The form of the linear equation is
[tex]y=mx+b[/tex]m is the slope
b is the y-intercept
For the 1st equation on the left
m = 1
b = -3
Substitute them in the form above
[tex]y=x-3[/tex]Add 3 to both sides
[tex]\begin{gathered} y+3=x-3+3 \\ y+3=x \end{gathered}[/tex]Subtract y from both sides
[tex]\begin{gathered} y-y+3=x-y \\ 3=x-y \\ x-y=3 \end{gathered}[/tex]This is the 4th equation on the right side
[tex]m=1,b=-3\rightarrow x-y=3[/tex]For the 2nd equation on the left
m = 1 and the line passes through the point (-1, 2)
Substitute m by 1 in the form
[tex]y=x+b[/tex]To find b substitute x by -1 and y by 2
[tex]2=-1+b[/tex]Add 1 to each side
[tex]\begin{gathered} 2+1=-1+1+b \\ 3=b \end{gathered}[/tex]Substitute it in the equation
[tex]y=x+3[/tex]Subtract 3 from each side
[tex]\begin{gathered} y-3=x+3-3 \\ y-3=x \end{gathered}[/tex]Subtract y from each side
[tex]\begin{gathered} y-y-3=x-y \\ -3=x-y \\ x-y=-3 \end{gathered}[/tex]This is the 3rd equation on the right side
[tex]m=1,(-1,2)\rightarrow x-y=-3[/tex]For the 3rd equation on the left
The line passes through points (-2, 3), (-3, 4)
We will use the rule of the slope to find m
[tex]\begin{gathered} m=\frac{y_2-y_1}{x_2-x_1} \\ \\ x_1=-2,x_2=-3 \\ y_1=3,y_2=4 \\ \\ m=\frac{4-3}{-3--2}=\frac{1}{-3+2}=\frac{1}{-1}=-1 \end{gathered}[/tex]Substitute it in the form of the equation
[tex]y=-x+b[/tex]To find b use one of the two given points
I will use the first point (-2, 3)
Substitute x by -2 and y by 3 in the equation to find b
[tex]\begin{gathered} 3=-(-2)+b \\ 3=2+b \end{gathered}[/tex]Subtract 2 from both sides
[tex]\begin{gathered} 3-2=2-2+b \\ 1=b \end{gathered}[/tex]Substitute it in the equation
[tex]y=-x+1[/tex]Add x to both sides
[tex]\begin{gathered} x+y=-x+x+1 \\ x+y=1 \end{gathered}[/tex]This is the 1st equation on the right side
[tex](-2,3),(-3,4)\rightarrow x+y=1[/tex]
