Answer :
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].
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].
