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Linear Relationships, SM-Bank 038

  1. Complete the table of values below for the rule  \( y=4x-7\).  (2 marks)
     
    \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. Use the table to list the coordinates of the points.  (2 marks)

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a.    \(\text{Table of values}\)

\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} & -11  & -7 & -3 & 1\\
\hline
\end{array}

b.    \((-1 , -11)\ \ (0 , -7)\ \ (1 , -3)\ \ (2 , 1)\)

Show Worked Solution

a.    \(\text{Table of values}\)

\begin{array} {|l|c|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} & 4\times (-1)-7=-11  & 4\times (0)-7=-7 & 4\times 1-7=-3 &4\times 2-7=1\\
\hline
\end{array}

b.    \((-1 , -11)\ \ (0 , -7)\ \ (1 , -3)\ \ (2 , 1)\)

Linear Relationships, SM-Bank 037

  1. Complete the table of values below for the rule  \( y=5-x\).  (2 marks)
     
    \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. Use the table to list the coordinates of the points.  (2 marks)

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a.    \(\text{Table of values}\)

\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} & 6  & 5 & 4 & 3\\
\hline
\end{array}

b.    \((-1 , 6)\ \ (0 , 5)\ \ (1 , 4)\ \ (2 , 3)\)

Show Worked Solution

a.    \(\text{Table of values}\)

\begin{array} {|l|c|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} & 5-(-1)=6  & 5-0=5 & 5-1=4 & 5-2=3\\
\hline
\end{array}

b.    \((-1 , 6)\ \ (0 , 5)\ \ (1 , 4)\ \ (2 , 3)\)

Linear Relationships, SM-Bank 036

Complete the table of values below for the given rule.  (2 marks)

\( v=4u-3\)

\begin{array} {|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\ \ u\ \ \rule[-1ex]{0pt}{0pt} &  -2  &  -1  &\ \ 0\ \ &\ \ 1\ \ &\ \ 2\ \ \\
\hline
\rule{0pt}{2.5ex} \ \ v\ \ \rule[-1ex]{0pt}{0pt} &   &   &  & \\
\hline
\end{array}

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\begin{array} {|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\ \ u\ \ \rule[-1ex]{0pt}{0pt} &  -2  &  -1  &\ \ 0\ \ &\ \ 1\ \ &\ \ 2\ \ \\
\hline
\rule{0pt}{2.5ex} \ \ v\ \ \rule[-1ex]{0pt}{0pt} & -11  & -7  & -3 & 1 & 5\\
\hline
\end{array}

Show Worked Solution

\begin{array} {|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\ \ u\ \ \rule[-1ex]{0pt}{0pt} &\ \ -2\ \ &\ \ -1\ \ &\ \ 0\ \ &\ \ 1\ \ &\ \ 2\ \ \\
\hline
\rule{0pt}{2.5ex} \ \ v\ \ \rule[-1ex]{0pt}{0pt} & 4\times -2-3=-11  & 4\times -1-3=-7 & 4\times 0-3=-3 & 4\times 1-3=1 & 4\times 2-3=5\\
\hline
\end{array}

Linear Relationships, SM-Bank 035

Complete the table of values below for the given rule.  (2 marks)

\( y=-x\)

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

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

Show Worked Solution

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

Linear Relationships, SM-Bank 034

Complete the table of values below for the given rule.  (2 marks)

\( y=2x+1\)

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

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

Show Worked Solution

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

Linear Relationships, SM-Bank 033

Complete the table of values below for the given rule.  (2 marks)

\( y=2+x\)

\begin{array} {|l|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\ \ x\ \ \rule[-1ex]{0pt}{0pt} &\ \ 1\ \ &\ \ 2\ \ &\ \ 3\ \ &\ \ 4\ \ \\
\hline
\rule{0pt}{2.5ex} \ \ y\ \ \rule[-1ex]{0pt}{0pt} &   &   &  & \\
\hline
\end{array}

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\begin{array} {|l|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\ \ x\ \ \rule[-1ex]{0pt}{0pt} &\ \ 1\ \ &\ \ 2\ \ &\ \ 3\ \ &\ \ 4\ \ \\
\hline
\rule{0pt}{2.5ex} \ \ y\ \ \rule[-1ex]{0pt}{0pt} & 3  & 4  & 5 & 6\\
\hline
\end{array}

Show Worked Solution

\begin{array} {|l|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\ \ x\ \ \rule[-1ex]{0pt}{0pt} &\ \ 1\ \ &\ \ 2\ \ &\ \ 3\ \ &\ \ 4\ \ \\
\hline
\rule{0pt}{2.5ex} \ \ y\ \ \rule[-1ex]{0pt}{0pt} & 2+1=3 & 2+2=4  & 2+3=5 & 2+4=6\\
\hline
\end{array}