Answer :
To determine which relation is a direct variation that contains the ordered pair [tex]\((2, 7)\)[/tex], we need to check each given equation to see if it holds true for [tex]\(x = 2\)[/tex] and [tex]\(y = 7\)[/tex].
### Equation 1: [tex]\(y = 4x - 1\)[/tex]
Substitute [tex]\(x = 2\)[/tex] and [tex]\(y = 7\)[/tex].
[tex]\[ y = 4(2) - 1 \][/tex]
[tex]\[ y = 8 - 1 \][/tex]
[tex]\[ y = 7 \][/tex]
This equation is satisfied, as [tex]\(7 = 7\)[/tex].
### Equation 2: [tex]\(y = \frac{7}{x}\)[/tex]
Substitute [tex]\(x = 2\)[/tex] and [tex]\(y = 7\)[/tex].
[tex]\[ y = \frac{7}{2} \][/tex]
[tex]\[ y = 3.5 \][/tex]
This equation is not satisfied, as [tex]\(3.5 \neq 7\)[/tex].
### Equation 3: [tex]\(y = \frac{2}{7}x\)[/tex]
Substitute [tex]\(x = 2\)[/tex] and [tex]\(y = 7\)[/tex].
[tex]\[ y = \frac{2}{7} \times 2 \][/tex]
[tex]\[ y = \frac{4}{7} \][/tex]
This equation is not satisfied, as [tex]\(\frac{4}{7} \neq 7\)[/tex].
### Equation 4: [tex]\(y = \frac{7}{2}x\)[/tex]
Substitute [tex]\(x = 2\)[/tex] and [tex]\(y = 7\)[/tex].
[tex]\[ y = \frac{7}{2} \times 2 \][/tex]
[tex]\[ y = 7 \][/tex]
This equation is satisfied, as [tex]\(7 = 7\)[/tex].
Both Equation 1 and Equation 4 satisfy the condition of containing the ordered pair [tex]\((2, 7)\)[/tex]. However, the direct variation form typically implies a linear relationship passing through the origin [tex]\( y = kx \)[/tex], not involving an additive constant.
Thus, the relation that is a direct variation and contains the ordered pair [tex]\((2, 7)\)[/tex] is:
[tex]\[ y = \frac{7}{2}x \][/tex]
So, Equation 4 is the correct relation.
### Equation 1: [tex]\(y = 4x - 1\)[/tex]
Substitute [tex]\(x = 2\)[/tex] and [tex]\(y = 7\)[/tex].
[tex]\[ y = 4(2) - 1 \][/tex]
[tex]\[ y = 8 - 1 \][/tex]
[tex]\[ y = 7 \][/tex]
This equation is satisfied, as [tex]\(7 = 7\)[/tex].
### Equation 2: [tex]\(y = \frac{7}{x}\)[/tex]
Substitute [tex]\(x = 2\)[/tex] and [tex]\(y = 7\)[/tex].
[tex]\[ y = \frac{7}{2} \][/tex]
[tex]\[ y = 3.5 \][/tex]
This equation is not satisfied, as [tex]\(3.5 \neq 7\)[/tex].
### Equation 3: [tex]\(y = \frac{2}{7}x\)[/tex]
Substitute [tex]\(x = 2\)[/tex] and [tex]\(y = 7\)[/tex].
[tex]\[ y = \frac{2}{7} \times 2 \][/tex]
[tex]\[ y = \frac{4}{7} \][/tex]
This equation is not satisfied, as [tex]\(\frac{4}{7} \neq 7\)[/tex].
### Equation 4: [tex]\(y = \frac{7}{2}x\)[/tex]
Substitute [tex]\(x = 2\)[/tex] and [tex]\(y = 7\)[/tex].
[tex]\[ y = \frac{7}{2} \times 2 \][/tex]
[tex]\[ y = 7 \][/tex]
This equation is satisfied, as [tex]\(7 = 7\)[/tex].
Both Equation 1 and Equation 4 satisfy the condition of containing the ordered pair [tex]\((2, 7)\)[/tex]. However, the direct variation form typically implies a linear relationship passing through the origin [tex]\( y = kx \)[/tex], not involving an additive constant.
Thus, the relation that is a direct variation and contains the ordered pair [tex]\((2, 7)\)[/tex] is:
[tex]\[ y = \frac{7}{2}x \][/tex]
So, Equation 4 is the correct relation.
