Let's break this down step-by-step:
1. Finding the equation of the first line:
You are given points for the first line: (-1, 1), (0, 3), (1, 5), (2, 7).
To find the slope [tex]\( m \)[/tex] of the first line, you can use any two points. Let's use (0, 3) and (1, 5):
[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - 3}{1 - 0} = \frac{2}{1} = 2 \][/tex]
The y-intercept is given directly from the point (0, 3), so the y-intercept [tex]\( c \)[/tex] is 3.
Thus, the first equation is:
[tex]\[ y = 2x + 3 \][/tex]
So, we fill in the blank for the first equation:
[tex]\[ y = \boxed{2}x + 3 \][/tex]
2. Finding the equation of the second line:
You are given points for the second line: (-2, -7), (0, -1), (2, 5), (4, 11).
To find the slope [tex]\( m \)[/tex] of the second line, you can use any two points. Let's use (0, -1) and (2, 5):
[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - (-1)}{2 - 0} = \frac{6}{2} = 3 \][/tex]
The y-intercept is given directly from the point (0, -1), so the y-intercept [tex]\( c \)[/tex] is -1.
Thus, the second equation is:
[tex]\[ y = 3x - 1 \][/tex]
So, we fill in the blank for the second equation:
[tex]\[ y = 3x - \boxed{1} \][/tex]
3. Finding the solution to the system of equations:
We need to find the point where the two lines intersect, i.e. the solution to the system:
[tex]\[
\begin{cases}
y = 2x + 3 \\
y = 3x - 1
\end{cases}
\][/tex]
Set the equations equal to each other to find [tex]\( x \)[/tex]:
[tex]\[ 2x + 3 = 3x - 1 \][/tex]
[tex]\[ 2x - 3x = -1 - 3 \][/tex]
[tex]\[ -x = -4 \][/tex]
[tex]\[ x = 4 \][/tex]
Substitute [tex]\( x = 4 \)[/tex] back into the first equation to find [tex]\( y \)[/tex]:
[tex]\[ y = 2(4) + 3 = 8 + 3 = 11 \][/tex]
So, the solution of the system is:
[tex]\[ ( \boxed{4} , \boxed{11} ) \][/tex]
