Question

Solve the system of equations.

3x + 4y + 6z = 3
4x + 3y + 3z - 4
5x + 6y + 7z = 5
a. (x = 0, y = 1,z = 1)
b. (x = 2, y = -1,z=-1)
c. (x=3, y=-2,z=2)
d. (x = 1, y = 0,z=0)

Answer 1

Value of x = 1, y = 0 and z = 0

Option D is correct.

Explanation:

3x + 4y + 6z = 3

4x + 3y + 3z = 4

5x + 6y + 7z = 5

We need to solve and find values of x, y and z.

I am using Elimination method.

Let

3x + 4y + 6z = 3 (1)

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

5x + 6y + 7z = 5 (3)

Multiply eq(1) with 4 and eq(2) with 3 and subtracting

12x+16y+24z = 12

12x + 9y +9z = 12

- - - -

______________

7y+15z = 0 (4)

Multiply eq(2) with 5 and eq(3) with 4 and subtracting

20x + 15y + 15z = 20

20x + 24y + 28z = 20

- - - -

_________________

-9y -13z = 0 (5)

Multiply eq(4) with 9 and eq(5) with 7 and add both equations

63y + 135 z = 0

-63y - 91 z = 0

_____________

44z = 0 => z =0

Putting value of z in eq(5)

-9y -13z = 0

-9y -13(0) = 0

-9y = 0

y =0

Putting value of y and z in eq(1)

3x + 4y + 6z = 3

3x + 4(0) +6(0)=3

3x = 3

x =1

So, Value of x = 1, y = 0 and z = 0

Option D is correct.