# Solution of Linear and Quadratic Equations Based on Triangular Linear Diophantine Fuzzy Numbers

Authors: Naveed Khan; Naveed Yaqoob; Mudassir Shams; Yaé Ulrich Gaba; Muhammad Riaz
Journal: Journal of Function Spaces
 (2021)

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## Abstract

This paper is introducing a new concept of triangular linear Diophantine fuzzy numbers (TLDFNs) in a generic way. We first introduce the concept of TLDFNs and then study the arithmetic operations on these numbers. We find a method for the ranking of these TLDFNs. At the end, we formulate the linear and quadratic equations of the typesA+X=B,A·X+B=C, and A·X2+B·X+C=D where the elements A,B,C, and D are TLDFNs. We provide a procedure for the solution of these equations using s,t,u,v-cut and also provide the examples.


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## Body

## 1. Introduction

In 1965, Zadeh [1] introduced a new notion of fuzzy set theory. Fuzzy set (FS) theory has been widely acclaimed as offering greater richness in applications than ordinary set theory. Zadeh popularized the concept of fuzzy sets for the first time. There is an area of FS theory, in which the arithmetic operations on FNs play an essential part known as fuzzy equations (FEQs). Fuzzy equations were studied by Sanchez [2], by using extended operations. Accordingly, a profuse number of researchers like Biacino and Lettieri [3], Buckley [4], and Wasowski [5] have studied several approaches to solve FEQs. In [6–9], Buckley and Qu introduced several techniques to evaluate the fuzzy equations of the type A·X+B=C and A·X2+B·X+C=D, where A,B,C,D, and X are fuzzy numbers (FNs). Jiang [10] studied an approach to solve simultaneous linear equations that coefficients are fuzzy numbers.Intuitionistic fuzzy sets [11, 12], neutrosophic sets [13, 14], and bipolar fuzzy sets [15] are the generalizations of the fuzzy sets. There are several mathematicians who solved linear and quadratic equations based on intuitionistic fuzzy sets, neutrosophic sets, and bipolar fuzzy sets. Banerjee and Roy [16] studied the intuitionistic fuzzy linear and quadratic equations, Chakraborty et al. [17] studied arithmetic operations on generalized intuitionistic fuzzy number and its applications to transportation problem, Rahaman et al. [18] introduced the solution techniques for linear and quadratic equations with coefficients as Cauchy neutrosophic numbers, and Akram et al. [19–23] introduced some methods for solving the bipolar fuzzy system of linear equations, also see [24–26].Linear Diophantine fuzzy set [27] is a new generalization of fuzzy set, intuitionistic fuzzy set, neutrosophic set, and bipolar fuzzy set which was introduced by Riaz and Hashmi in 2019. After the introduction of this concept, several mathematicians were attracted towards this concept and worked in this area. Riaz and others studied the decision-making problems related to linear Diophantine fuzzy Einstein aggregation operators [28], spherical linear Diophantine fuzzy sets [29], and linear Diophantine fuzzy relations [30]. Almagrabi et al. [31] introduced a new approach to q-linear Diophantine fuzzy emergency decision support system for COVID-19. Kamac [32] studied linear Diophantine fuzzy algebraic structures.Motivated by the work of Buckley and Qu [7], we solve the linear and quadratic equations with more generalized fuzzy numbers. As the linear Diophantine fuzzy set, [27] is the more generalized form of fuzzy sets so we studied the linear and quadratic equations based on linear Diophantine fuzzy numbers. In linear Diophantine fuzzy sets, we use the reference parameters, which allow us to choose the grades without any limitation; this helps us in obtaining better results.In Section2, we provided the fundamental definitions related to fuzzy sets and linear Diophantine fuzzy sets. In Section 3, we define linear Diophantine fuzzy numbers, in particular, triangular linear Diophantine fuzzy number. Also defined some basic operations on LDF numbers. In Section 4, we provide the ranking of LDF numbers, and in Section 5, we solved linear and quadratic equations based on LDF numbers.


## 2. Preliminaries and Basic Definitions

This section is devoted to review some indispensable concepts, which are very beneficial to develop the understanding of the prevalent model.Definition 1 (see [1]).
LetXbe a classical set,μMbe a function fromXto0,1.The MF (membership function)μMϑof a FS (fuzzy set)Mis defined by(1)M=ϑ,μMϑ∣ϑ∈XandμMϑ∈0,1.Definition 2 (see [33]).
LetMbe a fuzzy subset of universal setX. Then,Mis called convex FS if∀r,s∈Xandλ∈0,1we have(2)μMλr+1−λs≥minμMr,μMs.Definition 3 (see [1]).
A fuzzy setMis said to be normalized ifhM=1.Definition 4.
Anα-level set of a FSMis defined as(3)Mα=ϑ∈X:μMϑ≥αforeachα∈0,1.Definition 5 (see [33]).
A fuzzy subsetMdefined on a setℝ(of real numbers) is said to be a FN (fuzzy number) ifMsatisfies the following axioms:(a)
M is continuous: μMt is a continuous function from ℝ⟶0,1(b)
M is normalized: there exists t∈ℝ such that μMt=1(c)
Convexity ofM: i.e., ∀t,u,w∈ℝ, if t≤u≤w, then μMu≥minμMt,μMw(d)
Boundness of support: i.e.,∃S∈ℝ and ∀t∈ℝ, if ∣t∣≥S, then μMt=0We denote the set of all FNs byFnsℝ.Now, we study the idea of LDFSs (linear Diophantine fuzzy sets) and their fundamental operations.Definition 6 (see [27]).
LetXbe the universe. A LDFS£RonXis defined as follows:(4)£R=ϑ,MRτϑ,NRνϑ,αϑ,βϑ:ϑ∈Xwhere MRτϑ,NRνϑ,αϑ,βϑ∈0,1 such that
(5)0≤αϑMRτϑ+βϑNRνϑ≤1,∀ϑ∈X,0≤αϑ+βϑ≤1.
The hesitation part can be written as(6)ξπR=1−αϑMRτϑ+βϑNRνϑ,where ξ is the reference parameter.
We write in short£R=MRτ,NRν,α,β or £R=MRτ,NRν,α,β for
(7)£R=ϑ,MRτϑ,NRνϑ,αϑ,βϑ:ϑ∈X.Definition 7 (see [27]).
An absolute LDFS onXcan be written as(8)1£R=ϑ,1,0,1,0:ϑ∈X,and empty or null LDFS can be expressed as
(9)0£R=ϑ,0,1,0,1:ϑ∈X.Definition 8 (see [27]).
Let£R=MRτ,NRν,α,βand£P=MPτ,NPν,γ,δbe two LDFSs on the reference setXandϑ∈X. Then,(i)
£Rc=NRν,MRτ,β,α(ii)
£R=£P⇔MRτ=MPτ,NRν=NPν,α=γ,β=δ(iii)
£R⊆£P⇔MRτϑ≤MPτϑ,NRνϑ≥NPνϑ,αϑ≤γϑ,βϑ≥δϑ(iv)
£R∪£P=MR∪Pτ,NR∩Pν,α∨γ,β∧δ(v)
£R∩£P=MR∩Pτ,NR∪Pν,α∧γ,β∨δwhere
(10)MR∪Pτϑ=MRτϑ∨MPτϑ,MR∩Pτϑ=MRτϑ∧MPτϑ,NR∩Pνϑ=NRνϑ∧NPνϑ,NR∪Pνϑ=NRϑ∨NPνϑ.Definition 9 (see [27]).
Let£R=ϑ,MRτϑ,NRνϑ,αϑ,βϑ:ϑ∈Xbe an LDFS. For any constantss,t,u,v∈0,1such that0≤su+tv≤1with0≤u+v≤1,define thes,t,u,v-cut of£Ras follows:(11)£Ru,vs,t=ϑ∈X:MRτϑ≥s,NRνϑ≤t,αϑ≥u,βϑ≤v.


