How do you do this??

Question

x+2y+3z=5

3x+2y-2z=-13

5x+3y-z=-11

 

and what is this subject of Algebra 2 called? I need the name of these types of problems so I can look for help on the internet. 

 

Thanks!

Answer 1

SOLVING A SYSTEM OF 3 SIMULTANEOUS EQUATIONS:

 

You can use elimination to solve the system.

Note that if two equations are true then their

sum or difference is also true.  For example:

     (-2 + 5 = 3)

+   (7 - 10 = -3)   add them together

----------------

       5 - 5 = 0        the sum is also true

 

  (-2 + 5 = 3)

- (7-10 = -3)

----------------

   -9 + 15 = 6      The difference is also true

 

Also, for any true equation, if you multiply the whole

equation through by any number then the product

is also true.  For example:

11 - 7 = 4

multiply the whole equation by 2

2(11-7=4) gives  22 - 14 = 8 which is also true.

 

This includes dividing an equation by any number

which is merely multiplying by a fraction.

 

Those are the procedures we will use to solve the

system of equations.

***************************************

 

equation 1:    x+2y+3z=5
equation 2:   3x+2y-2z=-13
equation 3:    5x+3y-z=-11

 

We first want to create two new equations with 2 variables.

Note that equation 1 and equation 2 both contain 2y.

Let's subtract equation 2 from equation 1

 

  (x+2y+3z=5)

-(3x+2y-2z=-13)

--------------------

-2x + 0y + 5z = 18 

 

-2x + 5z = 18    (equation 1) - (equation 2)

 

Now we need to take another pair of equations

and again eliminate the y variable.  We can use

either equations 1 and 3 or equations 2 and 3

because we already did this with equations 1 and 2.

Look at equations 1 and 3.  Equation 1 has 2y and

equation 3 has 3y.  If we multiply equation 1 by

3 and equation 3 by -2 we will then have an

equation with 6y and an equation with -6y which

we can then add to eliminate y again.

 

  3(x+2y+3z=5) = 3x + 6y + 9z = 15    3(equation 1)

-2(5x+3y-z=-11)=-10x-6y + 2z = 22    -2(equation 3)

                          -----------------------

                           -7x + 11z = 37

 

We now have two equations with 2 unknowns which
we can use to solve for x and z.

 

-2x + 5z = 18      (equation 1)-(equation 2)

-7x + 11z = 37   3(equation 1)-2(equation 3)

 

We can use the same procedure to eliminate either x or z

from this system.  Let's eliminate x just because it's easier.

 

  7(-2x + 5z = 18) = -14x + 35z = 126

-2(-7x + 11z = 37) = 14x - 22z = -74

                               --------------------

                                       13z = 52

                                         z = 52/13

                                         z = 4

 

We can now use one of our two variable equations,

plug in the value z=4 and solve for x.

 

-2x + 5z = 18

-2x + 5(4) = 18

-2x + 20 = 18

-2x = -2

x = 1

 

We now have the value of x and z.

Plug those values back into one of the

original 3 equations to solve for y.

 

x+2y+3z=5

1 + 2y +3(4) = 5

2y + 13 = 5

2y = -8

y = -4

 

Final answer:  x=1, y=-4, z=4

 

To check your answer you can plug these values into the

starting equations to verify they are correct.

 

x+2y+3z=5

1+2(-4)+3(4)=5

1-8+12 = 5

13-8=5

5=5

 

3x+2y-2z=-13

3(1)+2(-4)-2(4)=-13

3-8-8 = -13

3-16=-13

-13=-13

 

5x+3y-z=-11

5(1)+3(-4)-4=-11

5-12-4=-11

5-16=-11

-11 = -11

 

as can be seen our answer satisfies all 3 given equations.

 

Hope this helps.  As Oscar said this can be done utilizing

matrices also.  You can look up an example of that and

determine which procedure you prefer.  Good luck.  Let

me know if there are any questions.

 

 

 

 

 

Answer 2

The thing you're looking at is:

"System of (three) Linear Equations"

Answer 3

  Try looking up Matrix Algebra..it should help