Question

Solve the system by elimination.

-2x+2y+3z=0
-2x-y+z=-3
2x+3y+3z=5

Answer 1

well you need to reduce it to 2 equations in 2 unknowns...

so to eliminate x add the 1st and 3rd equation which gives


5y + 6z = 5

and the 2nd and 3rd equations

3y + 4z = 2


so you are now looking at

5y + 6z = 5
3y + 4z = 2



now choose a letter to eliminate....

you'll find the answers will be mixed numbers...

Answer 2

I don't get how to explain the answer if I put it into a calculator.

Answer 3

well its a bit early for a calculator... unless it is an Nspire that will solve this

so using the equations

5y + 6z = 5 (4)
2y + 4x = 2 (5)

take the 5th equation and make y the subject

so

2y = 2 - 4z

or y = 1 - 2z

substitute it into the 4th equation

5(1 - 2z) + 6z = 5

so

5 - 10z + 6x = 5

which means

-4z + 5 = 5

or z = 0

substitute this into the 4th equation

5y + 6(0) = 5

so y = 1

now substiute z = 0 and y = 1 into the 1st equation and you get

-2x + 2(1) + 3(0) = 0

or

-2x + 2 = 0

solve for x

then you'll have the solutions

x = , y = 1 and z = 0