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

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 $x=[-b\pm sqrt(b^2-4ac)]/2a$: $x=[-1\pm sqrt((1{)}^2-4\ast 1\ast 1)]/2\ast 1$

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

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

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

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