Yup!

You can calculate the slope or gain (dVout / dVin) of the amp based on the R4 into R1 || R5 voltage divider. Don't forget that the input is attenuated by R6 into R2 || R3. Then, offset by following the voltage offsets around (if the R2-R3 divider has the same potential as R1-R5, then when R6 has no current flow, the output's at zero).

Further, if you actually have a negative supply handy, you only need one bias resistor (up or down as needed).

A good example and way-of-thinking about this:

Suppose you have a current shunt. So, zero referenced, no offset, low impedance. But too low voltage. Say you want 0-5V from a 100mV shunt. So you need a gain of 50. So, from op-amp output, 49k to -in, 1k to GND. Shunt to +in via 1k resistor (match input resistances -- helps to match the offsets due to input bias current). If the amp has +/-1mV input offset, it's 50mV at the output, which kind of sucks (1% error). So, suppose we bias the input by 1mV, and bias the other by 0-2mV from a trimmer, so you can calibrate it to zero offset. From a 10V supply, you'd use a 10Meg resistor to +in to guarantee an offset of 0-2mV, then a trimpot spanning GND to supply (0-10V) with the wiper to a 5Meg resistor to -in (which has a Thevenin equivalent of a little under 1k, so twice the offset needs half the dropping resistor). Or since Megs are kind of sucky, you could run the trimpot as a divider (say, a 1k trimmer with 100k from supply = 0.1V) and use smaller values (100k from the top of the trimmer, and 50k from the trimmer wiper).

Note that, in a case like this, you can approximate the voltage divider equation (R2 / R1+R2 and related) as linear (R2/R1) and parallel resistors as ignored (1k || 50k is barely less than 1k). Take advantage of whatever approximations you can, then fine tune it with real math (set up the full equations and tweak, or just play around in the simulator for the same effect).

Tim