Question

Solve for `x,y and z` given that: `2x-y+2z=-7 ;color(white)("d") -x+2y-4z=5 ;color(white)("d")x+4y-6z=-1 ?`

Answer 1

In general
Eliminate a variable with any 2 equations.
Then eliminate the same variable with 2 different equations.
Now you have 2 eq w 2 unknows ..... qed

Answer 2

Given set of equations is
`2x-y+2z=-7` ......(1)
`-x+2y-4z=5` .......(2)
`x+4y-6z=-1` .......(3)

Let us eliminate `x` by multiplying equation (2) with `2` and add to equation (1)

1). `2xx(-x+2y-4z=5)`
`=>-2x+4y-8z=10`
2). `(-2x+4y-8z=10)+(` `2x-y+2z=-7)`
we get
`3y-6z=3` ........(4)

Now adding equations (2) and (3) we get

`6y-10z=4` ......(5)

Now eliminating `y` from equations (4) and (5) by multiplying (4) with `2` and subtracting from (5)

1. `2xx(3y-6z=3)`
   `=>6y-12z=6`
2. `(6y-10z=4)-(6y-12z=6)`
   we get
   `2z=-2`

   `=>z=-1` .......(6)

Inserting this value in equation (4) we get

`3y-6(-1)=3`
`=>3y=-3`

`=>y=-1` .......(7)

Inserting values of `yand z` in equation (3) we get

`x+4(-1)-6(-1)=-1`
`x-4+6=-1`

`x=-3` ........(8)

Answer 3

Create an augmented matrix from the equations.
Perform elementary row operations, until the left side is an identity matrix.
The values for x, y, and z are in the column vector on the right.

Explanation:

Write the equation, `x+4y-6z=-1`, into the first row of an augmented matrix:

#[ (1,4,-6,|,-1) ]#

Note: I chose to write the last equation first because the x coefficient is 1 and the identity matrix

#[ (1,0,0), (0,1,0), (0,0,1) ]#

has a 1 in position, row 1, column 1; this will save us some work.

Write the equation, `-x+2y-4z=5`, into the second row of the augmented matrix:

#[ (1,4,-6,|,-1), (-1,2,-4,|,5) ]#

Write the equation, `2x-y+2z=-7`, into the third row of the augmented matrix:

#[ (1,4,-6,|,-1), (-1,2,-4,|,5), (2,-1,2,|,-7) ]#

Perform elementary row operations.

Make position row 2 column 1 become 0 by adding Row 1 to Row 2 and putting the result in Row 2:

`R_2+R_1toR_2`

#[ (1,4,-6,|,-1), (0,6,-10,|,4), (2,-1,2,|,-7) ]#

Make position row 3 column 1 become 0 by multiplying row 1 by -2 adding it to row 3 and putting the result in row 3:

`R_2-2R_1toR_2`

#[ (1,4,-6,|,-1), (0,6,-10,|,4), (0,-9,14,|,-5) ]#

Please observe that we have 1 in first row of the first column and 0s in all other rows of the first column; this looks like the first column in the identity matrix

#[ (1,0,0), (0,1,0), (0,0,1) ]#

Usually, we would make row 2 column 2 become 1 but not every element of the row is evenly divisible by 6 and we cannot swap in row 3 and make is coefficient become 1 for the same reason so we shall postpone this. Instead, will we put a 0 in position row 3 column 2 by multiplying row 2 by `3/2` adding it to row 3 and putting the result in row 3:

`R_3 + 3/2R_2toR_3`

#[ (1,4,-6,|,-1), (0,6,-10,|,4), (0,0,-1,|,1) ]#

We can make position row 3 column 1 become 1 by multiplying the row by -1:

`-1R_3toR_3`

#[ (1,4,-6,|,-1), (0,6,-10,|,4), (0,0,1,|,-1) ]#

Because we have a 1 in the third row of the third column, it is easy to make 0s in that column of the other 2 rows:

`R_2+10R_3toR_2`

#[ (1,4,-6,|,-1), (0,6,0,|,-6), (0,0,1,|,-1) ]#

`R_1+6R_3toR_1`

#[ (1,4,0,|,-7), (0,6,0,|,-6), (0,0,1,|,-1) ]#

Now Row 2 is evenly divisible by 6

`1/6R_2toR_2`

#[ (1,4,0,|,-7), (0,1,0,|,-1), (0,0,1,|,-1) ]#

The only thing that we have left is to make the 4 become 0:

`R_1-4R_2toR_1`

#[ (1,0,0,|,-3), (0,1,0,|,-1), (0,0,1,|,-1) ]#

We have an identity matrix on the left, therefore, the column vector on the right contains the solution values.

`x = -3, y = -1, and z = -1`

Answer 4

This is more about demonstrating the method. I have done some of the working out for you.

Explanation:

Have a look at https://socratic.org/help/symbols as a guide to formatting. Note in particular the hash symbol at the beginning and end of the maths part. That indicates to the site's system that it is to format that bit using mathematical structure.

Given:
`2x-y+2z=-7" ".....................Equation(1)`
`-x+2y-4z=5" "......................Equation(2)`
`+x+4y-6z=-1" "..................Equation(3)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(brown)("To solve for three unknown you need three relationships. We have")` `color(brown)("that so lets take a look.")`

Notice that `Equation(2) and Equation(3)` both have a single `x`. So we can use those to eliminate it.

`Eqn(2) + Eqn(3)`

`-x+2y-4z=5" ".................Eqn(2)`
`ul(+x+4y-6z=-1)" ".............Eqn(3)`
`color(white)("...")0+6y-10z=4color(white)("bbnb")larr" Added together"`

Add `10z` to both sides giving

`6y=10z+4`

Subtract 4 from both sides

`10z=6y-4`

Divide both sides by 10

`z=3/5y-2/5.........Equation(4)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Using `Eqn(4)` substitute for `z` in `Eqn's color(white)(.) 1,2and 3`

`2x-y+2(3/5y-2/5)=-7" ".....................Equation(1_a)`
`-x+2y-4(3/5y-2/5)=5" "....................Equation(2_a)`
`+x+4y-6(3/5y-2/5)=-1" "..................Equation(3_a)`

Now we have only 2 unknowns. The `x and y.`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Add `Eqn(2_a) and Eqn(3_a)` giving

`0+6y-10(3/5y-2/5)=4`

`6y-6y+4=4color(red)(larr "That one did not help")`

`4=4` is a well known fact and of no use to us at all.

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
So lets try another combination.

`Eqn(1_a)+[color(white)(2/2)2xxEqn(2_a)color(white)(.)]`

`color(white)(..)2x-y+2(3/5y-2/5)=-7" "........Eqn(1_a)`
`ul(-2x+4y-8(3/5y-2/5)=10)" "........2xxEqn(2_a)`
`color(white)("nn.")0 +3y-6(3/5y-2/5)= 3larr" Added together"`

`-3/5y+12/5=3`

`3/5y=12/5-3`

`y=-1`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Now that I have demonstrated the way of thinking about this type of question I decided to leave the rest for you to try.

By the way. I have not checked my numbers. So I would suggest that you run through them to make sure I have not 'dropped a clanger'.