Let R be a partially ordered linear space (See [3]) and operator T be defined on R. We shall find conditions in order that the equation Tv+r=v has a solution for a given r∈R. We formulate the results as follows:Theorem 1.If u_o,ω_o∈R(u_o≤ω_o) are two given elements and Tu_o≥u_o, Tω_o≤ω_o, furthermore if there exists a constant N<1 such that for u, ω(u≤ω) in [u_o, ω_o] Tω-Tu≥N(ω-u),then and the equation Tv=v has a solution in [u_n, ω_n] .Theorem 2. If u_o,ω_o∈R(u_o≤ω_o) are two giren elements and Tu_o≤u_o,Tω_o≥ω_o, furthermore if there exists a constant M>1 such that for u, ω(u≤ω) in [u_o,ω_o] Tω-Tu≤M(ω-u), then u_o≤u_1≤u_2≤…≤u_n≤…≤ω_n≤…≤ω_2≤ω_1≤ω_o and the equation Tv=v has a solution in [u_n, ω_n].Let operator T_1 be increasing and operator T_2 be decreasing. Define T=T_1+T_2. We haveTheorem 3. If u_o, ω_o ∈R(u_o≤ω_o) are two given elements and T_1u_o+T_2ω_o+γ≥ u_o, T_1ω_o + T_2u_o + r≤ω_o, furthermore if T is additive,then u_o≤u_1≤u_2≤…≤u_n≤…≤ω_n≤…≤ω_2≤ω_1≤ω_o and the equation Tv+γ=v has a solution in [u_n, ω_n].