Help with systems of equations?

Question

Here is the problem

3x-5y+2z=4
x-3y+2z=4
5x-11y+6z=12


I get , and i think im off here, y= z-2, x=3y-2z+4. and z?? I iknow i need to solve it like a matrix and swap the rows around 1st.? right ?Help..lol Thanks for any help

Thanks for the responses everybody. I am trying to solve the equation with Gaussian elimination method. I am starting to grasp it, But some of teh answers got me going 2 different directions. Can someonle please show me how to slove with gaussain. Thanks.everbody

Answer 1

The corresponding matrix (swapping rows 1 and 2) is

1 -3 2 | 4
3 -5 2 | 4
5 -11 6 |12

First, we want to obtain zeros below the main diagonal in the first column (where 3 and 5 are now)
=> subtract 3 rows 1 from row 2
=> subtract 5 rows 1 from row 3

1 -3 2 | 4
0 4 -4 | -8
0 4 -4 | -8

=> subtract row 2 from row 3

1 -3 2 | 4
0 4 -4 | -8
0 0 0 | 0

=> simplify by dividing row 2 by 4

1 -3 2 | 4
0 1 -1 | -2
0 0 0 | 0

There is a zero row. This means that one of the variables cannot be uniquely determined - it will just stay a parameter.
Just keep z as z. z=z.
Express x and y in terms of z.

Then (from row 2):
y - z = -2
y = z - 2

and from row 1
x = 4 + 3y - 2z
x = 4 + 3(z - 2) - 2z
x = z - 2

---> THIS IS AS FAR AS YOU GOT. ALL YOUR WORK IS CORRECT! NOW ALL THAT REMAINS TO BE DONE IS TO WRITE DOWN THE SOLUTION.

The solution is:
(x, y, z) = (z - 2, z - 2, z)
where z can take any real value.

Geometrically this means that the planes with equations
3x-5y+2z=4
x-3y+2z=4
5x-11y+6z=12
intersect in a line with equation
(x+2)/1 = (y+2)/1 = z/1

Hope this helps.

Answer 2

Eq. #1
3x-5y+2z=4

Eq. #2
x-3y+2z=4

Eq. #3
5x-11y+6z=12

From #2
x=4+3y-2z

If we replace it into #1 and #3, We Have
At Eq. #1:
3[ 4+3y-2z ] -5y + 2z = 4
12 + 9y -6z -5y + 2z = 4
(9-5)y+(2-6)z = 4 -12
4y-4z = -8

and dividing 4 into this, we have Eq. #1-2:
y - z = -2

At Eq. #3:
x=4+3y-2z
5 [ 4 + 3y -2z ] -11y +6z = 12
20 +15y -10z -11y + 6z = 12
(15-11)y + (6-10)z = 12-20
4y -4z = -8

and dividing 4 into this, we have Eq. #3-2:
y - z = -2

So we realize that the system is dependent.
It means that the meaning how the equations are expressed
is repeated twice in the system.

Therefore, there are infinite solutions to the system.

If we parameterize the z-variable like t-parameter,
we have:

z =t
y= z -2 = t - 2

Then, we take the equation #4:
x = 4 + 3y - 2z
x = 4 + 3(t-2) - 2t ==> x = 4 + 3t - 6 - 2t = -2 + t
x = t -2

The solution of the system is:
x = t -2
y = t -2
z = t

We prove it in Equation #1:
3 (t-2) -5 (t-2) + 2t = 4
3t -6 -5t +10 + 2t = 4
(3 -5 + 2) t + 10 - 6 = 4
0t + 4 = 4
4 = 4 [OK]

We prove it in Equation #2:
t-2 -3(t-2) + 2t = 4
t -2 -3t +6t + 2t = 4
(1 -3 + 6 +2)t + 6 -2 = 4
0t + 4 = 4
4 = 4 [OK]

We prove it in Equation #3:
5(t-2) -11(t-2) + 6t = 12
5t -10 -11t +22 +6t = 12
(5 -11 + 6)t + 22 -10 = 12
0t + 12 = 12
12 = 12 [OK]

So if t=0
x = -2
y = -2
z =0

So if t=2
x = 0
y = 0
z =2

There are as many solutions as t-parameter value can take.
Therefore, there are infinite solutions.

Answer 3

Let A be the matrix of coefficients. If |A|=0 then the matrix is singular meaning that the equations are linearly dependent.

|3 -5 2|
|1 -3 2|=A
|5 -11 6|

|A| = 3(-3*6 + 2*11) - 1(-5*6 + 2*11) + 5(-5*2 + 2*3)
|A| = 3*4 +1*8 - 5*4 = 0

Hence there is no unique solution.

Answer 4

take z in the third equation:

z = (-5x + 11y +12)/6
and replace z in equations 1 and 2

3x - 5y + 2/6 (-5x + 11y + 12) = 4
3x - 5y - 5/3x + 11/3y + 4 = 4
9x - 15y - 5x + 11y = 0
4x - 4y = 0
x = y
and z = 6 - 5/3 x + 11/3 y = 6 - 5/3x + 11/3x = 6 + 2x

replace y and z in equation 2
x - 3x + 2(6 + 2x) = 4
x - 3x + 12 + 4x = 4
2x = -12 + 4 = -8
x = -4
y = -4
z = 6 + 2x = 6 - 8 = -2

Answer 5

the place to start is to discover a thank you to combine the two equations into one ideally removing between the unknowns interior the approach. So: x-y=3 could be rearranged to x=3+y Substituting this into the 1st equation provides (3+y)squared + y squared = 29 remedy this to get an answer for y then substitute this selection for y interior the 2nd equation and remedy to get x. answer: y = 2 x = 5

Answer 6

The third equation is the first plus twice the second. There are an infinite number of solutions!

Answer 7

solve for x in the first equation by dividing everything by 3 and arriving at x value, then substitute what you obtain in equation number 2 and solve the two equations with two unknowns)left by solving for y and substituting in the last equation... simple...

Answer 8

hey do with determinants u'll get it.
use cramers rule..