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Sagot :
Certainly! Let's address each part of your task step-by-step. Make sure to have your graph sheet ready for visual representation and follow along with these detailed instructions.
### Part (a): Completing the Table of Values
Let's work with the equations given:
1. [tex]\( x - 2y = -2 \)[/tex]
2. [tex]\( x - 2y = 2 \)[/tex]
We need to find the [tex]\( y \)[/tex] values corresponding to [tex]\( x \)[/tex] values from [tex]\(-3\)[/tex] to [tex]\(3\)[/tex].
#### For [tex]\( x - 2y = -2 \)[/tex]
Rearranging the equation for [tex]\( y \)[/tex]:
[tex]\[ x - 2y = -2 \][/tex]
[tex]\[ -2y = -2 - x \][/tex]
[tex]\[ y = \frac{x + 2}{2} \][/tex]
Now, let’s determine the [tex]\( y \)[/tex] values for x [tex]\( = -3, -2, -1, 0, 1, 2, 3 \)[/tex]:
- For [tex]\( x = -3 \)[/tex]:
[tex]\[ y = \frac{-3 + 2}{2} = \frac{-1}{2} = -0.5 \][/tex]
- For [tex]\( x = -2 \)[/tex]:
Given: [tex]\( y = 1 \)[/tex] already provided in the table.
- For [tex]\( x = -1 \)[/tex]:
[tex]\[ y = \frac{-1 + 2}{2} = \frac{1}{2} = 0.5 \][/tex]
- For [tex]\( x = 0 \)[/tex]:
Given: [tex]\( y = 1 \)[/tex] already provided in the table.
- For [tex]\( x = 1 \)[/tex]:
[tex]\[ y = \frac{1 + 2}{2} = \frac{3}{2} = 1.5 \][/tex]
- For [tex]\( x = 2 \)[/tex]:
Given: [tex]\( y = 2 \)[/tex] already provided in the table.
- For [tex]\( x = 3 \)[/tex]:
[tex]\[ y = \frac{3 + 2}{2} = \frac{5}{2} = 2.5 \][/tex]
So, the completed table for [tex]\( x - 2y = -2 \)[/tex]:
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline
[tex]$x$[/tex] & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\
\hline
[tex]$x - 2y = -2$[/tex] & -0.5 & 1 & 0.5 & 1 & 1.5 & 2 & 2.5 \\
\hline
\end{tabular}
#### For [tex]\( x - 2y = 2 \)[/tex]
Rearranging the equation for [tex]\( y \)[/tex]:
[tex]\[ x - 2y = 2 \][/tex]
[tex]\[ -2y = 2 - x \][/tex]
[tex]\[ y = \frac{x - 2}{2} \][/tex]
Now, let’s determine the [tex]\( y \)[/tex] values for [tex]\(x\)[/tex] from [tex]\(-3\)[/tex] to [tex]\(3\)[/tex]:
- For [tex]\( x = -3 \)[/tex]:
[tex]\[ y = \frac{-3 - 2}{2} = \frac{-5}{2} = -2.5 \][/tex]
- For [tex]\( x = -2 \)[/tex]:
Given: [tex]\( y = -2 \)[/tex] already provided in the table.
- For [tex]\( x = -1 \)[/tex]:
[tex]\[ y = \frac{-1 - 2}{2} = \frac{-3}{2} = -1.5 \][/tex]
- For [tex]\( x = 0 \)[/tex]:
Given: [tex]\( y = -1 \)[/tex] already provided in the table.
- For [tex]\( x = 1 \)[/tex]:
[tex]\[ y = \frac{1 - 2}{2} = \frac{-1}{2} = -0.5 \][/tex]
- For [tex]\( x = 2 \)[/tex]:
Given: [tex]\( y = 0 \)[/tex] already provided in the table.
- For [tex]\( x = 3 \)[/tex]:
[tex]\[ y = \frac{3 - 2}{2} = \frac{1}{2} = 0.5 \][/tex]
So, the completed table for [tex]\( x - 2y = 2 \)[/tex]:
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline
[tex]$x$[/tex] & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\
\hline
[tex]$x - 2y = -2$[/tex] & -0.5 & 1 & 0.5 & 1 & 1.5 & 2 & 2.5 \\
[tex]$x - 2y = 2$[/tex] & -2.5 & -2 & -1.5 & -1 & -0.5 & 0 & 0.5 \\
\hline
\end{tabular}
### Part (b): Determining the Gradient
To determine the gradient (slope) of the line [tex]\( x - 2y = 2 \)[/tex]:
Rewriting the equation in slope-intercept form [tex]\( y = mx + b \)[/tex] where [tex]\( m \)[/tex] is the slope:
[tex]\[ x - 2y = 2 \][/tex]
[tex]\[ -2y = -x + 2 \][/tex]
[tex]\[ y = \frac{1}{2}x - 1 \][/tex]
Thus, the gradient [tex]\( m \)[/tex] of the line [tex]\( x - 2y = 2 \)[/tex] is:
[tex]\[ m = \frac{1}{2} \][/tex]
### Part (a): Completing the Table of Values
Let's work with the equations given:
1. [tex]\( x - 2y = -2 \)[/tex]
2. [tex]\( x - 2y = 2 \)[/tex]
We need to find the [tex]\( y \)[/tex] values corresponding to [tex]\( x \)[/tex] values from [tex]\(-3\)[/tex] to [tex]\(3\)[/tex].