## 3. Triangular LDF Numbers

Here, in this section, we provide definitions and arithmetic operations on LDF numbers (LDFNs).Definition 10.
A LDF number£Ris(i)
a LDF fuzzy subset of the real lineℝ(ii)
normal, i.e., there is anyϑ0∈ℝ such that MRτϑ0=1,NRνϑ0=0,αϑ0=1,βϑ0=0(iii)
convex for the membership functionsMRτ and α, i.e.,(12)MRτλϑ1+1−λϑ2≥minMRτϑ1,MRτϑ2∀ϑ1,ϑ2∈ℝ,λ∈0,1,αλϑ1+1−λϑ2≥minαϑ1,αϑ2∀ϑ1,ϑ2∈ℝ,λ∈0,1(iv)
concave for the nonmembership functionsNRν and β, i.e.,(13)NRνλϑ1+1−λϑ2≤maxNRνϑ1,NRνϑ2∀ϑ1,ϑ2∈ℝ,λ∈0,1,βλϑ1+1−λϑ2≤maxβϑ1,βϑ2∀ϑ1,ϑ2∈ℝ,λ∈0,1.We now provide the 4 types of triangular LDF numbers.Definition 11.
Let£Rbe a LDFS onℝwith the following membership functions (MRτandα) and nonmembership functions (NRνandβ)(14)MRτx=x−ϑ1ϑ3−ϑ1,ϑ1≤x≤ϑ3,ϑ5−xϑ5−ϑ3,ϑ3≤x≤ϑ5,0,otherwise,NRνx=ϑ3−xϑ3−ϑ2,ϑ2≤x≤ϑ3,x−ϑ3ϑ4−ϑ3,ϑ3≤x≤ϑ4,0,otherwise,αx=x−ϑ2′ϑ3′−ϑ2′,ϑ2′≤x≤ϑ3′,ϑ4′−xϑ4′−ϑ3′,ϑ3′≤x≤ϑ4′,0,otherwise,βx=ϑ3′−xϑ3′−ϑ1′,ϑ1′≤x≤ϑ3′,x−ϑ3′ϑ5′−ϑ3′,ϑ3′≤x≤ϑ5′,0,otherwise,where ϑ1′≤ϑ2′≤ϑ3′≤ϑ4′≤ϑ5′ for all x∈ℝ. Then, £R is called
(i)
a triangular LDFN of type-1 ifϑ3=ϑ3′ and ϑ1≤ϑ2≤ϑ3≤ϑ4≤ϑ5(ii)
a triangular LDFN of type-2 ifϑ3≠ϑ3′′ and ϑ1≤ϑ2≤ϑ3≤ϑ4≤ϑ5(iii)
a triangular LDFN of type-3 ifϑ3=ϑ3′ and ϑ2≤ϑ1≤ϑ3≤ϑ5≤ϑ4(iv)
a triangular LDFN of type-4 ifϑ3≠ϑ3′ and ϑ2≤ϑ1≤ϑ3≤ϑ5≤ϑ4Throughout the paper, we consider only triangular LDFN of type-1 and we write this type as triangular LDFN (TLDFN). This TLDFN is denoted by(15)£RTLDFN=ϑ1′,ϑ2′,ϑ3,ϑ4′,ϑ5′.ϑ1,ϑ2,ϑ3,ϑ4,ϑ5,The figure ofϑ1,ϑ2,ϑ3,ϑ4,ϑ5 is shown in Figure 1.Figure 1
The figure ofϑ1,ϑ2,ϑ3,ϑ4,ϑ5.The figure ofϑ1′,ϑ2′,ϑ3,ϑ4′,ϑ5′ is shown in Figure 2.Figure 2
The figure ofϑ1′,ϑ2′,ϑ3,ϑ4′,ϑ5′.The figure of£RTLDFN is shown in Figure 3.Remark 12.
If we takeϑ1′=ϑ2′=ϑ1=ϑ2andϑ4′=ϑ5′=ϑ1=ϑ2,then both type-1 and type-3 become the same.Definition 13.
Consider a TLDFN£RTLDFN=ϑ1′,ϑ2′,ϑ3,ϑ4′,ϑ5′ϑ1,ϑ2,ϑ3,ϑ4,ϑ5.Then,(i)
s-cut set of £RTLDFN is a crisp subset of ℝ, which is defined as follows:(16)£RTLDFNs=x∈X:MRτx≥s=MRτs¯,MRτs¯=ϑ1+sϑ3−ϑ1,ϑ5−sϑ5−ϑ3,(ii)
t-cut set of £RTLDFN is a crisp subset of ℝ, which is defined as follows:(17)£RTLDFNt=x∈X:NRνx≤t=NRνt¯,NRνt¯=ϑ3−tϑ3−ϑ2,ϑ3+tϑ4−ϑ3,(iii)
u-cut set of £RTLDFN is a crisp subset of ℝ, which is defined as follows:(18)£RTLDFNu=x∈X:αx≥u=αu¯,αu¯=ϑ2′+uϑ3−ϑ2′,ϑ4′−uϑ4′−ϑ3,(iv)
v-cut set of £RTLDFN is a crisp subset of ℝ, which is defined as follows:(19)£RTLDFNv=x∈X:βx≤v=βv¯,βv¯=ϑ3−vϑ3−ϑ1′,ϑ3+vϑ5′−ϑ3.Figure 3
The figure of£RTLDFN.We can denote thes,t,u,v-cut of £RTLDFN=ϑ1,ϑ2,ϑ3,ϑ4,ϑ5ϑ1′,ϑ2′,ϑ3,ϑ4′,ϑ5′ by
(20)£RTLDFNu,vs,t=MRτs¯,MRτs¯,NRνt¯,NRνt¯,αu¯,αu¯,βv¯,βv¯.We denote the set of all TLDFN onℝ by £RTLDFNℝ. The arithmetic operations based on extension principle are defined as follows.Definition 14.
Let£R=MRτ,NRν,α,βand£P=MPτ,NPν,γ,δbe two TLDFN onℝ. Then,(i)
£R+£P=supt=x+yminMRτx,MPτyinft=x+ymaxNRνx,NPνysupt=x+yminαx,γyinft=x+ymaxβx,δy(ii)
£R−£P=supt=x−yminMRτx,MPτyinft=x−ymaxNRνx,NPνysupt=x−yminαx,γyinft=x−ymaxβx,δy(iii)
£R×£P=supt=x×yminMRτx,MPτyinft=x×ymaxNRνx,NPνysupt=x×yminαx,γyinft=x×ymaxβx,δy(iv)
£R÷£P=supt=x÷yminMRτx,MPτyinft=x÷ymaxNRνx,NPνysupt=x÷yminαx,γyinft=x÷ymaxβx,δyDefinition 15.
A TLDFN£RTLDFN=ϑ1,ϑ2,ϑ3,ϑ4,ϑ5ϑ1′,ϑ2′,ϑ3,ϑ4′,ϑ5′is said to be positive if and only ifϑ1≥0andϑ1′≥0.Definition 16.
Two TLDFNs£RTLDFN=ϑ1,ϑ2,ϑ3,ϑ4,ϑ5ϑ1′,ϑ2′,ϑ3,ϑ4′,ϑ5′andξRTLDFN=δ1,δ2,δ3,δ4,δ5δ1′,δ2′,δ3,δ4′,δ5′are said to be equal if and only ifϑ1=δ1,ϑ2=δ2,ϑ3=δ3,ϑ4=δ4,ϑ5=δ5,ϑ1′=δ1′,ϑ2′=δ2′,ϑ4′=δ4′,andϑ5′=δ5′.We now define the arithmetic operations on TLDFNs using the concept of interval arithmetic.Definition 17.
Consider two positive TLDFNs£RTLDFN=ϑ1′,ϑ2′,ϑ3,ϑ4′,ϑ5′ϑ1,ϑ2,ϑ3,ϑ4,ϑ5andξRTLDFN=δ1′,δ2′,δ3,δ4′,δ5′δ1,δ2,δ3,δ4,δ5,then,(i)
£RTLDFN+ξRTLDFN=ϑ1′+δ1′,ϑ2′+δ2′,ϑ3+δ3,ϑ4′+δ4′,ϑ5′+δ5′ϑ1+δ1,ϑ2+δ2,ϑ3+δ3,ϑ4+δ4,ϑ5+δ5(ii)
£RTLDFN−ξRTLDFN=ϑ1′−δ5′,ϑ2′−δ4′,ϑ3−δ3,ϑ4′−δ2′,ϑ5′−δ1′ϑ1−δ5,ϑ2−δ4,ϑ3−δ3,ϑ4−δ2,ϑ5−δ1(iii)
£RTLDFN×ξRTLDFN=ϑ1′δ1′,ϑ2′δ2′,ϑ3δ3,ϑ4′δ4′,ϑ5′δ5′ϑ1δ1,ϑ2δ2,ϑ3δ3,ϑ4δ4,ϑ5δ5(iv)
£RTLDFN÷ξRTLDFN=ϑ1′/δ5′,ϑ2′/δ4′,ϑ3/δ3,ϑ4′/δ2′,ϑ5′/δ1′ϑ1/δ5,ϑ2/δ4,ϑ3/δ3,ϑ4/δ2,ϑ5/δ1(v)
k×£RTLDFN=kϑ1′,kϑ2′,kϑ3,kϑ4′,kϑ5′kϑ1,kϑ2,kϑ3,kϑ4,kϑ5ifk>0kϑ5′,kϑ4′,kϑ3,kϑ2′,kϑ1′kϑ5,kϑ4,kϑ3,kϑ2,kϑ1ifk<0


