$y$ varies jointly as $x$ and $z . y=48$ when $x=6$ and $z=4$. Find $y$ when $x=7$ and $z=3$. \[ y= \]

Step 1 ：The problem states that $y$ varies jointly as $x$ and $z$. This means that $y$ is directly proportional to both $x$ and $z$. We can express this relationship as $y = kxz$, where $k$ is the constant of variation.

Step 2 ：We can find the value of $k$ using the given values of $y$, $x$, and $z$. Given that $y=48$ when $x=6$ and $z=4$, we can substitute these values into the equation to get $48 = k(6)(4)$. Solving for $k$ gives us $k = 2.0$.

Step 3 ：Now that we have the value of $k$, we can find the value of $y$ when $x=7$ and $z=3$. Substituting these values into the equation gives us $y = 2.0(7)(3)$, which simplifies to $y = 42.0$.

Step 4 ：Final Answer: The value of $y$ when $x=7$ and $z=3$ is \(\boxed{42}\).