Solve the following system of equations.
\[
\left\{\begin{array}{l}
11 x^{2}+21 x-y=-17 \\
3 x^{2}-7 x-y=3
\end{array}\right.
\]

If therwis more than one solution, enter additional solutions with the "or" button. If there is no real solution, use the "No solution" button.
\[
(x, y)=\text { (1) }
\]

Step 1 ：Given the system of equations: \(11x^2 + 21x - y = -17\) and \(3x^2 - 7x - y = 3\)

Step 2 ：Add the two equations together to eliminate y: \((11x^2 + 21x - y) + (3x^2 - 7x - y) = -17 + 3\)

Step 3 ：This simplifies to: \(14x^2 + 14x = -14\)

Step 4 ：Divide the entire equation by 14 to simplify: \(x^2 + x = -1\)

Step 5 ：Rearrange the equation to form a quadratic equation: \(x^2 + x + 1 = 0\)

Step 6 ：Solve this quadratic equation using the quadratic formula \(x = [-b ± sqrt(b^2 - 4ac)] / 2a\): \(x = [-1 ± sqrt((1)^2 - 4*1*1)] / 2*1\)

Step 7 ：This simplifies to: \(x = [-1 ± sqrt(1 - 4)] / 2\)

Step 8 ：Further simplification gives: \(x = [-1 ± sqrt(-3)] / 2\)

Step 9 ：Since the square root of a negative number is not a real number, there are no real solutions for x.

Step 10 ：\(\boxed{\text{Therefore, there are no real solutions for the system of equations.}}\)