Let's go through the problem step-by-step to find the corresponding [tex]\( y \)[/tex]-values for the given [tex]\( x \)[/tex]-values using the formula [tex]\( y = 2x - 3 \)[/tex].
1. The given [tex]\( x \)[/tex]-values are [tex]\( 2 \)[/tex], [tex]\( 3 \)[/tex], [tex]\( 4 \)[/tex], [tex]\( 5 \)[/tex], [tex]\( 6 \)[/tex], and [tex]\( 28 \)[/tex].
2. We will substitute each [tex]\( x \)[/tex]-value into the expression [tex]\( y = 2x - 3 \)[/tex] and calculate the corresponding [tex]\( y \)[/tex]-value.
Let's substitute and calculate:
- For [tex]\( x = 2 \)[/tex]:
[tex]\[
y = 2(2) - 3 = 4 - 3 = 1
\][/tex]
- For [tex]\( x = 3 \)[/tex]:
[tex]\[
y = 2(3) - 3 = 6 - 3 = 3
\][/tex]
- For [tex]\( x = 4 \)[/tex]:
[tex]\[
y = 2(4) - 3 = 8 - 3 = 5
\][/tex]
- For [tex]\( x = 5 \)[/tex]:
[tex]\[
y = 2(5) - 3 = 10 - 3 = 7
\][/tex]
- For [tex]\( x = 6 \)[/tex]:
[tex]\[
y = 2(6) - 3 = 12 - 3 = 9
\][/tex]
- For [tex]\( x = 28 \)[/tex]:
[tex]\[
y = 2(28) - 3 = 56 - 3 = 53
\][/tex]
So the corresponding [tex]\( y \)[/tex]-values are:
[tex]\[
1, 3, 5, 7, 9, 53
\][/tex]
Updating the table with these [tex]\( y \)[/tex]-values:
[tex]\[
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
x & 2 & 3 & 4 & 5 & 6 & 28 \\
\hline
y & 1 & 3 & 5 & 7 & 9 & 53 \\
\hline
\end{tabular}
\][/tex]
To plot and draw the graph of [tex]\( y = 2x - 3 \)[/tex], you can use these points:
- Point (2, 1)
- Point (3, 3)
- Point (4, 5)
- Point (5, 7)
- Point (6, 9)
- Point (28, 53)
Plot these points on the coordinate plane and draw a straight line passing through them. The line you draw will represent the equation [tex]\( y = 2x - 3 \)[/tex].
When you plot these points correctly on a graph, you will see that they all lie on the line described by the equation [tex]\( y = 2x - 3 \)[/tex], confirming the correctness of our calculations.