## 4. Ranking Function of TLDFNs

There are many methods for defuzzification such as the centroid method, mean of interval method, and removal area method. In this paper, we have used the concept of the mean of interval method to find the value of the membership and nonmembership function of TLDFN.Consider a TLDFN(21)£RTLDFN=ϑ1′,ϑ2′,ϑ3,ϑ4′,ϑ5′.ϑ1,ϑ2,ϑ3,ϑ4,ϑ5,Thes,t,u,v-cut of £RTLDFN is
(22)£Ru,vs,t=ϑ∈X:MRτϑ≥s,NRνϑ≤t,αϑ≥u,βϑ≤v,where(23)MRτs¯=ϑ1+sϑ3−ϑ1,MRτs¯=ϑ5−sϑ5−ϑ3,NRνt¯=ϑ3−tϑ3−ϑ2,NRνt¯=ϑ3+tϑ4−ϑ3,αu¯=ϑ2′+uϑ3−ϑ2′,αu¯=ϑ4′−uϑ4′−ϑ3,βv¯=ϑ3−vϑ3−ϑ1′,βv¯=ϑ3+vϑ5′−ϑ3.Now, by the mean ofs,t,u,v-cut method, the representation of membership functions is
(24)RMRτ£RTLDFN=12∫01MRτs¯+MRτs¯ds=12∫01ϑ1+sϑ3−ϑ1+ϑ5−sϑ5−ϑ3ds=12ϑ1+12ϑ3−ϑ1+ϑ5−12ϑ5−ϑ3=ϑ1+2ϑ3+ϑ54,Rα£RTLDFN=12∫01αu¯+αu¯ds=12∫01ϑ2′+uϑ3−ϑ2′+ϑ4′−uϑ4′−ϑ3du=12ϑ2′+12ϑ3−ϑ2′+ϑ4′−12ϑ4′−ϑ3=ϑ2′+2ϑ3+ϑ4′4.Now, by the mean ofs,t,u,v-cut method, the representation of nonmembership functions is
(25)RNRν£RTLDFN=12∫01NRνt¯+NRνt¯dt=12∫01ϑ3−tϑ3−ϑ2+ϑ3+tϑ4−ϑ3dt=122ϑ3−12ϑ3−ϑ2+12ϑ4−ϑ3=ϑ2+2ϑ3+ϑ44,Rβ£RTLDFN=12∫01βv¯+βv¯dv=12∫01ϑ3−vϑ3−ϑ1′+ϑ3+vϑ5′−ϑ3′dv=122ϑ3−12ϑ3−ϑ1′+12ϑ5′−ϑ3=ϑ1′+2ϑ3+ϑ5′4.Now,(26)R£RTLDFN=RMRτ£RTLDFN+Rα£RTLDFN+RNRν£RTLDFN+Rβ£RTLDFN4=ϑ1+2ϑ3+ϑ5/4+ϑ2′+2ϑ3+ϑ4′/4+ϑ2+2ϑ3+ϑ4/4+ϑ1′+2ϑ3+ϑ5′/44=ϑ32+ϑ1+ϑ2+ϑ4+ϑ5+ϑ1′+ϑ2′+ϑ4′+ϑ5′16.Consider two positive TLDFNs£RTLDFN=ϑ1,ϑ2,ϑ3,ϑ4,ϑ5ϑ1′,ϑ2′,ϑ3,ϑ4′,ϑ5′ and ξRTLDFN=δ1,δ2,δ3,δ4,δ5δ1′,δ2′,δ3,δ4′,δ5′, then
(i)
£RTLDFN≺ξRTLDFN iff R£RTLDFN<RξRTLDFN(ii)
£RTLDFN≻ξRTLDFN iff R£RTLDFN>RξRTLDFN(iii)
£RTLDFN=ξRTLDFN iff R£RTLDFN=RξRTLDFN


