( x = 1 \)[/tex], [tex]\( y = 5 \)[/tex]; and when [tex]\( x = 2 \)[/tex], [tex]\( y = 7 \)[/tex]. The pattern in the y-values indicates the equation follows [tex]\( y = 2x + 3 \)[/tex].
2. For the second table of values:
[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
-2 & -7 \\
\hline
0 & -1 \\
\hline
2 & 5 \\
\hline
4 & 11 \\
\hline
\end{array}
\][/tex]
We observe that when [tex]\( x = -2 \)[/tex], [tex]\( y = -7 \)[/tex]; when [tex]\( x = 0 \)[/tex], [tex]\( y = -1 \)[/tex]; when [tex]\( x = 2 \)[/tex], [tex]\( y = 5 \)[/tex]; and when [tex]\( x = 4 \)[/tex], [tex]\( y = 11 \)[/tex]. The pattern in the y-values indicates the equation follows [tex]\( y = 3x - 1 \)[/tex].
3. Now, we solve the system of equations:
[tex]\[
y = 2x + 3
\][/tex]
[tex]\[
y = 3x - 1
\][/tex]
Setting the two equations equal to each other to find the intersection point (solution):
[tex]\[
2x + 3 = 3x - 1
\][/tex]
Subtract [tex]\( 2x \)[/tex] from both sides:
[tex]\[
3 = x - 1
\][/tex]
Add 1 to both sides to solve for [tex]\( x \)[/tex]:
[tex]\[
x = 4
\][/tex]
Substitute [tex]\( x = 4 \)[/tex] back into either equation to find [tex]\( y \)[/tex]. Using the first equation:
[tex]\[
y = 2(4) + 3 = 8 + 3 = 11
\][/tex]
The solution of the system is [tex]\((4, 11)\)[/tex].
So, filling in the blanks:
The first equation of this system is [tex]\( y = 2x + 3 \)[/tex].
The second equation of this system is [tex]\( y = 3x - 1 \)[/tex].
The solution of the system is [tex]\( (4, 11) \)[/tex].
