Question

{4x-2y+5z=6 {3x+3y+8z=4 {x-5y-3z=5

Answer 1

There are three possible outcomes that you may encounter when working with these system of equations:

•    one solution
•    no solution
•    infinite solutions

We are going to try and find values of x, y, and z that will satisfy all three equations at the same time. The following are the equations:

1. $4x-2y+5z = 6$
2. $3x+3y+8z = 4$
3. $x-5y-3z = 5$

We are going to use elimination(or addition) method

Step 1: Choose to eliminate any one of the variables from any pair of equations.

In this case it looks like if we multiply the third equation by 4 and  subtracting it from equation 1, it will be fairly simple to eliminate the x term from the first and third equation.

So multiplying Left Hand Side(L.H.S) and Right Hand Side(R.H.S) of 3rd equation with 4 gives us a new equation 4.:

4. 4x-20y-12z = 20      

Subtracting eq. 4 from Eq. 1:

(L.HS) : 4x-2y+5z-(4x-20y-12z) = 18y+17z

(R.H.S) : 20 - 6 = 14

5. 18y+17z=14

Step 2:  Eliminate the SAME variable chosen in step 2 from any other pair of equations, creating a system of two equations and 2 unknowns.

Similarly if we multiply 3rd equation with 3 and then subtract it from eq. 2 we get:

(L.HS) : 3x+3y+8z-(3x-15y-9z) = 18y+17z

(R.H.S) : 4 - 15 = -11

6. 18y+17z = -11

Step 3:  Solve the remaining system of equations 6 and 5 found in step 2 and 1.

Now if we try to solve equations 5 and 6 for the variables y and z. Subtracting eq 6 from eq. 5 we get:

(L.HS) : 18y+17z-(18y+17z) = 0

(R.HS) : 14-(-11) = 25

0 = 25

which is false, hence no solution exists