Basic thermodynamics, mass conservation

A hot steel strip is fed into rollers at 0.8 m/sec w/ height 0.03 m. If the strip leaves the rollers w/ height 0.012 m and the rollers increase its density by 10% and its width by 8%, at what speed does the rolled strip leave the rollers?

I thought,

velocity in= (rate of volume in)/(area in)
velocity out= (rate of volume out)/(area out)

where (rate of volume out) = (mass out)/(density out) = (mass)/(1.10*density in) because mass = mass in = mass out, and 10% density increase .... yields (rate of volume out) = 0.91*(rate of volume in)

Then,
area out = (0.012)(0.08w)
where w = width and 0.08w because 8% increase in density ... yields area out = 9.6*10^-4*w

Which would result,
velocity out = (0.8 velocity in)/(9.6*10^-4*w)

But I can't make any sense of it. Am I on the right track?

Steel is fairly incompressible, so I don't see how density can be increased by 10%. However , ignore that for this calculation.
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Call the density of the hot steel d.
I don't know what "w/ height 0.03 m" means. I assume height=0.03m
Call the width entering 'w'.

I assume you know that if X increases by, for example, 9%, then the new value is 1.09X.
So density of steel leaving roller = 1.10d.
Width of steel leaving roller = 1.08w.

Consider the mass entering and the mass leaving each second. They have to be equal.

Mass entering /second = volume/second x density = 0.8 x 0.03 x w x d = 0.024wd
Mass leaving /second = volume/second x density = v x 0.012 x (1.08w) x (1.10d) = 0.014256vwd

0.024wd = 0.014256vwd
v = 0.024/0.014256
= 1.68m/s

Your error is

"0.08w"

"Increase by 8%" means width out = 1.08 width in, not 0.08. That would be a large decrease.

I'm not sure what you mean by this:

"velocity out = (0.8 velocity in)/(9.6*10^-4*w)"

Maybe you meant to say

velocity out = 0.8 x area in / area out

Then you would be on the right track. Find an expression for area in, which will also involve w, so they will cancel out.