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solve for x, y, z.

0 votes

solve for x, y and z.

1/3x+1/2y-1/6z=4

1/4x-3/4y+1/2z=3/2

1/2x-2/3y-1/4z=-16/3

 

asked Feb 20, 2014 in ALGEBRA 2 by payton Apprentice

1 Answer

0 votes

Elimination method :

The system of equations are

image         → ( 1 )

image      → ( 2 )

image( 3 )

Change the equation from fraction form to linear equation.

The L.C.M of 3,2, and 6 in eq (1) is 6, so multiply eq (1) by 6.

image      → ( 4 )

The L.C.M of 4,4, and 2 in eq (2) is 4, so multiply eq (2) by 4.

image        → ( 5 )

The L.C.M of 2,3, and 4 in eq (3) is 12, so multiply eq (3) by 12.

image( 6 )

Write the eqn's (4) & (5) in column form and add the equations to eliminate y - variable.

image

The resultant equation is 3x + z = 30  → ( 7 )

To get the eq'ns (5) & (6) that contain opposite terms multiply the eq (5) by 8 and eq (6) by 3.

Write the equations in column form and subtract the equations to eliminate y - variable.

image

The resultant equation is -10x + 25z = 240 → - 2x + 5z = 48 → ( 8 )

To get the eq'ns (7) & (8) that contain opposite terms multiply the eq (7) by 5.

Write the equations in column form and subtract the equations to eliminate z - variable.

image

The resultant equation is 17x = 102 x = 102/17 = 6.

Substitute the value of x = 6 in eq (7), and solve for z.

3(6) + z = 30  → 18 + z = 30 → z = 30 - 18 = 12.

Substitute the value of x = 6 and z = 12 in eq (4), and solve for y.

image

2(6) + 3y - 12 = 24 ⇒  3y = 24 ⇒ y = 24/3 = 8.

Therefore, solution of the system is x = 6, y = 8, and z = 12.

 

answered Apr 7, 2014 by lilly Expert

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