1) Concept:

Step 1: Simplify and put all three equations in the form Ax+By+Cz=D, if needed.

Step 2: Use elimination method to eliminate any one of the variables. So that we get two equations with two unknowns.

Step 3: Solve the remaining system of equations just as solving a system of linear equations with two unknowns.

Step 4: Solve for the third variable.

Step 5: Check whether solved solutions satisfy the given system of equations by substituting the values obtained in the given system of linear equations.

2) Given:

13x−14y+z=−9 →(1)12x−13y−14z=−6 →(2)x−12y−z=−8 →(3)

Calculation:

From all the above three equations, we can observe that the coefficients of variables are in fractions.

To perform calculations easily, we can take L.C.M and make fraction as integers.

(I) Simplify all three equations

For simplifying (1) take L.C.M. By doing so, we get

13x−14y+z=−94x−3y+12z12=−94x−3y+12z=−108 →(1)

By simplifying (2), we get

12x−13y−14z=−66x−4y−3z12=−66x−4y−3z=12×−66x−4y−3z=−72      →(2)

Simplifying (3), we get

x−12y−z=−82x−y−2z2=−82x−y−2z=−16 →(3)  

Now the system of linear equations is

4x−3y+12z=−108 →(4)6x−4y−3z=−72      →(5)2x−y−2z=−16        →(6)

(II) We can choose to eliminate y

On multiplying (4) by 4 and (5) by 3 and by subtracting the resultant, we can eliminate y from equation (4) and (5)

Multiplying (3) by 4, we get

4(4x−3y+12z)=−10816x−12y+48z=−432, →(7)

Multiplying (5) by 3, we get

3(6x−4y−3z)=−7218x−12y−9z=−216 →(8)

Subtracting (7) from (8), we get

16x−12y+48z=−43218x−12y−9z=−216−2x+57z=−216 ¯→(9)

By multiplying (6) by 4 and subtracting from (4), we can eliminate y from (5) and (6)

By multiplying (5) by 4, we get

4(2x−y−2z=−16)8x−4y−8z=−64 →(10)

Subtracting (5) from (10), we get

6x−4y−3z=−728x−4y−8z=−64−2x+5z=−8 →(11)¯

Now we have two equations with two unknowns, the two equations are

−2x+57z=−216 →(9)−2x+5z=−8      →(11)

Solve for x and z

Subtracting (9) and (11), we get

−2x+57z=−216−2x+5z=−8     52z=−208z=−20852=−4                     ¯

Substitute z=−4 in (11), we get

−2x+5z=−8−2x+5(−4)=−8−2x−20=−8−2x=12x=12−2=−6x=−6

III) Solve for third variable

Substitute x=−6 and z=−4 in (4), we get

4x−3y+12z=−1084(−6)−3y+12(−4)=−108−24−3y−48=−108−3y=−108+48+24−3y=−36y=−36−3=12

IV) Verify:

Substitute x=−6, y=12 and z=−4 in (4), we get

4x−3y+12z=−1084(−6)−3(12)+12(−4)=−108−24−36−48=−108−108=−108L.H.S=R.H.S

Conclusion:

(−6,12,−4) is the solution.