Complete the ordered pairs so that each is a solution of the given linear equation. Then graph the equation.

[tex]y = x + 3[/tex]

First, complete each line of the table so that the ordered pair [tex]\((x, y)\)[/tex] is a solution to the equation.

[tex]\[
\begin{tabular}{|c|c|}
\hline
$x$ & $y$ \\
\hline
0 & $\square$ \\
\hline
\end{tabular}
\][/tex]



Answer :

Certainly! Let's go through the solution step-by-step.

We are given the linear equation:
[tex]\[ y = x + 3 \][/tex]

To complete the ordered pair [tex]\((x, y)\)[/tex] in the table where [tex]\( x = 0 \)[/tex], we need to find the value of [tex]\( y \)[/tex] that satisfies the equation.

1. Substitute [tex]\( x = 0 \)[/tex] into the equation [tex]\( y = x + 3 \)[/tex]:

[tex]\[ y = 0 + 3 \][/tex]

2. Solve for [tex]\( y \)[/tex]:

[tex]\[ y = 3 \][/tex]

So, when [tex]\( x = 0 \)[/tex], [tex]\( y = 3 \)[/tex].

Therefore, the completed ordered pair is:
[tex]\[ (0, 3) \][/tex]

The table should now look like this:
\begin{tabular}{|c|c|}
\hline
[tex]\( x \)[/tex] & [tex]\( y \)[/tex] \\
\hline
0 & 3 \\
\hline
\end{tabular}

Next, you can graph the linear equation [tex]\( y = x + 3 \)[/tex]. To do this, you will need to plot the points that lie on the line.

To get a few more points to make the graph clear, let's calculate [tex]\( y \)[/tex] for some other values of [tex]\( x \)[/tex].

1. For [tex]\( x = 1 \)[/tex]:
[tex]\[ y = 1 + 3 = 4 \][/tex]
Ordered pair: [tex]\((1, 4)\)[/tex]

2. For [tex]\( x = -1 \)[/tex]:
[tex]\[ y = -1 + 3 = 2 \][/tex]
Ordered pair: [tex]\((-1, 2)\)[/tex]

3. For [tex]\( x = 2 \)[/tex]:
[tex]\[ y = 2 + 3 = 5 \][/tex]
Ordered pair: [tex]\((2, 5)\)[/tex]

4. For [tex]\( x = -2 \)[/tex]:
[tex]\[ y = -2 + 3 = 1 \][/tex]
Ordered pair: [tex]\((-2, 1)\)[/tex]

Now you can plot these points on a Cartesian coordinate system:
[tex]\[ (0, 3), (1, 4), (-1, 2), (2, 5), (-2, 1) \][/tex]

Once you've plotted these points, draw a straight line through them. This line represents the equation [tex]\( y = x + 3 \)[/tex].

Your graph should look roughly like this:

```
y
|
6
5
4

3
2

1 *
0------------------ x
-2 -1 0 1 2
```

There you have it, a detailed step-by-step solution to complete the table and graph the given linear equation [tex]\( y = x + 3 \)[/tex].