Let's break down the problem step-by-step to identify the correct equations that represent the situation and then find the valid solutions.
First, we interpret the problem statement to form the system of equations.
1. "The square of the first number is equal to the sum of the second number and 16."
This translates to the equation:
[tex]\[
x^2 = y + 16
\][/tex]
2. "The difference of 4 times the second number and 1 is equal to the first number multiplied by 7."
This translates to the equation:
[tex]\[
4y - 1 = 7x
\][/tex]
Now, we identify these equations in the given choices:
- \( x^2 = y + 16 \) — This is present in the second row of the "Equations" section.
- \( 4y - 1 = 7x \) — This is present in the third row of the "Equations" section.
So, the correct system of equations is:
[tex]\[
x^2 = y + 16 \quad \text{and} \quad 4y - 1 = 7x
\][/tex]
Next, we need to find the solution(s) of this system from the given options.
Given the solutions to check are:
- \((1, 15)\)
- \((5, 9)\)
- \((2, -12)\)
- \((8, 48)\)
- \((4, -7)\)
- \((9, 3)\)
By validating these pairs with the correct system of equations:
1. For \((1, 15)\):
- \(1^2 = 15 + 16 \Rightarrow 1 \neq 31\)
- \(4 \times 15 - 1 = 7 \times 1 \Rightarrow 59 \neq 7\)
2. For \((5, 9)\):
- \(5^2 = 9 + 16 \Rightarrow 25 = 25\)
- \(4 \times 9 - 1 = 7 \times 5 \Rightarrow 35 = 35\)
3. For \((2, -12)\):
- \(2^2 = -12 + 16 \Rightarrow 4 = 4\)
- \(4 \times -12 - 1 = 7 \times 2 \Rightarrow -49 \neq 14\)
4. For \((8, 48)\):
- \(8^2 = 48 + 16 \Rightarrow 64 = 64\)
- \(4 \times 48 - 1 = 7 \times 8 \Rightarrow 191 \neq 56\)
5. For \((4, -7)\):
- \(4^2 = -7 + 16 \Rightarrow 16 = 9\)
- \(4 \times -7 - 1 = 7 \times 4 \Rightarrow -29 \neq 28\)
6. For \((9, 3)\):
- \(9^2 = 3 + 16 \Rightarrow 81 \neq 19\)
- \(4 \times 3 - 1 = 7 \times 9 \Rightarrow 11 \neq 63\)
Therefore, the correct and valid solution is:
[tex]\[
(5, 9)
\][/tex]
The tabular match:
[tex]\[
\begin{tabular}{|l|l|l|l|}
\hline \multicolumn{2}{|c|}{ Equations } & \multicolumn{2}{|c|}{ Solutions } \\
\hline[tex]$y^2+16=x$[/tex] & [tex]$(2 x)^2=y+16$[/tex] & [tex]$(1,15)$[/tex] & [tex]$(5,9)$[/tex] \\
\hline[tex]$x^2=y+16$[/tex] & [tex]$7 y-1=4 x$[/tex] & [tex]$(2,-12)$[/tex] & [tex]$(8,48)$[/tex] \\
\hline [tex]$1-4 y=7 x$[/tex] & [tex]$4 y-1=7 x$[/tex] & [tex]$(4,-7)$[/tex] & [tex]$(9,3)$[/tex] \\
\hline
\end{tabular}
\][/tex]
The correct responses are:
- The system of equations:
[tex]\[
\text{Row 2:} \quad x^2 = y + 16 \quad \text{and from Row 3:} \quad 4y - 1 = 7x
\][/tex]
- The valid solution:
[tex]\[
(5, 9)
\][/tex]
