Question

Solve the system by elimination.
-2x+2y+3z=0
-2x-y+z=-3
2x+3y+3z=5

Solve the system by substitution.
-x-y-z=-8
-4+4y+5z=7
2x+2z=4

Answer 1

To solve by elimination add the first equation and second equation together which will give you only 1 equation with x in it. Then you should be able to solve.
For subsitiution, solve for one variable (x would be easiest and then substitute for x into the other equations. Does that make sense?

Answer 2

It does kinda make sense. Can you show me the steps so I could understand it more? I've heard what you said before but I never get it. I'm really new to these systems, and I need to know how to do them by tomorrow, because I have a test.

Answer 3

And I really appreciate you replying to my question. thank you.(:

Answer 4

timashorty have you done system of equations by elimination with only 2 variables?

Answer 5

yes I have but I still struggle with that too. There's so many steps, different systems. its complicated to remember all that. And I just started algebra2 a week ago, and I'm late in class the teacher skipped alot of lessons for me, which is why I'm lost with these two equations.

Answer 6

-2x + 2y + 3z = 0
-2x - y + z = -3--->(-1)
--------------------
-2x + 2y + 3z = 0
2x + y - z = 3 (result of multiplying by -1)
-------------------add
3y + 2z = 3

we eliminated the x out of the equation, so with the last two equations x has to also be eliminated. Remember, when doing problems like this, the same variable has to be eliminated.

-2x - y + z = -3
2x + 3y + 3z = 5
---------------
2y + 4z = 2

now we take our answers and eliminate another variable..
3y + 2z = 3 -->(-2)
2y + 4z = 2
-----------
-6y - 4z = -6 (result of multiplying by -2)
2y + 4z = 2
-----------add
-4y = -4
y = 1

now we sub 1 in for y in one of the equations that does not have an x term.
2y + 4z = 2
2(1) + 4z = 2
2 + 4z = 2
4z = 2 - 2
4z = 0
z = 0/4 = 0

now we sub 1 in for y and 0 in for z in one of the original equations
2x + 3y + 3z = 5
2x + 3(1) + 3(0) = 5
2x + 3 = 5
2x = 5 - 3
2x = 2
x = 1

check...
-2x + 2y + 3z = 0
-2(1) + 2(1) + 3(0) = 0
-2 + 2 + 0 = 0
0 = 0 (correct)

x = 1, y = 1, and z = 0

Answer 7

Thank you so much. You helped alot!

Answer 8

Can you help me with the second one two?