To solve the given system of equations, we need to find the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] that satisfy both equations simultaneously. The system is:
[tex]\[
\left\{\begin{aligned}
10x^2 - y &= 48 \\
2y &= 16x^2 + 48
\end{aligned}\right.
\][/tex]
### Step 1: Express [tex]\( y \)[/tex] from the second equation
The second equation is:
[tex]\[ 2y = 16x^2 + 48 \][/tex]
Solving for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{16x^2 + 48}{2} \][/tex]
[tex]\[ y = 8x^2 + 24 \][/tex]
### Step 2: Substitute [tex]\( y \)[/tex] in the first equation
Substitute [tex]\( y = 8x^2 + 24 \)[/tex] into the first equation:
[tex]\[ 10x^2 - (8x^2 + 24) = 48 \][/tex]
Simplifying this:
[tex]\[ 10x^2 - 8x^2 - 24 = 48 \][/tex]
[tex]\[ 2x^2 - 24 = 48 \][/tex]
[tex]\[ 2x^2 = 72 \][/tex]
[tex]\[ x^2 = 36 \][/tex]
[tex]\[ x = \pm 6 \][/tex]
### Step 3: Calculate [tex]\( y \)[/tex] for [tex]\( x = 6 \)[/tex] and [tex]\( x = -6 \)[/tex]
For [tex]\( x = 6 \)[/tex]:
[tex]\[ y = 8(6^2) + 24 \][/tex]
[tex]\[ y = 8(36) + 24 \][/tex]
[tex]\[ y = 288 + 24 \][/tex]
[tex]\[ y = 312 \][/tex]
For [tex]\( x = -6 \)[/tex]:
[tex]\[ y = 8((-6)^2) + 24 \][/tex]
[tex]\[ y = 8(36) + 24 \][/tex]
[tex]\[ y = 288 + 24 \][/tex]
[tex]\[ y = 312 \][/tex]
### Step 4: List the solutions
The solutions are thus:
[tex]\[ (x, y) = (6, 312) \][/tex]
[tex]\[ (x, y) = (-6, 312) \][/tex]
Therefore, the solutions to the system of equations are:
[tex]\[ \boxed{(6, 312) \text{ and } (-6, 312)} \][/tex]
