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It is seems impossible to give a better answer than Pete L. Clark's but those passing by might appreciate a sketchy proof.

If $X \subset \mathbb P^n$ is a projective variety then it is the intersection of finitely many hypersurfaces of degree at most $k$. If we consider the natural morphism from $\mathbb P^n$ to $\mathbb P H^0(\mathbb P^n,\mathcal O_{\mathbb P^n}(k))^{\ast}$ then the image of $X$ will be the intersection of the image of $\mathbb P^n$ with a finite number of hyperplanes.

Being the image of $\mathbb P^n$ itself a intersection of quadrics, it follows that any projective variety can be expressed as the intersection of quadrics and hyperplanes.

A more condensed version of the argument above can also be found here.

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It is impossible to give a better answer than Pete L. Clark's but those passing by might appreciate a sketchy proof.

If $X \subset \mathbb P^n$ is a projective variety then it is the intersection of finitely many hypersurfaces of degree at most $k$. If we consider the natural morphism from $\mathbb P^n$ to $\mathbb P H^0(\mathbb P^n,\mathcal O_{\mathbb P^n}(k))^{\ast}$ then the image of $X$ will be the intersection of the image of $\mathbb P^n$ with a finite number of hyperplanes.

Being the image of $\mathbb P^n$ itself a intersection of quadrics, it follows that any projective variety can be expressed as the intersection of quadrics and hyperplanes.

A more condensed version of the argument above can also be found here.