Solve for the following system of equations algebraically for all values of x,y and z:

Question

2x+3y-4z=-1  

x-2y+5z=3  

-4x+y+z=16

Answer 1

Start with the following system:

2x + 3y - 4z = -1  

x - 2y + 5z = 3

-4x + y + z = 16

We'll solve this system using elimination. First let's eliminate the x in the second and third equations.

1) Let's multiply the second equation by -2 and add the first equation to it.

2x + 3y - 4z = -1

+ -2x + 4y - 10z = -6

----------------------------------

7y - 14z = -7

2) Let's multiply the second equation by 4 and add it to the third equation.

4x - 8y + 20z = 12

+ -4x + y + z = 16

---------------------------------

-7y + 21z = 28

We'll replace the original second and third equations with the new ones we just found. The system is now

2x + 3y - 4z = -1  

7y - 14z = -7

-7y +21z = 28

3) Divide both the second and third equations by 7 since every coefficient is divisible by 7

2x + 3y - 4z = -1  

y - 2z = -1

- y +3z = 4

4) Add the second equation to the third equation

y - 2z = -1

+ - y +3z = 4

-------------------

z = 3

5) We solved for z = 3. Now plug this into the second equation from step 3 to solve for y.

y - 2z = -1

y - 2(3) = -1

y - 6 = -1

y = 5

6) We solved for y = 5 and z = 3. Plug both of these into the first equation from step 3 to solve for x.

2x + 3y - 4z = -1  

2x + 3(5) - 4(3) = -1

2x + 15 - 12 = -1

2x + 3 = -1

2x = -4

x = -2

The final answer is x = -2, y = 5, z = 3

Be sure to plug these values into the original system of equations to check your answer.

2(-2) + 3(5) -4(3) = -4 + 15 - 12 = -1 √

(-2) - 2(5) + 5(3) = -2 - 10 + 15 = 3 √

-4(-2) + 5 + 3 = 8 + 5 + 3 = 16 √

Answer 2

Assuming that you have not yet covered matrix manipulation, let's use the Elimination method (which looks much like matrix manipulation).

To help, let's number the equations:

2x+3y-4z=-1 [eq1]

x-2y+5z=3 [eq2]

-4x+y+z=16 [eq3]

Eliminate one of the variables and get two equations:

2x+3y-4z=-1 [eq1 again

2x-4y+10z=6 [2*eq2]

---------------------------------- [elimination; subtract equations]

7y-14z=-7 [eq4]

4x+6y-8z=-2 [2*eq1]

-4x+y+z = 16 [eq3 again]

------------------------------------ [elimination; add equations]

7y -7z = 14 [eq5]

7y-14z=-7 [eq4 again]

------------------------------------ [elimination' subtract equations]

7z = 21 [eq6]

z = 3

Now, the three equations are simplified to become:

2x+3y=11 [eq1a]

x-2y=-12 [eq2a]

-4x+y=13 [eq3a]

Any two of those may now be used to eliminate another variable:

-4x+y=13 [eq3a again]

4x-8y=-48 [4*eq2a]

--------------------------- [elimination; add equations]

-7y = -35

y = 5

Now, the original three equations are simplified to become:

2x+3(5)-4(3)=-1 

2x = -4

x = -2

x-2(5)+5(3)=3 [eq2]

x = -2

-4x+(5)+(3)=16 [eq3]

-4x = 8

x = -2

Any of the original three equations [or some of the others] could have been used to determine the value of x.

x = -2

y = 5

z = 3