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Linear Relationships, SM Bank 045

  1. Complete the tables of values below for each given rule.  (3 marks)

    \(y=2x+1\)
    \begin{array} {|l|c|c|c|c|}
    \hline
    \rule{0pt}{2.5ex}\ \ x\ \ \rule[-1ex]{0pt}{0pt} & -1  &\ \ 0\ \ &\ \ 1\ \ &\ \ 2\ \  \\
    \hline
    \rule{0pt}{2.5ex} \ \ y\ \ \rule[-1ex]{0pt}{0pt} &   &   &  & \\
    \hline
    \end{array}

    		  

    \(y=x-2\)
    \begin{array} {|l|c|c|c|c|}
    \hline
    \rule{0pt}{2.5ex}\ \ x\ \ \rule[-1ex]{0pt}{0pt} & -1  &\ \ 0\ \ &\ \ 1\ \ &\ \ 2\ \  \\
    \hline
    \rule{0pt}{2.5ex} \ \ y\ \ \rule[-1ex]{0pt}{0pt} &   &   &  & \\
    \hline
    \end{array}


  2. On the number plane below, graph the equations from part (a).  (2 marks)
     
  3. Using the graph, find the point of intersection of the two lines.  (1 mark)

Show Answers Only

a.

\(y=2x+1\)
\begin{array} {|l|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\ \ x\ \ \rule[-1ex]{0pt}{0pt} & -1  &\ \ 0\ \ &\ \ 1\ \ &\ \ 2\ \  \\
\hline
\rule{0pt}{2.5ex} \ \ y\ \ \rule[-1ex]{0pt}{0pt} & -1  & 1  & 3 & 5\\
\hline
\end{array}

  

\(y=x-2\)
\begin{array} {|l|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\ \ x\ \ \rule[-1ex]{0pt}{0pt} & -1  &\ \ 0\ \ &\ \ 1\ \ &\ \ 2\ \  \\
\hline
\rule{0pt}{2.5ex} \ \ y\ \ \rule[-1ex]{0pt}{0pt} & -3  & -2  & -1 & 0\\
\hline
\end{array}

 b.   

c.     \((-3 , -5)\)

Show Worked Solution

a.

\(y=2x+1\)
\begin{array} {|l|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\ \ x\ \ \rule[-1ex]{0pt}{0pt} & -1  &\ \ 0\ \ &\ \ 1\ \ &\ \ 2\ \  \\
\hline
\rule{0pt}{2.5ex} \ \ y\ \ \rule[-1ex]{0pt}{0pt} & -1  & 1  & 3 & 5\\
\hline
\end{array}

  

\(y=x-2\)
\begin{array} {|l|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\ \ x\ \ \rule[-1ex]{0pt}{0pt} & -1  &\ \ 0\ \ &\ \ 1\ \ &\ \ 2\ \  \\
\hline
\rule{0pt}{2.5ex} \ \ y\ \ \rule[-1ex]{0pt}{0pt} & -3  & -2  & -1 & 0\\
\hline
\end{array}

 b.   

c.     \(\text{Point of intersection:   }(-3 , -5)\)