## 5. Solution of LDF Equations

### 5.1. Solution ofA+X=B by Using the Method of s,t,u,v-Cut

LetA,B, and X be the LDFNs and let A=ϑ1,ϑ2,ϑ3,ϑ4,ϑ5ϑ1′,ϑ2′,ϑ3,ϑ4′,ϑ5′ and B=δ1,δ2,δ3,δ4,δ5δ1′,δ2′,δ3,δ4′,δ5′. Then,
(27)A+X=Bis a LDF equation (LDFE). Let X≈x1,x2,x3,x4,x5x1′,x2′,x3,x4′,x5′. Then, X=B−A in general is not the solution of Equation (27).Let(28)Au,vs,t=MAτs¯,MAτs¯,NAνt¯,NAνt¯,αAu¯,αAu¯,βAv¯,βAv¯,Bu,vs,t=MBτs¯,MBτs¯,NBνt¯,NBνt¯,αBu¯,αBu¯,βBv¯,βBv¯,Xu,vs,t=MXτs¯,MXτs¯,NXνt¯,NXνt¯,αXu¯,αXu¯,βXv¯,βXv¯represent the s,t,u,v-cuts of A,B, and X, respectively, in the given (27). Substituting these into Equation (27), we get
(29)Au,vs,t+Xu,vs,t=Bu,vs,t.By comparing thes,t,u,v-cuts of A,B, and X, we get
(30)MAτs¯,MAτs¯+MXτs¯,MXτs¯=MBτs¯,MBτs¯,NAνt¯,NAνt¯+NXνt¯,NXνt¯=NBνt¯,NBνt¯,αAu¯,αAu¯+αXu¯,αXu¯=αBu¯,αBu¯,βAv¯,βAv¯+βXv¯,βXv¯=βBv¯,βBv¯.Now,(31)MXτs¯=MBτs¯−MAτs¯,MXτs¯=MBτs¯−MAτs¯,NXνt¯=NBνt¯−NAνt¯,NXνt¯=NBνt¯−NAνt¯,αXu¯=αBu¯−αAu¯,αXu¯=αBu¯−αAu¯,βXv¯=βBv¯−βAv¯,βXv¯=βBv¯−βAv¯.Then, the solution of the equationA+X=B exists iff
(1)
MXτs¯ is monotonically increasing in 0≤s≤1(2)
MXτs¯ is monotonically decreasing in 0≤s≤1(3)
NXνt¯ is monotonically decreasing in 0≤t≤1(4)
NXνt¯ is monotonically increasing in 0≤t≤1(5)
αXu¯ is monotonically increasing in 0≤u≤1(6)
αXu¯ is monotonically decreasing in 0≤u≤1(7)
βXv¯ is monotonically decreasing in 0≤v≤1(8)
βXv¯ is monotonically increasing in 0≤v≤1(9)
MXτ1¯=MXτ1¯=NXν0¯=NXν0¯=αX1¯=αX1¯=βX0¯=βX0¯.Example 1.
Consider the equationA+X=B,where(32)A=3,5,7,10,15,2,6,7,8,11,B=1,6,11,15,24,3,9,11,13,22.
Thes,t,u,v-cuts of A,B, and X are
(33)Au,vs,t=3+4s,15−8s,7−2t,7+3t,6+u,8−u,7−5v,7+4v,Bu,vs,t=1+10s,24−13s,11−5t,11+4t,9+2u,13−2u,11−8v,11+11v,Xu,vs,t=MXτs¯,MXτs¯,NXνt¯,NXνt¯,αXu¯,αXu¯,βXv¯,βXv¯,respectively. The s,t,u,v-cut equation is
(34)Au,vs,t+Xu,vs,t=Bu,vs,t.
By comparing thes,t,u,v-cuts of A,B, and X, we get
(35)MXτs¯=MBτs¯−MAτs¯=−2+6s,MXτs¯=MBτs¯−MAτs¯=9−5s,NXνt¯=NBνt¯−NAνt¯=4−3t,NXνt¯=NBνt¯−NAνt¯=4+t,αXu¯=αBu¯−αAu¯=3+u,αXu¯=αBu¯−αAu¯=5−u,βXv¯=βBv¯−βAv¯=4−3v,βXv¯=βBv¯−βAv¯=4+7v.
It is easy to see thatMXτs¯,NXνt¯,αXu¯, and βXv¯ are increasing and MXτs¯,NXνt¯,αXu¯, and βXv¯ are decreasing in 0≤s,t,u,v≤1. Also,
(36)MXτ1¯=MXτ1¯=NXν0¯=NXν0¯=αX1¯=αX1¯=βX0¯=βX0¯=4.
This shows that the solution ofA+X=B exists with s,t,u,v-cut. The solution is
(37)X=1,3,4,5,11.−2,1,4,5,9,
The solution in continuous form is(38)MRτx=2+x6,−2≤x≤4,9−x5,4≤x≤9,0,otherwise,NRνx=4−x3,1≤x≤4,−4+x,4≤x≤5,0,otherwise,αx=−3+x,3≤x≤4,5−x,4≤x≤5,0,otherwise,βx=4−x3,1≤x≤4,−4+x7,4≤x≤11,0,otherwise.
The graph of the solution is given in Figure4.Figure 4
The graph of the solution obtained in Example1.


### 5.2. Solution ofA·X+B=C by Using the Method of s,t,u,v-Cut

LetA,B,C, and X be the LDFNs and let A=ϑ1,ϑ2,ϑ3,ϑ4,ϑ5ϑ1′,ϑ2′,ϑ3,ϑ4′,ϑ5′, B=δ1,δ2,δ3,δ4,δ5δ1′,δ2′,δ3,δ4′,δ5′, and C=η1,η2,η3,η4,η5η1′,η2′,η3,η4′,η5′. Then,
(39)A·X+B=Cis a LDF equation (LDFE). Let X≈x1,x2,x3,x4,x5x1′,x2′,x3,x3′,x5′. Then, X=C−B/A in general is not the solution of Equation (39).Let(40)Au,vs,t=MAτs¯,MAτs¯,NAνt¯,NAνt¯,αAu¯,αAu¯,βAv¯,βAv¯,Bu,vs,t=MBτs¯,MBτs¯,NBνt¯,NBνt¯,αBu¯,αBu¯,βBv¯,βBv¯,Cu,vs,t=MCτs¯,MCτs¯,NCνt¯,NCνt¯,αCu¯,αCu¯,βCv¯,βCv¯,Xu,vs,t=MXτs¯,MXτs¯,NXνt¯,NXνt¯,MXτs¯,MXτs¯,NXνt¯,NXνt¯represent the s,t,u,v-cuts of A,B,C, and X, respectively, in the given (39). Substituting these into Equation (39), we get
(41)Au,vs,t·Xu,vs,t+Bu,vs,t=Cu,vs,t.By comparing thes,t,u,v-cuts of A,B,C, and X, we get
(42)MAτs¯,MAτs¯·MXτs¯,MXτs¯+MBτs¯,MBτs¯=MCτs¯,MCτs¯,NAνt¯,NAνt¯·NXνt¯,NXνt¯+NBνt¯,NBνt¯=NCνt¯,NCνt¯,αAu¯,αAu¯·αXu¯,αXu¯+αBu¯,αBu¯=αCu¯,αCu¯,βAv¯,βAv¯·βXv¯,βXv¯+βBv¯,βBv¯=βCv¯,βCv¯.Now,(43)MXτs¯=MCτs¯−MBτs¯MAτs¯,MXτs¯=MCτs¯−MBτs¯MAτs¯,NXνt¯=NCνt¯−NBνt¯NAνt¯,NXνt¯=NCνt¯−NBνt¯NAνt¯,αXu¯=αCu¯−αBu¯αAu¯,αXu¯=αCu¯−αBu¯αAu¯,βXv¯=βCv¯−βBv¯βAv¯,βXv¯=βCv¯−βBv¯βAv¯.Then, the solution of the equationA·X+B=C exists iff
(1)
MXτs¯ is monotonically increasing in 0≤s≤1(2)
MXτs¯ is monotonically decreasing in 0≤s≤1(3)
NXνt¯ is monotonically decreasing in 0≤t≤1(4)
NXνt¯ is monotonically increasing in 0≤t≤1(5)
αXu¯ is monotonically increasing in 0≤u≤1(6)
αXu¯ is monotonically decreasing in 0≤u≤1(7)
βXv¯ is monotonically decreasing in 0≤v≤1(8)
βXv¯ is monotonically increasing in 0≤v≤1(44)MXτ1¯=MXτ1¯=NXν0¯=NXν0¯=αX1¯=αX1¯=βX0¯=βX0¯.Example 2.
Consider the equationA·X+B=C,where(45)A=1,3,5,6,11,1,2,5,7,10,B=4,5,8,11,19,4,6,8,10,15,C=1,5,18,29,85.1,4,18,38,65,
Thes,t,u,v-cuts of A,B,C, and X are
(46)Au,vs,t=3+2u,6−u,5−4v,5+6v,1+4s,10−5s,5−3t,5+2t,Bu,vs,t=5+3u,11−3u,8−4v,8+11v,4+4s,15−7s,8−2t,8+2t,Cu,vs,t=5+13u,29−11u,18−17v,18+67v,1+17s,65−47s,18−14t,18+20t,Xu,vs,t=αXu¯,αXu¯,βXv¯,βXv¯,MXτs¯,MXτs¯,NXνt¯,NXνt¯,respectively. The s,t,u,v-cut equation is
(47)Au,vs,t·Xu,vs,t+Bu,vs,t=Cu,vs,t.
By comparing thes,t,u,v-cuts of A,B,C, and X, we get
(48)MXτs¯=1+17s−4+4s1+4s=−3+13s1+4s,MXτs¯=65−47s−15−7s10−5s=10−8s2−s,NXνt¯=18−14t−8−2t5−3t=10−12t5−3t,NXνt¯=18+20t−8+2t5+2t=10+18t5+2t,αXu¯=5+13u−5+3u3+2u=10u3+2u,αXu¯=29−11u−11−3u6−u=18−8u6−u,βXv¯=18−17v−8−4v5−4v=10−13v5−4v,βXv¯=18+67v−8+11v5+6v=10+56v5+6v.
It is easy to see thatMXτs¯,NXνt¯,αXu¯, and βXv¯ are increasing and MXτs¯,NXνt¯,αXu¯, and βXv¯ are decreasing in 0≤s,t,u,v≤1. Also,
(49)MXτ1¯=MXτ1¯=NXν0¯=NXν0¯=αX1¯=αX1¯=βX0¯=βX0¯=2.
This shows that the solution ofA·X+B=C exists with s,t,u,v-cut. The solution is
(50)X=−3,0,2,3,6.−3,−1,2,4,5,
The solution in continuous form is(51)MRτx=3+x13−4x,−3≤x≤2,−10+2x−8+x,2≤x≤5,0,otherwise,NRνx=−10+5x−12+3x,−1≤x≤2,−10+5x18−2x,2≤x≤4,0,otherwise,αx=3x10−2x,0≤x≤2,−18+6x−8+x,2≤x≤3,0,otherwise,βx=−10+5x−13+4x,−3≤x≤2,−10+5x56−6x,2≤x≤6,0,otherwise.
The graph of the solution is given in Figure5.Figure 5
The graph of the solution obtained in Example2.


