In this page linear equation solution3 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.

**Question3:**

Solve the equations

3 x - 2 y + z = 0

4 x + 6 y - 3 z = 13

x - 2 y + 2 z = -4

**Solution:**

3 x - 2 y + z = 0 --------(1)

4 x + 6 y - 3 z = 13 --------(2)

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

Consider the equations (1) and (2)

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

(1) x 3 => 9 x - 6 y + 3 z = 0

4 x + 6 y - 3 z = 13

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

13 x = 13

x =13/13

x = 1

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

4 x + 6 y - 3 z = 13

(3) x 3=> 3 x - 6 y + 6 z = -12

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

7 x + 3 z = 1 ------- (4)

Now we are going to apply the value of x in this 4th equation

So that we will get =>7(1) + 3 z =1

7 + 3 z = 1

3 z = 1 – 7

3 z = -6

z = -6/3

z = -2

Substitute x = 1, z = -2 in (3) we get

1 - 2 y + 2 (-2) = -4

1 - 2 y - 4 = -4

- 2 y -3 = -4

- 2 y = -4 + 3

- 2 y = -1

Y = 1/2

Therefore solution is x = 1,y = 1/2 and z = -2

linear equation solution3 linear equation solution3

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