If y is partly constant and partly varies inversely as the square of x, and y=11 when x=1, and y=5 when x=2, find y when x=4.

The quantity $y$ is partly constant and partly varies inversely as the square of $x$. This can be expressed as:

where $A$ and $B$ are constants.

• When $x=1$, $y=11$: $$11 = A + \frac{B}{1^2}$$ $$11 = A + B \quad (*)$$
• When $x=2$, $y=5$: $$5 = A + \frac{B}{2^2}$$ $$5 = A + \frac{B}{4} \quad (**)$$

From equation $(*)$, we can express $A$ as $A = 11 - B$.
Substitute this expression for $A$ into equation $(**)$:

Subtract 11 from both sides:

Multiply both sides by $-\frac{4}{3}$:

Now substitute $B=8$ back into equation $(*)$ to find $A$:

Substitute the values of $A=3$ and $B=8$ into the general relationship:

Substitute $x=4$ into the relationship found in Step 4: