Answered

Using a scale of [tex]$2 \text{ cm}$[/tex] to 1 unit on both axes, draw two perpendicular axes [tex]\(ox\)[/tex] and [tex]\(oy\)[/tex] on a graph sheet for [tex]\(-5 \leq x \leq 5\)[/tex] and [tex]\(-5 \leq y \leq 5\)[/tex].

a) Copy and complete the table of values for the relations [tex]\(x - 2y = -2\)[/tex] and [tex]\(x - 2y = 2\)[/tex] for [tex]\(x\)[/tex] from -3 to 3.

[tex]\[
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline
$x$ & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\
\hline
$x - 2y = -2$ & 1 & 0 & -1 & -2 & -3 & -4 & -5 \\
\hline
$x - 2y = 2$ & -2.5 & -1.5 & -0.5 & 0.5 & 1.5 & 2.5 & 3.5 \\
\hline
\end{tabular}
\][/tex]

b) Determine the gradient of the line [tex]\(x - 2y = 2\)[/tex] using the graph.



Answer :

GinnyAnswer
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]

Other Questions