Solve the system of 3 equations by elimination. State as an ordered triple

Question

Solve the system by elimination

x+y+2z=2

3x+4y-z=25

4x-y+3z=-7

Answer 1

x + y + 2z = 2

3x + 4y - z = 25

4x - y + 3z = -7

Multiply the first equation by -3 and add the result to the second equation.

Multiply the first equation by -4 and add the result to the third equation.

x + y + 2z = 2

y - 7z = 19

-5y - 5z = -15

Multiply the second equation by 5 and add the result to the third equation.

x + y + 2z = 2

y - 7z = 19

-40z = 80

So, from the third equation, we see that z = -2.

Back substitute into the second equation: y - 7(-2) = 19. So, y = 5.

Back substitute into the first equation: x + 5 + 2(-2) = 2. So, x = 1.

Solution: (1, 5, -2)

Answer 2

x+y+2z=2

3x+4y-z=25

4x-y+3z=-7

Pick any two equations to start with. You'll elimination one variable with them. I will start with the first and last equation. Add then together to eliminate the y.

x+y+2z=2

4x-y+3z=-7

5x +5z=-5

Pick any other combination of equations, but know that you have to eliminate the SAME variable again. So I will multiply the bottom equation by 4 and add it to the 2nd equation, to again eliminate the y.

3x+4y-z=25

16x-4y+12z=-28

19x + 11z = -3

Take the two new equations and work those as a system. You can do either elimination or substitution, but I would go with substitution this time, because it's easy to take the 5x +5z=-5 and solve for x (or do the same with y).

5x +5z=-5

5x = -5z - 5

x =- z -1

Substitute the x = -z-1 into the other equation.

19(-z-1)+11z=-3

-19z-19+11z=-3

-8z = 16

z = -2.

Substitute z = -2 into x =- z -1

x =-(-2)-1 = 1

Substitute both of those into x+y+2z=2 to get

1+y+2(-2)=2

-3+y=2

y=5

Solution: (1, 5, -2)

Answer 3

SOLUTION IN VIDEO!