Solve 5x + 6y + z = 9, 6x + 3 y = 6, 2x + 6y + z =4

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krishna-agrawala's profile pic

krishna-agrawala | College Teacher | (Level 3) Valedictorian

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Given:

5x + 6y + z = 9   ...   (1)

6x + 3y = 6   ...   (2)

2x + 6y + z = 4   ...   (3)

Subtracting equation (3) from (1):

5x - 2x + 6y - 6y + z - z = 9 - 4

==> 3x = 5

==> x = 5/3

Substituting this value of x in equation (2):

6*5/3 + 3y = 6

==> 10 + 3y = 6

3y = 6 - 10 = - 4

y = -4/3

Substituting the above values of x and y in equation (1):

5*5/3 + 6*(-4/3) + z = 9

==> 25/3 - 24/3 + z = 9

1/3 + z = 9

z = 9 - 1/3 = 26/3

Answer:

x = 5/3

y = -4/3

z = 26/3

william1941's profile pic

william1941 | College Teacher | (Level 3) Valedictorian

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We have 3 equations to solve for x, y and z

5x + 6y + z = 9…(1)

6x + 3 y = 6…(2)

2x + 6y + z =4…(3)

(1) – (3)

=> 3x = 5

=> x = 5/3

Use this in (2)

6x + 3 y = 6

=> 6*(5/3) + 3y = 6

=> 10 + 3y =6

=> 3y = -4

=> y = -4/3

substitute x and y in (1) to get z

5x + 6y + z = 9

=> 25/3 – 24/3-9=-z

=> (25 – 24 -27)/3 = -z

=> z= 26/3

Therefore x = 5/3 , y= -4/3 and z= 26/3

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll use the matrix to calculate the solution of the system:

We'll calculate the determinant of the system. The determinant is formed from the coefficients of the variables: x,y,z.

            5   6   1

detA = 6   3    0 

            2   6    1

det A = 5*3*1 + 6*6*1 + 6*0*2 - 2*3*1 - 6*0*5 - 6*6*1

We'll eliminate like terms and we'll get:

det A = 15 - 6

det A = 9

x = detX/detA

y = dety/detA

z = detz/detA

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

5x+6y+z=9...........(1)

6x+3y =6................(2)

2x+6y+z=4 .................(3).

The 2nd equation is free from z.

So we eleminate z between equations (1) and (3) by Eq(1)-eq(3) which also eliminates y. So we solve x :

(5x+6y+z) - (2x+6y+z) = 9 - 4 = 5

3x = 5.

x = 5/3.

Substitute x = 5/3 in eq(2):

6x+3y = 6. Or 3y = 6-6x = 6-6(5/3) = 6-10 = -4.

3y = -4.

y= -4/3.

Substitute x = 5/3 and y =-4/3 in (1):

5x+6y+z = 9

5(5/3)+6(-4/3)+z = 9

25/3 -24/3 +z = 9

1/3 +z = 9

z = 9-1/3 = 26/3

Therefore x = 5/3 , y =-4/3 and z =28/3

 

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