Answer :
To determine the solutions, we need to verify which pairs [tex]\((x, y)\)[/tex] satisfy both equations in the given system:
[tex]\[ \left\{\begin{aligned} 10x^2 - y &= 48 \\ 2y &= 16x^2 + 48 \end{aligned}\right. \][/tex]
The steps we need to follow are to substitute each potential solution into the system of equations and check if both equations hold true.
### Checking the first pair [tex]\((2\sqrt{3}, 120)\)[/tex]:
1. For the first equation [tex]\(10x^2 - y = 48\)[/tex]:
[tex]\[ x = 2\sqrt{3}, \quad y = 120 \\ 10(2\sqrt{3})^2 - 120 = 10(4 \cdot 3) - 120 = 120 - 120 = 0 \neq 48 \][/tex]
The first equation is not satisfied.
2. For the second equation [tex]\(2y = 16x^2 + 48\)[/tex]:
[tex]\[ x = 2\sqrt{3}, \quad y = 120 \\ 2(120) = 240 \\ 16(2\sqrt{3})^2 + 48 = 16(4 \cdot 3) + 48 = 192 + 48 = 240 \][/tex]
The second equation is satisfied.
Since the first equation is not satisfied, [tex]\((2\sqrt{3}, 120)\)[/tex] is not a solution.
### Checking the second pair [tex]\((-2\sqrt{3}, 120)\)[/tex]:
1. For the first equation [tex]\(10x^2 - y = 48\)[/tex]:
[tex]\[ x = -2\sqrt{3}, \quad y = 120 \\ 10(-2\sqrt{3})^2 - 120 = 10(4 \cdot 3) - 120 = 120 - 120 = 0 \neq 48 \][/tex]
The first equation is not satisfied.
2. For the second equation [tex]\(2y = 16x^2 + 48\)[/tex]:
[tex]\[ x = -2\sqrt{3}, \quad y = 120 \\ 2(120) = 240 \\ 16(-2\sqrt{3})^2 + 48 = 16(4 \cdot 3) + 48 = 192 + 48 = 240 \][/tex]
The second equation is satisfied.
Since the first equation is not satisfied, [tex]\((-2\sqrt{3}, 120)\)[/tex] is not a solution.
### Checking the third pair [tex]\((6, 312)\)[/tex]:
1. For the first equation [tex]\(10x^2 - y = 48\)[/tex]:
[tex]\[ x = 6, \quad y = 312 \\ 10(6)^2 - 312 = 10(36) - 312 = 360 - 312 = 48 \][/tex]
The first equation is satisfied.
2. For the second equation [tex]\(2y = 16x^2 + 48\)[/tex]:
[tex]\[ x = 6, \quad y = 312 \\ 2(312) = 624 \\ 16(6)^2 + 48 = 16(36) + 48 = 576 + 48 = 624 \][/tex]
The second equation is satisfied.
Since both equations are satisfied, [tex]\((6, 312)\)[/tex] is a valid solution.
### Checking the fourth pair [tex]\((-6, 312)\)[/tex]:
1. For the first equation [tex]\(10x^2 - y = 48\)[/tex]:
[tex]\[ x = -6, \quad y = 312 \\ 10(-6)^2 - 312 = 10(36) - 312 = 360 - 312 = 48 \][/tex]
The first equation is satisfied.
2. For the second equation [tex]\(2y = 16x^2 + 48\)[/tex]:
[tex]\[ x = -6, \quad y = 312 \\ 2(312) = 624 \\ 16(-6)^2 + 48 = 16(36) + 48 = 576 + 48 = 624 \][/tex]
The second equation is satisfied.
Since both equations are satisfied, [tex]\((-6, 312)\)[/tex] is a valid solution.
### Checking the fifth pair [tex]\((6, -264)\)[/tex]:
1. For the first equation [tex]\(10x^2 - y = 48\)[/tex]:
[tex]\[ x = 6, \quad y = -264 \\ 10(6)^2 - (-264) = 10(36) + 264 = 360 + 264 = 624 \neq 48 \][/tex]
The first equation is not satisfied.
2. For the second equation [tex]\(2y = 16x^2 + 48\)[/tex]:
[tex]\[ x = 6, \quad y = -264 \\ 2(-264) = -528 \\ 16(6)^2 + 48 = 16(36) + 48 = 576 + 48 = 624 \][/tex]
The second equation is not satisfied.
Since neither equation is satisfied, [tex]\((6, -264)\)[/tex] is not a solution.