### 5.3. Solution ofA·X2+B·X+C=D by Using the Method of α-Cut.

LetA,B,C,D, and X be the LDFNs and let
(52)A=ϑ1,ϑ2,ϑ3,ϑ4,ϑ5,ϑ1′,ϑ2′,ϑ3,ϑ4′,ϑ5′,B=δ1,δ2,δ3,δ4,δ5,δ1′,δ21,δ3,δ4′,δ5′,C=η1,η2,η3,η4,η5,η1′,η2′,η3,η4′,η5′,D=ζ1,ζ2,ζ3,ζ4,ζ5,ζ1′,ζ2′,ζ3,ζ4′,ζ5′.Then,(53)A·X2+B·X+C=Dis a LDF equation (LDFE). Let X≈x1,x2,x3,x4,x5x1′,x2′,x3,x4′,x5′. Let
(54)Au,vs,t=MAτs¯,MAτs¯,NAνt¯,NAνt¯,αAu¯,αAu¯,βAv¯,βAv¯,Bu,vs,t=MBτs¯,MBτs¯,NBνt¯,NBνt¯,αBu¯,αBu¯,βBv¯,βBv¯,Cu,vs,t=MCτs¯,MCτs¯,NCνt¯,NCνt¯,αCu¯,αCu¯,βCv¯,βCv¯,Du,vs,t=MDτs¯,MDτs¯,NDνt¯,NDνt¯,αDu¯,αDu¯,βDv¯,βDv¯,Xu,vs,t=MXτs¯,MXτs¯,NXνt¯,NXνt¯,αXu¯,αXu¯,βXv¯,βXv¯represent the s,t,u,v-cuts of A,B,C,D, and X, respectively, in the given (53). Substituting these into Equation (53), we get
(55)Au,vs,t·Xu,vs,t2+Bu,vs,t·Xu,vs,t+Cu,vs,t=Du,vs,t.By comparing thes,t,u,v-cuts of A,B,C,D, and X, we get
(56)MAτs¯,MAτs¯·MXτs¯,MXτs¯2+MBτs¯,MBτs¯·MXτs¯,MXτs¯=MDτs¯,MDτs¯+MCτs¯,MCτs¯,NAνt¯,NAνt¯·NXνt¯,NXνt¯2+NBνt¯,NBνt¯·NXνt¯,NXνt¯=NDνt¯,NDνt¯+NCνt¯,NCνt¯,αAu¯,αAu¯·αXu¯,αXu¯2+αBu¯,αBu¯·αXu¯,αXu¯=αDu¯,αDu¯+αCu¯,αCu¯,βAv¯,βAv¯·βXv¯,βXv¯2+βBv¯,βBv¯·βXv¯,βXv¯=βDv¯,βDv¯+βCv¯,βCv¯.Now,(57)MXτs¯=−MBτs¯±MBτs¯2−4MAτs¯MCτs¯−MDτs¯2MAτs¯,MXτs¯=−MBτs¯±MBτs¯2−4MAτs¯MCτs¯−MDτs¯2MAτs¯,NXνt¯=−NBνt¯±NBνt¯2−4NAνt¯NCνt¯−NDνt¯2NAνt¯,NXνt¯=−NBνt¯±NBνt¯2−4NAνt¯NCνt¯−NDνt¯2NAνt¯,αXu¯=−αBu¯±αBu¯2−4αAu¯αCu¯−αDu¯2αAu¯,αXu¯=−αBu¯±αBu¯2−4αAu¯αCu¯−αDu¯2αAu¯,βXv¯=−βBv¯±βBv¯2−4βAv¯βCv¯−βDv¯2βAv¯,βXv¯=−βBv¯±βBv¯2−4βAv¯βCv¯−βDv¯2βAv¯.Then, the solution of the equationA·X2+B·X+C=D exists iff
(1)
MXτs¯ is monotonically increasing in 0≤s≤1(2)
MXτs¯ is monotonically decreasing in 0≤s≤1(3)
NXνt¯ is monotonically decreasing in 0≤t≤1(4)
NXνt¯ is monotonically increasing in 0≤t≤1(5)
αXu¯ is monotonically increasing in 0≤u≤1(6)
αXu¯ is monotonically decreasing in 0≤u≤1(7)
βXv¯ is monotonically decreasing in 0≤v≤1(8)
βXv¯ is monotonically increasing in 0≤v≤1(9)
MXτ1¯=MXτ1¯=NXν0¯=NXν0¯=αX1¯=αX1¯=βX0¯=βX0¯Example 3.
Consider the equationA·X2+B·X+C=D,where(58)A=4,5,7,9,10,2,6,7,8,13,B=2,4,5,6,8,4,4,5,6,7,C=1,2,4,5,7,1,3,4,5,7,D=1,3,6,8,12,1,4,6,8,11.Thes,t-cuts of A,B,C,D, and X are given in Table 1.Table 1
s,t-cuts of A,B,C,D, and X.
s,t-cutsXABCDC−Dϑ1+sϑ3−ϑ1MXτs¯4+3s2+3s1+3s1+5s−2sϑ5−sϑ5−ϑ3MXτs¯10−3s8−3s7−3s12−6s−5+3sϑ3−tϑ3−ϑ2NXνt¯7−2t5−t4−2t6−3t−2+tϑ3+tϑ4−ϑ3NXνt¯7+2t5+t4+t6+2t−2−tBy comparing thes,t-cuts of A,B,C,D, and X, we get
(59)MXτs¯=−2+3s+2+3s2−44+3s−2s24+3s,MXτs¯=−8−3s+8−3s2−410−3s−5+3s210−3s,NXνt¯=−5−t+5−t2−47−2t−2+t27−2t,NXνt¯=−5+t+5+t2−47+2t−2−t27+2t.The graph obtained bys,t-cut is shown in Figure 6.Figure 6
Solution bys,t-cut.Theu,v-cuts of A,B,C,D, and X are given in Table 2.Table 2
u,v-cuts of A,B,C,D, and X.
u,v-cutsXABCDC−Dϑ2′+uϑ3−ϑ2′αu¯6+u4+u3+u4+2u−1−uϑ4′−uϑ4′−ϑ3αu¯8−u6−u5−u8−2u−3+uϑ3−vϑ3−ϑ1′βv¯7−5v5−v4−3v6−5v−2+2vϑ3+vϑ5′−ϑ3βv¯7+6v5+2v4+3v6+5v−2−2vBy comparing theu,v-cuts of A,B,C,D, and X, we get
(60)αu¯=−4+u+4+u2−46+u−1−u26+u,αu¯=−6−u+6−u2−48−u−3+u28−u,βv¯=−5−v+5−v2−47−5v−2+2v27−5v,βv¯=−5+2v+5+2v2−47+6v−2−2v27+6v.The graph obtained byu,v-cut is shown in Figure 7.Figure 7
Solution byu,v-cut.It is easy to see thatMXτs¯,NXνt¯,αXu¯, and βXv¯ are increasing and MXτs¯,NXνt¯,αXu¯, and βXv¯ are decreasing in 0≤s,t,u,v≤1. Also,
(61)MXτ1¯=MXτ1¯=NXν0¯=NXν0¯=αX1¯=αX1¯=βX0¯=βX0¯=0.2857.This shows that the solution ofA·X2+B·X+C=D exists with s,t,u,v-cut. The solution is
(62)X=0,−2+106,27,−3+338,−7+257260,15,27,13,−4+6610=0,0.1937,0.2857,0.3431,0.3474.0,0.2,0.2857,0.3333,0.4124,The solution in continuous form is(63)MRτx=−2x2x+13x2+3x−2,0≤x≤0.2857,10x2+8x−53x2+x−1,0.2857≤x≤0.4124,0,otherwise,NRνx=7x−22x−1,0.2≤x≤0.2857,−7x−22x−1,0.2857≤x≤0.3333,0,otherwise,αx=−6x2−4x+1x2+x−1,0.1937≤x≤0.2857,8x2+6x−3x2+x−1,0.2857≤x≤0.3431,0,otherwise,βx=7x2+5x−25x2+x−2,0≤x≤0.2857,−7x2−5x+223x2+x−1,0.2857≤x≤0.3474,0,otherwise.The graph of the solution is given in Figure8.Figure 8
The graph of the solution obtained in Example3.


