solve for the following system of 3 equations. Find x and z for: x+2y-2z=1; x-2y+3z=4; 2X+y+Z=3

Question

 I tried making this into a system of 2 equations and solving but I keep getting the answer incorrect. Can someone please help?

Answer 1

We can use the matrix method also

 

We have from the coefficients

 

r1:    1    2     -2     1

r2:    1   -2      3     4

r3:    2    1      1     3

 

What want is a matrix that looks like:

 

r1:   1   ...     ...   ...   

       0     1     ...   ...   

       0     0       1  ...       The blanks will be other numbers

 

Let's get a zero in the 1st column of row 2

Multiply r1 by -1 and add to r2

 

1    2    -2    1

0   -4     5    3       r1(-1) +r2

2    1     1    3

 

Let's get a zero in the 1st column of r3

Multiply r1 by -2 and add to r3

 

1    2    -2    1

0   -4     5    3

0   -3     5    1      -2(r1) + r3

 

Let's get a 1 in the 2nd column of r2

Divide r2 by -4

 

1    2     -2       1

0    1   -5/4   -3/4      r2/(-4)

0   -3      5       1

 

Let's get a zero in column 2 of r3

Multiply r2 by 3 and add to r3

 

1    2     -2       1

0    1   -5/4   -3/4

0    0    5/4    -5/4     3r2 + r3

 

Now all we need is a 1 in column3 of r3

Multiply r3 by 4/5

 

1    2    -2      1

0    1   -5/4   -3/4

0    0     1     -1         (4/5)r3

 

From row 3 we have z = -1

 

Substituting into row 2 we have:

y -5/4(-1) = -3/4

y + 5/4 = -3/4

y = -8/4

y = -2

 

Substituting those values into row 1 we have:

x + 2(-2) -2(-1) = 1

x -4 + 2 = 1

x-2 = 1

x = 3

 

Final answer:   x=3, y=-2, z=-1

 

 

Answer 2

So, as you can see by the fact that two people came up with different answer, it is important to take x,y, and z and plug them into the original equations and check to see if the answer you derived actually works.

 

I leave it up yo the student to figure out which answer is correct and which isn't.

Answer 3

x + 2y - 2z = 1             eq1

x - 2y + 3z = 4             eq2

2x + y + z = 3               eq3

 

 

Since we want to solve for x and z, we need to use the substitution and elimination methods to reduce the number of equations to work with.  We want these equations to only have x and z variables.  The y variables is not that important to us.

 

We can substitute eq3 into the first two equations.  From eq3.

 

y = 3 - 2x - z

 

 

Notice that the variable y is expressed in terms of only x and z.

 

x + 2(3 - 2x - z) - 2z = 1

x + 6 - 4x - 2z - 2z = 1

-3x - 4z = -5                   new eq1

 

 

x - 2(3 - 2x - z) + 3z = 4

x - 6 - 4x - 2z + 3z = 4

-3x + z = 10                    new eq2

 

 

Now we only have two equations with only two variables x and z.

 

-3x - 4z = -5                new eq1

-3x + z = 10                new eq2

 

 

Subtract new eq2 from new eq1 to eliminate the x terms.

 

-5z = -15

 

   z = 3

 

 

Substitute this value of z into new eq1 to solve for x.  I leave this part to you finish off.

 

Answer 4

Solving the third equation for y, we get y = 3-2x-z

 

Replace y by 3-2x-z in each of the other two equations to obtain:          

 

-3x - 4z = -5

 5x +5z = 10

 

Divide the second equation by 5:   -3x - 4z = -5

                                                   x + z  =  2

 

Multiply the second equation by 3;  -3x - 4z = -5

                                                   3x +3z = 6

 

Add the equations to get z = -1

 

So, x = 3 and y = -2