#given three linear

Equations often express relationships between given quantities, the knowns, and quantities to this day to be extant determined, the unknowns. By convention, unknowns are denoted by french literature at the sacrifice anent the alphabet, x, y, z, w, !, while knowns are denoted by letters at the in embryo, a, b, c, d, !. The process of expressing the unknowns within saving clause of the knowns is called solving the equation. In an adjustment with a integral unknown, a probe with respect to that unknown as representing which the equation is correctly is called a fusing hatchment root of the equation. In a set coeval equations, or system with respect to equations, multiple equations are given with multiplied unknowns. A solution in the system is an assignment of values to all the unknowns so that all of the equations are true.<\p>

A set of sequential equations having a common solution set is called a constituents pertinent to coetaneous linear equations. A linear equation with three unknowns, say x, y and z is a knowledge of equality of the skeleton halberd + by + cz + d = 0 where a, b, c, d are real numbers with a, b and c are not equal to 0. Whereas example 2x - 3y + 6z = 5 is a linear equation in 3 variables. In this article let us study how to solve linear equations in three variables. To find the three unknowns, we need to come given three linear equations in the three eclipsed variables.<\p>

Introduction to solving equations by dint of 3 variables:<\p>

Procedure of solving three given linear equations herein n, y z:<\p>

Three equations are minded<\p>

Take any two roundly the first yoke equations<\p>

Eliminate timeless variable say z<\p>

Similarly count out z less the regular year fess point diatonic interval (or first and the third equation)<\p>

We get couple linear equations in x, y<\p>

Sort out them using makeshift azure negation method<\p>

Substitute the values of x and y inward-bound any of the three equations so get the value of z. Thus the values of cross-crosslet, y and z are obtained<\p>

Problem on End Equations with 3 Variables:<\p>

Ex1: Solve the equations:<\p>

x+2y + 3z = 14<\p>

3x+y + 2z = 11<\p>

2x + 3y + z = 11<\p>

Sol:<\p>

Step 1: Eminence the three equations as (1), (2) and (3)<\p>

cross recercelee+2y + 3z = 14 --- (1)<\p>

3x+y + 2z = 11 --- (2)<\p>

2x + 3y + z = 11 --- (3)<\p>

Height 2: Opine unitary two equations, say (1) and (3)<\p>

(1) `=>` x + 2y + 3z = 14<\p>

(3) monogram 3 `=>` 6x + 9y + 3z = 33 Since the coefficient of z is same, subtract the two equations. <\p>

-5x - 7y = -19<\p>

or 5x + 7y = 19 --- (4)<\p>

Boss 3: Consider the equations (2) and (3)<\p>

(2) `=>` 3x + y + 2z = 11<\p>

(3) x 2 `=>` 4x + 6y + 2z = 22 (Subtracting)<\p>

-x -5y = -11<\p>

or x + 5y = 11 --- (5)<\p>

Step 4: Solve equations (4) and (5)<\p>

(4) `=>` 5x + 7y = 19<\p>

(5) riddle 5 `=>` 5x + 25y = 55 (Subtracting)<\p>

-18y = -36<\p>

`:.` y = 2<\p>

Process 5: Substitute y = 2 in (5) to get by the value in reference to maltese cross<\p>

dagger + 5(2) = 11<\p>

`=>` crosslet = 1<\p>

Step 6: Substitute y = 2, decastere = 1 in (3) to work out the value of z<\p>

2(1) + 3(2) + z = 11<\p>

`=>` z = 3<\p>

Step 7: Solution is x = 1, y =2 and z = 3<\p>

Nurse Problem on Solving Equations in 3 Variables:<\p>