Let's solve the given problem step-by-step.
### Part (a)
First, we need to complete the tables of values for the given simultaneous equations [tex]\( y + 9x = 0 \)[/tex] and [tex]\( 2y - x = 0 \)[/tex].
#### For the equation [tex]\( y + 9x = 0 \)[/tex]:
Given:
[tex]\[
\begin{array}{c|ccc}
x & 0 & 1 & 2 \\
\hline
y & A & B & -18 \\
\end{array}
\][/tex]
To find [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
1. When [tex]\( x = 0 \)[/tex]:
[tex]\[ y + 9(0) = 0 \][/tex]
So,
[tex]\[ y = 0 \][/tex]
Thus,
[tex]\[ A = 0 \][/tex]
2. When [tex]\( x = 1 \)[/tex]:
[tex]\[ y + 9(1) = 0 \][/tex]
So,
[tex]\[ y + 9 = 0 \][/tex]
[tex]\[ y = -9 \][/tex]
Thus,
[tex]\[ B = -9 \][/tex]
Therefore, the completed table for [tex]\( y + 9x = 0 \)[/tex] is:
[tex]\[
\begin{array}{c|ccc}
x & 0 & 1 & 2 \\
\hline
y & 0 & -9 & -18 \\
\end{array}
\][/tex]
#### For the equation [tex]\( 2y - x = 0 \)[/tex]:
Given:
[tex]\[
\begin{array}{c|ccc}
x & 0 & 1 & 2 \\
\hline
y & C & D & 1 \\
\end{array}
\][/tex]
To find [tex]\( C \)[/tex] and [tex]\( D \)[/tex]:
1. When [tex]\( x = 0 \)[/tex]:
[tex]\[ 2y - 0 = 0 \][/tex]
So,
[tex]\[ 2y = 0 \][/tex]
[tex]\[ y = 0 \][/tex]
Thus,
[tex]\[ C = 0 \][/tex]
2. When [tex]\( x = 1 \)[/tex]:
[tex]\[ 2y - 1 = 0 \][/tex]
So,
[tex]\[ 2y = 1 \][/tex]
[tex]\[ y = \frac{1}{2} \][/tex]
Thus,
[tex]\[ D = 0.5 \][/tex]
Therefore, the completed table for [tex]\( 2y - x = 0 \)[/tex] is:
[tex]\[
\begin{array}{c|ccc}
x & 0 & 1 & 2 \\
\hline
y & 0 & 0.5 & 1 \\
\end{array}
\][/tex]
### Part (b)
Do you need to plot their graphs to work out their solution?
No, you do not need to plot their graphs to find the solution to the simultaneous equations. The solution can be found algebraically. By setting the equations equal to each other, you can solve for the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]. This allows for determining the intersection point of the two lines, which is the solution to the system of equations.
