Solve the Systems of Linear Equation

Question

3x + 7y + 8z = 41

x - 2y - 7z = -24

x + y - 2z = -3

Answer 1

Question: Solve the Systems of Linear Equation

3x + 7y + 8z = 41 ------- Equation 1

x - 2y - 7z = -24 ------- Equation 2

x + y - 2z = -3 ------- Equation 3

There are many different ways of solving systems of linear equations. Elimination method, Substitution method, Equal Values method, and Matrix are some of the common solving techniques. However, some methods are easier than others in different situations. For this problem, the ellimination method would be the easiest.

Elimination Method:

Step 1: Write all the equations given in standard form (Ax+By=C). However, all of our equations are already in this form, so we can skip this step.

Step 2: Pick a variable to eliminate. We will choose x because the coefficient (number before x or the number being multipled to x) is the same for both Equation 2 and 3.

Step 3: Subtract Equation 3 from Equation 2 in order to elliminate the variable x.

x - 2y - 7z = -24 ------- Equation 2

(-) x + y - 2z = -3 ------- Equation 3

–––––––––––––––––––––

-3y - 5z = -21 -------- Equation 4

Step 4: Eliminate the variable x from another pair of equations. We will multiply Equation 2 by 3 so the x has the same coefficient as the Equation 1.

3(x - 2y - 7z = -24)

3x - 6y - 21z = -72 -------- 3 Times Equation 2

Step 5: Subtract 3 times Equation 2 from Equation 1 to elliminate the x.

3x + 7y + 8z = 41 -------- Equation 1

(-) 3x - 6y - 21z = -72 -------- 3 Times Equation 2

———————————

13y + 29z = 113 -------- Equation 5

Step 6: Choose another variable to eliminate. Let's pick y. We will cancel the variable y from Equation 4 and Equation 5. Equation 4 has y with the coefficient -3 and Equation 5 has y with the coefficient 13. The least common multiple of these numbers is 13*-3=-39. So we will multiply 13 to Equation 4 and -3 to Equation 5 so they have the same coefficient of y.

13 (-3y - 5z = -21)

-39y - 65z = -273 -------- 13 Times Equation 4

-3 (13y + 29z = 113)

-39y - 87z = -339 -------- -3 Times Equation 5

Step 7: Subtract -3 times Equation 5 from 13 times Equation 4. Then solve for z.

-39y - 65z = -273 -------- 13 Times Equation 4

(-) -39y - 87z = -339 -------- -3 Times Equation 5

———————————

22z = 66

––– –––

22 22

z = 3

Step 8: Plug in the value of z into Equation 4 or Equation 5 to solve for y.

-3y - 5z = -21 -------- Equation 4

-3y - 5(3)= -21

-3y - 15= -21

+ 15 +15

–––––––––––

-3y = -6

––– ––

-3 -3

y = 2

Step 9: Plug in the values of y and z into either of the original equations (Equation 1, Equation 2 or Equation 3) to find the value of x.

3x + 7y + 8z = 41 ------- Equation 1

3x +7(2) +8(3) =41

3x +14 + 24 =41

3x + 38 = 41

-38 -38

––––––––––

3x = 3

–– ––

3 3

x = 1

Answer 2

Multiply the second and third equations by -3 to obtain:

3x + 7y + 8z = 41

-3x + 6y + 21z = 72

-3x - 3y + 6z = 9

Add the first equation to the second and the first equation to the third:

13y + 29z = 113

4y + 14z = 50

Multiply the first equation by 4 and the second by -13:

52y + 116z = 452

-52y -182z = -650

Add the equations: -66z = -198

So, z = 3

4y + 14(3) = 50

4y = 8

y = 2

3x + 7(2) + 8(3) = 41

3x + 38 = 41

3x = 3

x = 1

Solution: x = 1, y = 2, z = 3

Answer 3

This is a set with no obvious easy solutions...and if I were asked to do it, I would use a computer to invert the

3 x 3 matrix. In my opinion it is counterproductive to ask students to solve problems which are just onerous!

To solve it by hand you might try it this way:

Multiply the 2nd and 3rd equations by 3 and subtract the first equation from each of these.

This will eliminate x from the 2nd and 3rd equations, yielding a 2 x 2 set in y and z.

The 2 x 2 set is still messy and I would use determinants to get y and z directly.

Then plug the values of y and z into the first equation to find x.