Question

what phases are present in a slow cooled 1090 stell and how much of each of these phases are present?

so the phases are cementite and pearlite and it's hypereutectoid.

The formula is hypereutectoid=cementite(6.67%)+pearlite(0.8%)

I know in a 1090 steel, there is .9%C.
I know the amount of cementite and pearlite is 100%
so if i find pearlite, then amount of cementite=100-pearlite.
the equation now becomes

.9%C=(100-pearlite)(6.67%)+pearlite(.8%)

from here, it's basic algebra and i just have to solve for pearlite.
apparently, the answer is 98.3% pearlite and cementite is 1.7%.

My question is how do I solve this equation because I keep getting 111% for pearlite.

Answer 1

The reason you did not get the right answer was because you used percentages on the right hand side and not fractions.

Let x = fraction of pearlite (1.0=100%)
and (1-x) = fraction of cementite,
using your equation:
0.9% = (1-x)*6.67% + x*0.8%
Solving gives
x=0.983 (pearlite) and
1-x=0.017 (cementite)