Question 4

Use the elimination method to find all solutions of the system:
\[
\left\{\begin{array}{l}
3 x^{2}-y^{2}=11 \\
x^{2}+4 y^{2}=8
\end{array}\right.
\]

The four solutions of the system are: $(-a,-b),(-a, b),(a,-b)$, and $(a, b)$.
Using positive numbers find:
\[
a=
\]
and $b=$
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Step 1 ：Add the two equations together to get \(4x^2 + 3y^2 = 19\)

Step 2 ：Isolate \(x^2\) in the first equation to get \(x^2 = \frac{11 + y^2}{3}\)

Step 3 ：Substitute \(x^2\) into the second equation to get \(\frac{11 + y^2}{3} + 4y^2 = 8\)

Step 4 ：Multiply through by 3 to clear the fraction to get \(11 + y^2 + 12y^2 = 24\)

Step 5 ：Combine like terms to get \(13y^2 + 11 = 24\)

Step 6 ：Subtract 11 from both sides to get \(13y^2 = 13\)

Step 7 ：Divide by 13 to get \(y^2 = 1\), so \(y = 1\) or \(y = -1\)

Step 8 ：Substitute \(y = 1\) into the first equation to get \(3x^2 - 1 = 11\)

Step 9 ：Add 1 to both sides to get \(3x^2 = 12\)

Step 10 ：Divide by 3 to get \(x^2 = 4\), so \(x = 2\) or \(x = -2\)

Step 11 ：Substitute \(y = -1\) into the first equation to get \(3x^2 - (-1) = 11\)

Step 12 ：Add 1 to both sides to get \(3x^2 = 12\)

Step 13 ：Divide by 3 to get \(x^2 = 4\), so \(x = 2\) or \(x = -2\)

Step 14 ：Therefore, the four solutions of the system are \((-2,-1),(-2, 1),(2,-1)\), and \((2, 1)\)

Step 15 ：So, \(a = 2\) and \(b = 1\)

Step 16 ：The final answer is \(\boxed{a = 2}\) and \(\boxed{b = 1}\)