Solve the system of linear questions below using a calulator?

U can solve this using matrices. u put each equation into a matrix and then solve for x,y,z. This can be done with a graphing calculator or a 'GDC'

If you want me to take u throught the process step-by-step

IM at

timmyyassa

First, write this as a system of equations so it looks nice.
2x+y+z=4
-x+2y+z=3
9x+2y+3z=13

To use a calculator, use the MATRIX function. I assume you have a TI-83 or something like it.
In the matrix menu, edit a 3x4 matrix (perhaps matrix "A") and input the augmented coeffecient matrix as (you see how the numbers come from the equations above?):

[[2 1 1 4]
 [-1 2 1 3]
 [9 2 3 13]]

Then exit. Now go back to the matrix menu, go to MATH (or wherever there are operations on matrices), and select rref. Enter. Then go back to the matrix menu, and select matrix A (if you used A).

Your screen should now look like:
rref(A
Close parantheses and hit enter.

My result looked like this
[[ 1 0 .2 1 ]
 [ 0 1 .6 2 ]
 [ 0 0 0 0 ]]

This means that there is no unique solution, but an infinite number of solutions, such that
x + .2z = 1
y + .6z = 2

These can be written as, in general (1 - .2z, 2 - .6z, z)

There is not a set of 3 independant equations

{x =   1 - z/5, y =   2 - (3 z)/5}

Okay, those top 3 answers are taking you down what sounds to be a difficult road, the way we learned to do this is just by elimination, you eliminate one variable, for example take your first equation and your second equation, now lets eliminate the x variable, so lets make the x's equal,

2x+y+z=4(this one is fine so lets just leave it alone)
-x+2y+z=3(we need to make this x the opposite of the top x, so we need to make it a -2, so lets multiply it by 2, you cannot just multiply the x by 2 though, you need to multiply that entire equation by 2, so it turns out to be, -2x+4y+2z=6)

now that you have your 2 equations, just add them together

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

The x's elimate each other, y+4y=5y,z+2z=3z, and 4+6=10, so your new equation looks like this 5y+3z=10 (now keep this in mind, we will come back to it, but leave it alone for now)

now we need one more equation so lets to the second and the third one, one again we need to eliminate the x because we did that in the first one, so we need the x's to be equal, so we need to multiply 9 to -x+2y+z=3, so now it is -9x+18y+9z=27 and now we add.

-9x+18y+9z=27
9x+2y+3z=13

once again x's elimate each other, 18y+2y=20y, 9z+3z=12z,27+13=40.
now we have 20y+12z=40, now we bring back the 5y+3z=10

Okay, now lets eliminate the y's, so lets multiply the second equation by -4 so the y's are opposite one another (-20y-12z= -40)

Now we add again.

20y+12z= 40
-20y-12z= -40

Now here is the tricky point, everything cancels out, 20y+(-20y)=0, 12z+(-12z)=0, and 40+(-40)=0, so that means 0=0, which means that for this question, the answer is,

infinitely many, this just means that whatever number you plug in for y or z works.(as long as its the same number)

now we need to find x, lets just plug in some random number for both x and y,5 for example, now lets do the first equation 2x+5+5=4, first subtract both 5's from the 4, so we get -6, now just solve, so x=-3, now we are still now done, now take your -3 and divide it by the original 5, so you get -3/5

so here it is, x= -3/5 of y or z, so -3/5y=x or -3/5z=x and that is your answer.

this sounds like it may take a while and it does, but if you dont have a calculator and one of these problems comes up then this is how you would do it.