## 5.1. Solution ofA+X=B by Using the Method of s,t,u,v-Cut

LetA,B, and X be the LDFNs and let A=ϑ1,ϑ2,ϑ3,ϑ4,ϑ5ϑ1′,ϑ2′,ϑ3,ϑ4′,ϑ5′ and B=δ1,δ2,δ3,δ4,δ5δ1′,δ2′,δ3,δ4′,δ5′. Then,
(27)A+X=Bis a LDF equation (LDFE). Let X≈x1,x2,x3,x4,x5x1′,x2′,x3,x4′,x5′. Then, X=B−A in general is not the solution of Equation (27).Let(28)Au,vs,t=MAτs¯,MAτs¯,NAνt¯,NAνt¯,αAu¯,αAu¯,βAv¯,βAv¯,Bu,vs,t=MBτs¯,MBτs¯,NBνt¯,NBνt¯,αBu¯,αBu¯,βBv¯,βBv¯,Xu,vs,t=MXτs¯,MXτs¯,NXνt¯,NXνt¯,αXu¯,αXu¯,βXv¯,βXv¯represent the s,t,u,v-cuts of A,B, and X, respectively, in the given (27). Substituting these into Equation (27), we get
(29)Au,vs,t+Xu,vs,t=Bu,vs,t.By comparing thes,t,u,v-cuts of A,B, and X, we get
(30)MAτs¯,MAτs¯+MXτs¯,MXτs¯=MBτs¯,MBτs¯,NAνt¯,NAνt¯+NXνt¯,NXνt¯=NBνt¯,NBνt¯,αAu¯,αAu¯+αXu¯,αXu¯=αBu¯,αBu¯,βAv¯,βAv¯+βXv¯,βXv¯=βBv¯,βBv¯.Now,(31)MXτs¯=MBτs¯−MAτs¯,MXτs¯=MBτs¯−MAτs¯,NXνt¯=NBνt¯−NAνt¯,NXνt¯=NBνt¯−NAνt¯,αXu¯=αBu¯−αAu¯,αXu¯=αBu¯−αAu¯,βXv¯=βBv¯−βAv¯,βXv¯=βBv¯−βAv¯.Then, the solution of the equationA+X=B exists iff
(1)
MXτs¯ is monotonically increasing in 0≤s≤1(2)
MXτs¯ is monotonically decreasing in 0≤s≤1(3)
NXνt¯ is monotonically decreasing in 0≤t≤1(4)
NXνt¯ is monotonically increasing in 0≤t≤1(5)
αXu¯ is monotonically increasing in 0≤u≤1(6)
αXu¯ is monotonically decreasing in 0≤u≤1(7)
βXv¯ is monotonically decreasing in 0≤v≤1(8)
βXv¯ is monotonically increasing in 0≤v≤1(9)
MXτ1¯=MXτ1¯=NXν0¯=NXν0¯=αX1¯=αX1¯=βX0¯=βX0¯.Example 1.
Consider the equationA+X=B,where(32)A=3,5,7,10,15,2,6,7,8,11,B=1,6,11,15,24,3,9,11,13,22.
Thes,t,u,v-cuts of A,B, and X are
(33)Au,vs,t=3+4s,15−8s,7−2t,7+3t,6+u,8−u,7−5v,7+4v,Bu,vs,t=1+10s,24−13s,11−5t,11+4t,9+2u,13−2u,11−8v,11+11v,Xu,vs,t=MXτs¯,MXτs¯,NXνt¯,NXνt¯,αXu¯,αXu¯,βXv¯,βXv¯,respectively. The s,t,u,v-cut equation is
(34)Au,vs,t+Xu,vs,t=Bu,vs,t.
By comparing thes,t,u,v-cuts of A,B, and X, we get
(35)MXτs¯=MBτs¯−MAτs¯=−2+6s,MXτs¯=MBτs¯−MAτs¯=9−5s,NXνt¯=NBνt¯−NAνt¯=4−3t,NXνt¯=NBνt¯−NAνt¯=4+t,αXu¯=αBu¯−αAu¯=3+u,αXu¯=αBu¯−αAu¯=5−u,βXv¯=βBv¯−βAv¯=4−3v,βXv¯=βBv¯−βAv¯=4+7v.
It is easy to see thatMXτs¯,NXνt¯,αXu¯, and βXv¯ are increasing and MXτs¯,NXνt¯,αXu¯, and βXv¯ are decreasing in 0≤s,t,u,v≤1. Also,
(36)MXτ1¯=MXτ1¯=NXν0¯=NXν0¯=αX1¯=αX1¯=βX0¯=βX0¯=4.
This shows that the solution ofA+X=B exists with s,t,u,v-cut. The solution is
(37)X=1,3,4,5,11.−2,1,4,5,9,
The solution in continuous form is(38)MRτx=2+x6,−2≤x≤4,9−x5,4≤x≤9,0,otherwise,NRνx=4−x3,1≤x≤4,−4+x,4≤x≤5,0,otherwise,αx=−3+x,3≤x≤4,5−x,4≤x≤5,0,otherwise,βx=4−x3,1≤x≤4,−4+x7,4≤x≤11,0,otherwise.
The graph of the solution is given in Figure4.Figure 4
The graph of the solution obtained in Example1.


