One of the reaction that takes place in producing steel from iron ore is the reduction of iron (II) oxide by carbon monoxide to give iron metal and ${CO}_2$ .
$FeO\left(s\right)+CO\left(g\right)\rightleftharpoons Fe\left(s\right)+{CO}_2\left(g\right);K_p=0.265atm$ at 105 K
What are the equilibrium partial pressures of $CO$ and ${CO}_2$ at 1050 K if the initial pressure are  $P_{CO}=1.4atm$ and $P_{{CO}_2}=0.80atm$?

The initial partial pressures of CO and $C{O}_2$ are 1.40 atm and 0.8. atm respectively.
The expression for the reaction quotient is :
$Q_p=\frac{P_{C{O}_2}}{P_{CO}}=\frac{0.80}{1.4}=0.571$
As the value of the reaction quotient is greater than the value of the equilibrium constant, the reaction will move in the backward direction.
To attain the equilibrium, the partial pressure of $C{O}_2$ will decrease and the partial pressure of CO will increase.
Let p atm be the decrease in the partial pressure of $C{O}_2$ . The increase in the partial pressure of CO will be p atm.
The equilibrium partial pressure of $C{O}_2$ is 0.80-p atm.
The equilibrium partial pressure of CO is 1.4 + p atm.
The equilibrium constant is: 
$K_p=\frac{P_{C{O}_2}}{P_{CO}}=\frac{0.80-p}{1.4+p}=0.265$
$0.265(1.4+p)=0.80-p$
$0.371+0.265p=0.80-p$
$p=0.339$ atm
The equilibrium partial pressure of CO is $1.4+0.339=1.739$.
The equilibrium partial pressure of $C{O}_2$ is $0.80-0.339=0.461$ atm.