#### For [tex]\( x - 2y = -2 \)[/tex]
Rearranging the equation for [tex]\( y \)[/tex]:
[tex]\[ x - 2y = -2 \][/tex]
[tex]\[ -2y = -2 - x \][/tex]
[tex]\[ y = \frac{x + 2}{2} \][/tex]
Now, let’s determine the [tex]\( y \)[/tex] values for x [tex]\( = -3, -2, -1, 0, 1, 2, 3 \)[/tex]:
- For [tex]\( x = -3 \)[/tex]:
[tex]\[ y = \frac{-3 + 2}{2} = \frac{-1}{2} = -0.5 \][/tex]
- For [tex]\( x = -2 \)[/tex]:
Given: [tex]\( y = 1 \)[/tex] already provided in the table.
- For [tex]\( x = -1 \)[/tex]:
[tex]\[ y = \frac{-1 + 2}{2} = \frac{1}{2} = 0.5 \][/tex]
- For [tex]\( x = 0 \)[/tex]:
Given: [tex]\( y = 1 \)[/tex] already provided in the table.
- For [tex]\( x = 1 \)[/tex]:
[tex]\[ y = \frac{1 + 2}{2} = \frac{3}{2} = 1.5 \][/tex]
- For [tex]\( x = 2 \)[/tex]:
Given: [tex]\( y = 2 \)[/tex] already provided in the table.
- For [tex]\( x = 3 \)[/tex]:
[tex]\[ y = \frac{3 + 2}{2} = \frac{5}{2} = 2.5 \][/tex]
So, the completed table for [tex]\( x - 2y = -2 \)[/tex]:
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline
[tex]$x$[/tex] & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\
\hline
[tex]$x - 2y = -2$[/tex] & -0.5 & 1 & 0.5 & 1 & 1.5 & 2 & 2.5 \\
\hline
\end{tabular}
#### For [tex]\( x - 2y = 2 \)[/tex]
Rearranging the equation for [tex]\( y \)[/tex]:
[tex]\[ x - 2y = 2 \][/tex]
[tex]\[ -2y = 2 - x \][/tex]
[tex]\[ y = \frac{x - 2}{2} \][/tex]
Now, let’s determine the [tex]\( y \)[/tex] values for [tex]\(x\)[/tex] from [tex]\(-3\)[/tex] to [tex]\(3\)[/tex]:
- For [tex]\( x = -3 \)[/tex]:
[tex]\[ y = \frac{-3 - 2}{2} = \frac{-5}{2} = -2.5 \][/tex]
- For [tex]\( x = -2 \)[/tex]:
Given: [tex]\( y = -2 \)[/tex] already provided in the table.
- For [tex]\( x = -1 \)[/tex]:
[tex]\[ y = \frac{-1 - 2}{2} = \frac{-3}{2} = -1.5 \][/tex]
- For [tex]\( x = 0 \)[/tex]:
Given: [tex]\( y = -1 \)[/tex] already provided in the table.
- For [tex]\( x = 1 \)[/tex]:
[tex]\[ y = \frac{1 - 2}{2} = \frac{-1}{2} = -0.5 \][/tex]
- For [tex]\( x = 2 \)[/tex]:
Given: [tex]\( y = 0 \)[/tex] already provided in the table.
- For [tex]\( x = 3 \)[/tex]:
[tex]\[ y = \frac{3 - 2}{2} = \frac{1}{2} = 0.5 \][/tex]
So, the completed table for [tex]\( x - 2y = 2 \)[/tex]:
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline
[tex]$x$[/tex] & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\
\hline
[tex]$x - 2y = -2$[/tex] & -0.5 & 1 & 0.5 & 1 & 1.5 & 2 & 2.5 \\
[tex]$x - 2y = 2$[/tex] & -2.5 & -2 & -1.5 & -1 & -0.5 & 0 & 0.5 \\
\hline
\end{tabular}
### Part (b): Determining the Gradient
To determine the gradient (slope) of the line [tex]\( x - 2y = 2 \)[/tex]:
Rewriting the equation in slope-intercept form [tex]\( y = mx + b \)[/tex] where [tex]\( m \)[/tex] is the slope:
[tex]\[ x - 2y = 2 \][/tex]
[tex]\[ -2y = -x + 2 \][/tex]
[tex]\[ y = \frac{1}{2}x - 1 \][/tex]
Thus, the gradient [tex]\( m \)[/tex] of the line [tex]\( x - 2y = 2 \)[/tex] is:
[tex]\[ m = \frac{1}{2} \][/tex]
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