### Conclusion
The correct and valid solutions are:
[tex]\[ (6, 312) \quad \text{and} \quad (-6, 312) \][/tex]
[tex]\[ \left\{\begin{aligned} 10x^2 - y &= 48 \\ 2y &= 16x^2 + 48 \end{aligned}\right. \][/tex]
The steps we need to follow are to substitute each potential solution into the system of equations and check if both equations hold true.
### Checking the first pair [tex]\((2\sqrt{3}, 120)\)[/tex]:
1. For the first equation [tex]\(10x^2 - y = 48\)[/tex]:
[tex]\[ x = 2\sqrt{3}, \quad y = 120 \\ 10(2\sqrt{3})^2 - 120 = 10(4 \cdot 3) - 120 = 120 - 120 = 0 \neq 48 \][/tex]
The first equation is not satisfied.
2. For the second equation [tex]\(2y = 16x^2 + 48\)[/tex]:
[tex]\[ x = 2\sqrt{3}, \quad y = 120 \\ 2(120) = 240 \\ 16(2\sqrt{3})^2 + 48 = 16(4 \cdot 3) + 48 = 192 + 48 = 240 \][/tex]
The second equation is satisfied.
Since the first equation is not satisfied, [tex]\((2\sqrt{3}, 120)\)[/tex] is not a solution.
### Checking the second pair [tex]\((-2\sqrt{3}, 120)\)[/tex]:
1. For the first equation [tex]\(10x^2 - y = 48\)[/tex]:
[tex]\[ x = -2\sqrt{3}, \quad y = 120 \\ 10(-2\sqrt{3})^2 - 120 = 10(4 \cdot 3) - 120 = 120 - 120 = 0 \neq 48 \][/tex]
The first equation is not satisfied.
2. For the second equation [tex]\(2y = 16x^2 + 48\)[/tex]:
[tex]\[ x = -2\sqrt{3}, \quad y = 120 \\ 2(120) = 240 \\ 16(-2\sqrt{3})^2 + 48 = 16(4 \cdot 3) + 48 = 192 + 48 = 240 \][/tex]
The second equation is satisfied.
Since the first equation is not satisfied, [tex]\((-2\sqrt{3}, 120)\)[/tex] is not a solution.
### Checking the third pair [tex]\((6, 312)\)[/tex]:
1. For the first equation [tex]\(10x^2 - y = 48\)[/tex]:
[tex]\[ x = 6, \quad y = 312 \\ 10(6)^2 - 312 = 10(36) - 312 = 360 - 312 = 48 \][/tex]
The first equation is satisfied.
2. For the second equation [tex]\(2y = 16x^2 + 48\)[/tex]:
[tex]\[ x = 6, \quad y = 312 \\ 2(312) = 624 \\ 16(6)^2 + 48 = 16(36) + 48 = 576 + 48 = 624 \][/tex]
The second equation is satisfied.
Since both equations are satisfied, [tex]\((6, 312)\)[/tex] is a valid solution.
### Checking the fourth pair [tex]\((-6, 312)\)[/tex]:
1. For the first equation [tex]\(10x^2 - y = 48\)[/tex]:
[tex]\[ x = -6, \quad y = 312 \\ 10(-6)^2 - 312 = 10(36) - 312 = 360 - 312 = 48 \][/tex]
The first equation is satisfied.
2. For the second equation [tex]\(2y = 16x^2 + 48\)[/tex]:
[tex]\[ x = -6, \quad y = 312 \\ 2(312) = 624 \\ 16(-6)^2 + 48 = 16(36) + 48 = 576 + 48 = 624 \][/tex]
The second equation is satisfied.
Since both equations are satisfied, [tex]\((-6, 312)\)[/tex] is a valid solution.
### Checking the fifth pair [tex]\((6, -264)\)[/tex]:
1. For the first equation [tex]\(10x^2 - y = 48\)[/tex]:
[tex]\[ x = 6, \quad y = -264 \\ 10(6)^2 - (-264) = 10(36) + 264 = 360 + 264 = 624 \neq 48 \][/tex]
The first equation is not satisfied.
2. For the second equation [tex]\(2y = 16x^2 + 48\)[/tex]:
[tex]\[ x = 6, \quad y = -264 \\ 2(-264) = -528 \\ 16(6)^2 + 48 = 16(36) + 48 = 576 + 48 = 624 \][/tex]
The second equation is not satisfied.
Since neither equation is satisfied, [tex]\((6, -264)\)[/tex] is not a solution.
### Conclusion
The correct and valid solutions are:
[tex]\[ (6, 312) \quad \text{and} \quad (-6, 312) \][/tex]
