Solve System, any help would be greatly appreciated.
{x+y+z=3
x+y-z
2x-3y+z=-1

- anonymous

Solve System, any help would be greatly appreciated.
{x+y+z=3
x+y-z
2x-3y+z=-1

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- schrodinger

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- whpalmer4

\[x+y+z=3\]\[x+y-z = ???\]\[2x-3y+z=-1\]
What's the right hand side of the second equation?

- anonymous

=9

- anonymous

{x+y+z=3
x+y-z=9
2x-3y+z=-1

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## More answers

- whpalmer4

\[x+y+z=3\]\[x+y-z=9\]\[2x-3y+z=-1\]
I would observe that adding the first two equations will eliminate \(z\). Can you see another way to add or subtraction equations or multiples thereof to eliminate \(z\)?

- anonymous

-3y-(y1+y2)??

- whpalmer4

how about adding the second and third equations? \(z\) has coefficients which are equal in magnitude but opposite in sign...
So what do you get if you add the first and second equations together? What about the second and third equations?

- anonymous

Im sorry add them how as in Factor?

- whpalmer4

No, just add them like columns of numbers:
\[x+y+z=3\]\[x+y-z=9\]-------------\[2x+2y+0z=12\]

- whpalmer4

See how the \(z\) term vanished?

- whpalmer4

Do that with another combination, so you get two new equations in terms only of \(x, y\).

- whpalmer4

Then we'll do the same process again, to get an equation in terms of just one of the variables, which we solve. With the value of that variable in hand, we work backwards to find the other variables.

- anonymous

k just a sec

- anonymous

so i got to get x and y =0?

- whpalmer4

No. What do you get if you add the second and third equations in the manner that I did with the first two?

- anonymous

2x-3y+z=-1
x+ y- z= 9
3x-2y+0x=8

- whpalmer4

Let me try again. Here I'll add the first two equations:
\[x+y+z=3\]\[x+y-z=9\]-------------\[2x+2y+0z=12\]
Now I'll add the last two equations:
\[x+y-z=9\]\[2x-3y+z=-1\]------------------\[3x-2y+0z=8\]
Now I have a system of 2 equations in 2 unknowns:
\[2x+2y=12\]\[3x-2y=8\]
Add those together, and what do you get?

- anonymous

-x+0y=4

- whpalmer4

\[2x+2y=12\]\[3x-2y=8\]------------\[5x=20\]\[x=\]
Now use the known value of \(x\) in one of the two equations you just added to find \(y\).

- anonymous

x=4

- anonymous

I do not get this so x=4 but it is asking me to solve for y and z.... so how do i get those? Also i apologize Im really dull with these....

- whpalmer4

Okay, you know that \(x=4\). You also know that \(2x+2y=12\) and \(3x-2y=8\). Plug \(x=4\) into one of those two equations and solve for \(y\).

- anonymous

y=4

- whpalmer4

Let's use the first one.
\[2x+2y=12\]\[2(4)+2y=12\]\[8+2y=12\]\[2y=12-8\]\[2y=4\]\[y=2\]
Now check using the other equation:
\[3(4)-2(2) = 8\]\[12-4=8\checkmark\]
So, we now know \(x=4\) and \(y=2\).

- whpalmer4

Use one of the original equations + those two values to find \(z\).

- anonymous

2x-3y+z=-1
2(4)-3(2)+z=1
8-6+z=-1
2+z=-1
z=-3

- whpalmer4

Okay, now you need to try those 3 values in each of the original equations and make sure they are all correct. It is not sufficient to try just one or two of the equations.

- whpalmer4

Of course, if one of them turns out to be incorrect, you don't have to try the rest!

- texaschic101

good job @whpalmer4 .....you did it again......patience is a virtue

- anonymous

dang i think i got it....

- anonymous

God thank you so much for bearing with me.... and thank so much for helping me understand.

- whpalmer4

I'm a little worried that you don't seem to grasp the concept of adding the equations together. You can add them (like we did), you can subtract them (I prefer adding, few mistakes with - signs!), and you can also multiply an equation by a number before adding or subtracting. For example, you can subtract by multiplying one of the equations by -1, then adding.
Practice with this stuff helps! And remember (especially when learning) to check ALL of the equations when you think you have a solution...

- texaschic101

he is good isn't he....he should be a teacher :)

- whpalmer4

Glad I could help! but now it's time to go watch a movie :-)

- anonymous

enjoy it sir and have a wonderful day.

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