To determine which relation is a direct variation that contains the ordered pair [tex]\((2, 7)\)[/tex], we will check each of the given equations by substituting [tex]\(x = 2\)[/tex] and [tex]\(y = 7\)[/tex] into them.
1. Checking the equation [tex]\( y = 4x - 1 \)[/tex]:
Substitute [tex]\( x = 2 \)[/tex]:
[tex]\[
y = 4(2) - 1 = 8 - 1 = 7
\][/tex]
So, when [tex]\(x = 2\)[/tex], [tex]\( y \)[/tex] is indeed 7.
2. Checking the equation [tex]\( y = \frac{7}{x} \)[/tex]:
Substitute [tex]\( x = 2 \)[/tex]:
[tex]\[
y = \frac{7}{2} = 3.5
\][/tex]
Here, when [tex]\(x = 2\)[/tex], [tex]\( y \)[/tex] is 3.5, not 7.
3. Checking the equation [tex]\( y = \frac{2}{7}x \)[/tex]:
Substitute [tex]\( x = 2 \)[/tex]:
[tex]\[
y = \frac{2}{7} \times 2 = \frac{4}{7}
\][/tex]
In this case, when [tex]\(x = 2\)[/tex], [tex]\( y \)[/tex] is [tex]\(\frac{4}{7}\)[/tex], not 7.
4. Checking the equation [tex]\( y = \frac{7}{2}x \)[/tex]:
Substitute [tex]\( x = 2 \)[/tex]:
[tex]\[
y = \frac{7}{2} \times 2 = 7
\][/tex]
Here, when [tex]\(x = 2\)[/tex], [tex]\( y \)[/tex] is indeed 7, similar to the first equation.
After checking all the given equations, we see that the first equation [tex]\( y = 4x - 1 \)[/tex] satisfies the ordered pair [tex]\((2, 7)\)[/tex].
Therefore, the relation is [tex]\( y = 4x - 1 \)[/tex].
