SOLVING LINEAR EQUATIONS IN THREE VARAIBLES PRACTICE PROBLEMS

Problem 1 :

Solve the equations

x+2y+3z  =  14, 3x+y+2z  =  11, 2x+3y+z  =  11

Solution :

x+2y+3z  =  14  --------(1)

3x+y+2z  =  11 --------(2)

2x+3y+z  =  11  --------(3)

(1)-3(3)

x+2y+3z - 3(2x+3y+z)  =  14 - 3(11)

x+2y+3z - 6x-9y-3z  =  14 - 33

-5x-7y  =  -19 ----(4)

(2)-2(3)

3x+y+2z-2(2x+3y+z)  =  11-2(11)

3x+y+2z-4x-6y-2z  =  11-22

-x-5y  =  -11 ----(5)

(4) - 5(5)

-5x-7y-5(-x-5y)  =  -19-5(-11)

-5x-7y+5x+25y  =  -19+55

-7y+25y  =  36

18y  =  36

y  =  2

By applying the value of y in (5), we get

-x-5(2)  =  -11

-x-10  =  -11

-x  =  -1

x  =  1

By applying the value of x and y in (1), we get

1+2(2)+3z  =  14

5+3z =  14

3z  =  9

z  =  3

So, the solution is x  =  1, y  =  2 and z  =  3.

Problem 2 :

Solve the equations

3x-3y+4z  =  14, -9x-6y+2z  =  1, 6x+3y+z  =  5

Solution :

3x-3y+4z  =  14 ----(1)

-9x-6y+2z  =  1  ----(2)

6x+3y+z  =  5  ----(3)

2(1) - (2)

2(3x-3y+4z)-(-9x-6y+2z)  =  2(14) - 1

6x-6y+8z+9x+6y-2z  =  28-1

15x+6z  =  27  -----(4)

(2) + 2(3)

-9x-6y+2z+2(6x+3y+z)  =  1 + 2⋅5

-9x-6y+2z+12x+6y+2z  =  1 + 10

3x+4z  =  11 -----(5)

2(4) - 3(5)

2(15x+6z)-3(3x+4z)  =  2⋅27 - 3⋅11

30x+12z-9x-12z  =  54-33

21x  =  21

x  =  1

By applying the value of x in (5), we get

3(1)+4z  =  11

3+4z  =  11

4z  =  8

z  =  2

By applying the value of x and z in (1), we get

3(1)-3y+4(2)  =  14

3-3y+8  =  14

-3y+11  =  14

-3y  =  3

y  =  -1

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

Problem 3 :

Solve the equations

3x-2y+z  =  0, 4x+6y-3z  =  13, x-2y+2z  =  -4

Solution :

3x-2y+z  =  0  -----(1)

4x+6y-3z  =  13  -----(2)

x-2y+2z  =  -4  -----(3)

3(1) + (2)

3(3x-2y+z)+(4x+6y-3z)  =  0+13

9x-6y+3z+4x+6y-3z  =  13

9x+4x  =  13

13x  =  13

x  =  1

(2) + 3(3)

(4x+6y-3z)+3(x-2y+2z)  =  13+3⋅(-4)

4x+6y-3z+3x-6y+6z  =  13-12

7x+3z  =  1 ----(4)

By applying the value of x in (4), we get

7(1)+3z  =  1

3z  =  -6

z  =  -2

By applying the value of x and z in (1), we get

3(1)-2y-2  =  0 

3-2-2y  =  0

-2y  =  -1

y  =  1/2

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

Problem 4 :

Solve the equations

2x-2y+4z  =  -12, 3x+2y+2z  =  19, -x+y-z  =  3

Solution :

2x-2y+4z  =  -12 -----(1)

3x+2y+2z  =  19 -----(2)

-x+y-z  =  3  -----(3)

(1)+(2)

(2x-2y+4z)+(3x+2y+2z)  =  -12+19

5x+6z  =  7 ------(4)

(2)-2(3)

(3x+2y+2z)-2(-x+y-z)  =  19-2⋅ 3

3x+2y+2z+2x-2y+2z  =  19-6

5x+4z  =  13 ------(5)

(4)-(5)

5x+6z - (5x+4z)  =  7-13

2z  =  -6

z  =  -3

By applying the value of z in (4), we get

5x+4(-3)  =  13

5x  =  13+12

5x  =  25

x  =  5

By applying the value of x and z in (3), we get

-5+y-(-3)  =  3

-5+y+3  =  3

-2+y  =  3

y  =  5

Therefore solution is x  =  5, y  =  5 and z  =  -3.

Problem 5 :

Solve the equations

2x+3y-z  =  5, 4x+y+3z  =  5, 3x+2y+2z  =  5

Solution :

2x+3y-z  =  5 ----(1)

4x+y+3z  =  5 ----(2)

3x+2y+2z  =  5----(3)

(1)-3(2)

(2x+3y-z)-3(4x+y+3z)  =  5-3⋅5

2x-12x+3y-3y-z-9z  =  5 - 15

-10x-10z  =  -10

x+z  =  1----(4)

2(2)-(3)

2(4x+y+3z)-(3x+2y+2z)  =  2⋅5-5

8x+2y+6z -3x-2y-2z  =  5

5x+4z  =  5 ---(5)

4(4)-(5)

4x+4z-5x-4z  =  4-5

-x  =  -1

x  =  1

By applying the value of x in (1), we get

5(1)+4z  =  5

4z  =  0

z  =  0

By applying the value of z and x in (1), we get

2(1)+3y-0  =  5

3y  =  3

y  =  1

Therefore solution is x  =  1, y  =  1 and z  =  0.

Problem 6 :

Solve the equations

x+2y+3z  =  10, x-2y+4z  =  3, x+y–3z  =  2

Solution :

x+2y+3z  =  10  ----(1)

x-2y+4z  =  3  ----(2)

x+y–3z  =  2   ----(3)

(1) + (2)

(x+2y+3z)+(x-2y+4z)  =  10+3

2x+7z  =  13  ----(4)

(2)+2(3)

(x-2y+4z)+2(x+y–3z)  =  3+4

x+2x-2y+2y+4z-6z  =  7

3x-2z  =  7  ----(5)

2(4)+7(5)

2(2x+7z)+7(3x-2z)  =  26+7⋅7

4x+14z+21x-14z  =  26+49

25x  =  75

x  =  3

By applying the value of x in (3), we get

3(3)-2z  =  7

9-2z  =  7

-2z  =  7-9

-2z  =  -2

z  =  1

By applying the value of z and x in (1), we get

3+2y+3(1)  =  10

6+2y  =  10

2y  =  4

y  =  2

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

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