## Linear Equation solution3

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