jmc

The line $y=b-x$ intersects the $x$-axis at the point where $0=b-x$, or $x=b$. So, we seek the $x$-coordinate of point $Q$.

Since the $y$-axis is parallel to the line $x=4$, we see that $\angle QSR=\angle QPO$. Also $QOP=QRS=90$. Thus $△QOP\sim △QRS$, so $\frac{[QRS]}{[QOP]}={\left(\frac{QR}{QO}\right)}^2$, which means we have ${\left(\frac{QR}{QO}\right)}^2=\frac{9}{25}$, so $\frac{QR}{QO}=\frac{3}{5}$. Since $QR+QO=4$, we have $\frac{3}{5}QO+QO=4$, and $QO=4\cdot \frac{5}{8}=\frac{5}{2}$. Therefore, the $x$-coordinate of $Q$ is $\frac{5}{2}=\boxed{2.5}$.