## 5.2. Solution ofA·X+B=C by Using the Method of s,t,u,v-Cut

LetA,B,C, and X be the LDFNs and let A=ϑ1,ϑ2,ϑ3,ϑ4,ϑ5ϑ1′,ϑ2′,ϑ3,ϑ4′,ϑ5′, B=δ1,δ2,δ3,δ4,δ5δ1′,δ2′,δ3,δ4′,δ5′, and C=η1,η2,η3,η4,η5η1′,η2′,η3,η4′,η5′. Then,
(39)A·X+B=Cis a LDF equation (LDFE). Let X≈x1,x2,x3,x4,x5x1′,x2′,x3,x3′,x5′. Then, X=C−B/A in general is not the solution of Equation (39).Let(40)Au,vs,t=MAτs¯,MAτs¯,NAνt¯,NAνt¯,αAu¯,αAu¯,βAv¯,βAv¯,Bu,vs,t=MBτs¯,MBτs¯,NBνt¯,NBνt¯,αBu¯,αBu¯,βBv¯,βBv¯,Cu,vs,t=MCτs¯,MCτs¯,NCνt¯,NCνt¯,αCu¯,αCu¯,βCv¯,βCv¯,Xu,vs,t=MXτs¯,MXτs¯,NXνt¯,NXνt¯,MXτs¯,MXτs¯,NXνt¯,NXνt¯represent the s,t,u,v-cuts of A,B,C, and X, respectively, in the given (39). Substituting these into Equation (39), we get
(41)Au,vs,t·Xu,vs,t+Bu,vs,t=Cu,vs,t.By comparing thes,t,u,v-cuts of A,B,C, and X, we get
(42)MAτs¯,MAτs¯·MXτs¯,MXτs¯+MBτs¯,MBτs¯=MCτs¯,MCτs¯,NAνt¯,NAνt¯·NXνt¯,NXνt¯+NBνt¯,NBνt¯=NCνt¯,NCνt¯,αAu¯,αAu¯·αXu¯,αXu¯+αBu¯,αBu¯=αCu¯,αCu¯,βAv¯,βAv¯·βXv¯,βXv¯+βBv¯,βBv¯=βCv¯,βCv¯.Now,(43)MXτs¯=MCτs¯−MBτs¯MAτs¯,MXτs¯=MCτs¯−MBτs¯MAτs¯,NXνt¯=NCνt¯−NBνt¯NAνt¯,NXνt¯=NCνt¯−NBνt¯NAνt¯,αXu¯=αCu¯−αBu¯αAu¯,αXu¯=αCu¯−αBu¯αAu¯,βXv¯=βCv¯−βBv¯βAv¯,βXv¯=βCv¯−βBv¯βAv¯.Then, the solution of the equationA·X+B=C exists iff
(1)
MXτs¯ is monotonically increasing in 0≤s≤1(2)
MXτs¯ is monotonically decreasing in 0≤s≤1(3)
NXνt¯ is monotonically decreasing in 0≤t≤1(4)
NXνt¯ is monotonically increasing in 0≤t≤1(5)
αXu¯ is monotonically increasing in 0≤u≤1(6)
αXu¯ is monotonically decreasing in 0≤u≤1(7)
βXv¯ is monotonically decreasing in 0≤v≤1(8)
βXv¯ is monotonically increasing in 0≤v≤1(44)MXτ1¯=MXτ1¯=NXν0¯=NXν0¯=αX1¯=αX1¯=βX0¯=βX0¯.Example 2.
Consider the equationA·X+B=C,where(45)A=1,3,5,6,11,1,2,5,7,10,B=4,5,8,11,19,4,6,8,10,15,C=1,5,18,29,85.1,4,18,38,65,
Thes,t,u,v-cuts of A,B,C, and X are
(46)Au,vs,t=3+2u,6−u,5−4v,5+6v,1+4s,10−5s,5−3t,5+2t,Bu,vs,t=5+3u,11−3u,8−4v,8+11v,4+4s,15−7s,8−2t,8+2t,Cu,vs,t=5+13u,29−11u,18−17v,18+67v,1+17s,65−47s,18−14t,18+20t,Xu,vs,t=αXu¯,αXu¯,βXv¯,βXv¯,MXτs¯,MXτs¯,NXνt¯,NXνt¯,respectively. The s,t,u,v-cut equation is
(47)Au,vs,t·Xu,vs,t+Bu,vs,t=Cu,vs,t.
By comparing thes,t,u,v-cuts of A,B,C, and X, we get
(48)MXτs¯=1+17s−4+4s1+4s=−3+13s1+4s,MXτs¯=65−47s−15−7s10−5s=10−8s2−s,NXνt¯=18−14t−8−2t5−3t=10−12t5−3t,NXνt¯=18+20t−8+2t5+2t=10+18t5+2t,αXu¯=5+13u−5+3u3+2u=10u3+2u,αXu¯=29−11u−11−3u6−u=18−8u6−u,βXv¯=18−17v−8−4v5−4v=10−13v5−4v,βXv¯=18+67v−8+11v5+6v=10+56v5+6v.
It is easy to see thatMXτs¯,NXνt¯,αXu¯, and βXv¯ are increasing and MXτs¯,NXνt¯,αXu¯, and βXv¯ are decreasing in 0≤s,t,u,v≤1. Also,
(49)MXτ1¯=MXτ1¯=NXν0¯=NXν0¯=αX1¯=αX1¯=βX0¯=βX0¯=2.
This shows that the solution ofA·X+B=C exists with s,t,u,v-cut. The solution is
(50)X=−3,0,2,3,6.−3,−1,2,4,5,
The solution in continuous form is(51)MRτx=3+x13−4x,−3≤x≤2,−10+2x−8+x,2≤x≤5,0,otherwise,NRνx=−10+5x−12+3x,−1≤x≤2,−10+5x18−2x,2≤x≤4,0,otherwise,αx=3x10−2x,0≤x≤2,−18+6x−8+x,2≤x≤3,0,otherwise,βx=−10+5x−13+4x,−3≤x≤2,−10+5x56−6x,2≤x≤6,0,otherwise.
The graph of the solution is given in Figure5.Figure 5
The graph of the solution obtained in Example2.


