In this page linear equation solution6 we are going to see solution of first problem.

**Procedure of solving three given linear equations in x,y,z.**

(i)Three equations are given

(ii) Take any two say first to equations

(iiii)Eliminate one variable z

(iv) Similarly eliminate z from the second and third(or first and the third equations)

(v) We get two linear equations in x,y

(vi) Solve them in a usual way of solving linear equations in two variable

(vii) Substitute the values of x and y in any one of the three equations to get the values of z. Thus the values of x,y and z are obtained.

**Question 6:**

Solve the equations

x + 2 y + 3 z = 10

x - 2 y + 4 z = 3

x + y – 3 z = 2

**Solution:**

x + 2 y + 3 z = 10 --------(1)

x - 2 y + 4 z = 3 --------(2)

x + y – 3 z = 2 --------(3)

Consider the equations (1) and (2)

Now we are going to eliminate the variable y in the equations (1) and (2)

x + 2 y + 3 z = 10

x - 2 y + 4 z = 3

---------------------

2 x + 7 z = 13 ------- (4)

Now let us consider the equations (2) and (3)

x - 2 y + 4 z = 3

(3) x 2=> 2 x + 2 y - 6 z = 4

--------------------

3 x - 2 z = 7 ------- (5)

(4) x 2 => 4 x + 14 z = 26

(5) x 7 => 21 x - 14 z = 49

------------------

25 x = 75

x = 75/25

x = 3

Substitute x = 3 in (5) we get

3 (3) - 2 z = 7

9 - 2 z = 7

- 2 z = 7 - 9

- 2 z = - 2

z = (-2)/(-2)

z = 1

Substitute x = 3, z = 1 in (3) we get

x + y – 3 z = 2

3 + y – 3 (1) = 2

3 + y - 3 = 2

y = 2

Therefore solution
is x = 3,y = 2 and z = 1

linear equation solution6 linear equation solution6

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