Question

Which system of linear equations best represents the tables shown below?

Answer 1

From the first table, we can use the points (-1,3) and (5,-3) to find the first equation:

$\begin{gathered} \text{slope:} \\ m=\frac{y_2-y_1}{x_2-x_1} \\ \Rightarrow m=\frac{-3-3}{5-(-1)}=\frac{-6}{5+1}=\frac{-6}{6}=-1 \\ m=-1 \end{gathered}$

then we can use the point to find the equation in slope-point form:

$\begin{gathered} y-y_1=m(x-x_1) \\ \Rightarrow y-3=-1(x-(-1))=-x-1 \\ \Rightarrow y=-x-1+3=-x+2 \\ y=-x+2 \end{gathered}$

we can do the same with the next table, using the points (1,-1) and (7,11):

$\begin{gathered} \text{slope:} \\ m=\frac{11-(-1)}{7-1}=\frac{11+1}{6}=\frac{12}{6}=2 \\ m=2 \\ \text{Equation:} \\ y-(-1)=2(x-1)=2x-2 \\ \Rightarrow y+1=2x-2 \\ \Rightarrow y=2x-2-1=2x-3 \\ y=2x-3 \end{gathered}$

therefore, the system of equations is:

$\begin{cases}y=-x+2 \\ y=2x-3\end{cases}$