## 5.3. Solution ofA·X2+B·X+C=D by Using the Method of α-Cut.

LetA,B,C,D, and X be the LDFNs and let
(52)A=ϑ1,ϑ2,ϑ3,ϑ4,ϑ5,ϑ1′,ϑ2′,ϑ3,ϑ4′,ϑ5′,B=δ1,δ2,δ3,δ4,δ5,δ1′,δ21,δ3,δ4′,δ5′,C=η1,η2,η3,η4,η5,η1′,η2′,η3,η4′,η5′,D=ζ1,ζ2,ζ3,ζ4,ζ5,ζ1′,ζ2′,ζ3,ζ4′,ζ5′.Then,(53)A·X2+B·X+C=Dis a LDF equation (LDFE). Let X≈x1,x2,x3,x4,x5x1′,x2′,x3,x4′,x5′. Let
(54)Au,vs,t=MAτs¯,MAτs¯,NAνt¯,NAνt¯,αAu¯,αAu¯,βAv¯,βAv¯,Bu,vs,t=MBτs¯,MBτs¯,NBνt¯,NBνt¯,αBu¯,αBu¯,βBv¯,βBv¯,Cu,vs,t=MCτs¯,MCτs¯,NCνt¯,NCνt¯,αCu¯,αCu¯,βCv¯,βCv¯,Du,vs,t=MDτs¯,MDτs¯,NDνt¯,NDνt¯,αDu¯,αDu¯,βDv¯,βDv¯,Xu,vs,t=MXτs¯,MXτs¯,NXνt¯,NXνt¯,αXu¯,αXu¯,βXv¯,βXv¯represent the s,t,u,v-cuts of A,B,C,D, and X, respectively, in the given (53). Substituting these into Equation (53), we get
(55)Au,vs,t·Xu,vs,t2+Bu,vs,t·Xu,vs,t+Cu,vs,t=Du,vs,t.By comparing thes,t,u,v-cuts of A,B,C,D, and X, we get
(56)MAτs¯,MAτs¯·MXτs¯,MXτs¯2+MBτs¯,MBτs¯·MXτs¯,MXτs¯=MDτs¯,MDτs¯+MCτs¯,MCτs¯,NAνt¯,NAνt¯·NXνt¯,NXνt¯2+NBνt¯,NBνt¯·NXνt¯,NXνt¯=NDνt¯,NDνt¯+NCνt¯,NCνt¯,αAu¯,αAu¯·αXu¯,αXu¯2+αBu¯,αBu¯·αXu¯,αXu¯=αDu¯,αDu¯+αCu¯,αCu¯,βAv¯,βAv¯·βXv¯,βXv¯2+βBv¯,βBv¯·βXv¯,βXv¯=βDv¯,βDv¯+βCv¯,βCv¯.Now,(57)MXτs¯=−MBτs¯±MBτs¯2−4MAτs¯MCτs¯−MDτs¯2MAτs¯,MXτs¯=−MBτs¯±MBτs¯2−4MAτs¯MCτs¯−MDτs¯2MAτs¯,NXνt¯=−NBνt¯±NBνt¯2−4NAνt¯NCνt¯−NDνt¯2NAνt¯,NXνt¯=−NBνt¯±NBνt¯2−4NAνt¯NCνt¯−NDνt¯2NAνt¯,αXu¯=−αBu¯±αBu¯2−4αAu¯αCu¯−αDu¯2αAu¯,αXu¯=−αBu¯±αBu¯2−4αAu¯αCu¯−αDu¯2αAu¯,βXv¯=−βBv¯±βBv¯2−4βAv¯βCv¯−βDv¯2βAv¯,βXv¯=−βBv¯±βBv¯2−4βAv¯βCv¯−βDv¯2βAv¯.Then, the solution of the equationA·X2+B·X+C=D exists iff
(1)
MXτs¯ is monotonically increasing in 0≤s≤1(2)
MXτs¯ is monotonically decreasing in 0≤s≤1(3)
NXνt¯ is monotonically decreasing in 0≤t≤1(4)
NXνt¯ is monotonically increasing in 0≤t≤1(5)
αXu¯ is monotonically increasing in 0≤u≤1(6)
αXu¯ is monotonically decreasing in 0≤u≤1(7)
βXv¯ is monotonically decreasing in 0≤v≤1(8)
βXv¯ is monotonically increasing in 0≤v≤1(9)
MXτ1¯=MXτ1¯=NXν0¯=NXν0¯=αX1¯=αX1¯=βX0¯=βX0¯Example 3.
Consider the equationA·X2+B·X+C=D,where(58)A=4,5,7,9,10,2,6,7,8,13,B=2,4,5,6,8,4,4,5,6,7,C=1,2,4,5,7,1,3,4,5,7,D=1,3,6,8,12,1,4,6,8,11.Thes,t-cuts of A,B,C,D, and X are given in Table 1.Table 1
s,t-cuts of A,B,C,D, and X.
s,t-cutsXABCDC−Dϑ1+sϑ3−ϑ1MXτs¯4+3s2+3s1+3s1+5s−2sϑ5−sϑ5−ϑ3MXτs¯10−3s8−3s7−3s12−6s−5+3sϑ3−tϑ3−ϑ2NXνt¯7−2t5−t4−2t6−3t−2+tϑ3+tϑ4−ϑ3NXνt¯7+2t5+t4+t6+2t−2−tBy comparing thes,t-cuts of A,B,C,D, and X, we get
(59)MXτs¯=−2+3s+2+3s2−44+3s−2s24+3s,MXτs¯=−8−3s+8−3s2−410−3s−5+3s210−3s,NXνt¯=−5−t+5−t2−47−2t−2+t27−2t,NXνt¯=−5+t+5+t2−47+2t−2−t27+2t.The graph obtained bys,t-cut is shown in Figure 6.Figure 6
Solution bys,t-cut.Theu,v-cuts of A,B,C,D, and X are given in Table 2.Table 2
u,v-cuts of A,B,C,D, and X.
u,v-cutsXABCDC−Dϑ2′+uϑ3−ϑ2′αu¯6+u4+u3+u4+2u−1−uϑ4′−uϑ4′−ϑ3αu¯8−u6−u5−u8−2u−3+uϑ3−vϑ3−ϑ1′βv¯7−5v5−v4−3v6−5v−2+2vϑ3+vϑ5′−ϑ3βv¯7+6v5+2v4+3v6+5v−2−2vBy comparing theu,v-cuts of A,B,C,D, and X, we get
(60)αu¯=−4+u+4+u2−46+u−1−u26+u,αu¯=−6−u+6−u2−48−u−3+u28−u,βv¯=−5−v+5−v2−47−5v−2+2v27−5v,βv¯=−5+2v+5+2v2−47+6v−2−2v27+6v.The graph obtained byu,v-cut is shown in Figure 7.Figure 7
Solution byu,v-cut.It is easy to see thatMXτs¯,NXνt¯,αXu¯, and βXv¯ are increasing and MXτs¯,NXνt¯,αXu¯, and βXv¯ are decreasing in 0≤s,t,u,v≤1. Also,
(61)MXτ1¯=MXτ1¯=NXν0¯=NXν0¯=αX1¯=αX1¯=βX0¯=βX0¯=0.2857.This shows that the solution ofA·X2+B·X+C=D exists with s,t,u,v-cut. The solution is
(62)X=0,−2+106,27,−3+338,−7+257260,15,27,13,−4+6610=0,0.1937,0.2857,0.3431,0.3474.0,0.2,0.2857,0.3333,0.4124,The solution in continuous form is(63)MRτx=−2x2x+13x2+3x−2,0≤x≤0.2857,10x2+8x−53x2+x−1,0.2857≤x≤0.4124,0,otherwise,NRνx=7x−22x−1,0.2≤x≤0.2857,−7x−22x−1,0.2857≤x≤0.3333,0,otherwise,αx=−6x2−4x+1x2+x−1,0.1937≤x≤0.2857,8x2+6x−3x2+x−1,0.2857≤x≤0.3431,0,otherwise,βx=7x2+5x−25x2+x−2,0≤x≤0.2857,−7x2−5x+223x2+x−1,0.2857≤x≤0.3474,0,otherwise.The graph of the solution is given in Figure8.Figure 8
The graph of the solution obtained in Example3.


## 6. Conclusion

In this paper, we have defined the linear Diophantine fuzzy numbers, in particular triangular linear Diophantine fuzzy number, and present some properties related to them. After finding the ranking function of triangular linear Diophantine fuzzy number, our study has focussed on the linear Diophantine fuzzy equations. We used the more general approach to solve LDF equations that is the method ofs,t,u,v-cut. In LDF sets, there is no limitation to take the grades like in intuitionistic fuzzy sets, Pythagorean fuzzy sets, and q-rung orthopair fuzzy sets. The linear Diophantine fuzzy numbers may have several applications, like in linear programming, transportation problems, assignment problems, and shortest route problems. Our future work may be on the following topics:
(i)
LDF linear programming problems(ii)
LDF assignment problems and transportation problems(iii)
LDF shortest path problems(iv)
Numerical solutions of linear and nonlinear LDF equations



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