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mariarad [96]
3 years ago
6

Pls do the question on the bottom of the page

Mathematics
1 answer:
djverab [1.8K]3 years ago
7 0
48% free-throws 24/25
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Solve the system <br> 2x+2y+z=-2 <br> -x-2y+2z=-5<br> 2x+4y+z=0
drek231 [11]

The values of (x,y,z) are (3,-1,-2) , if the given are equations 2x+2y+z=- 2,-x-2y+2z=-5 and 2x+4y+z=0.

Step-by-step explanation:

The given is,

                          2x+2y+z=- 2.......................................(1)

                         -x-2y+2z=-5......................................(2)

                            2x+4y+z=0.........................................(3)

Step:1

           Equation (2) is multiplied by -1            ( Eqn(2) × -1 )

                                         x+2y-2z=5.............................(4)

          Subtract the equation (1) and (4)

                                        2x+2y+z=- 2

                                         x+2y-2z=5

                 ( - )

           (2x-x)+(2y-2y)+(z+2z)=(-2-5)

                                                  x+3z=-7......................(5)

Step:2

          Equation (2) is multiplied by -2                 ( Eqn(2) × -2)

                                        2x+4y-4z=10........................(6)

         Subtract equation (6) and (3),                  

                                        2x+4y-4z=10

                                         2x+4y+z=0

                   ( - )

       (2x-2x)+(4y-4y)+(-4z-z)= (10-0)

                                                     -5z=10

                                                         z = - \frac{10}{5}

                                                         z = -2

         From the equation (5),

                                                  x+3z=-7  

                                          x+(3)(-2)=-7

                                                           x = -7+6

                                                            x = -1

         From equation (1),

                                            2x+2y+z=- 2

          Substitute x and z values,

                               (2)(-1)+2y+(-2)=-2

                                                     2y - 4=2

                                                           2y=4-2

                                                           2y=2

                                                            y=\frac{2}{2}

                                                             y = 1

Step:3

                Check for solution,

                                  -x-2y+2z=-5

                Substitute x,y and z values,

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

                                         1-2-4=-5

                                                    -5 = -5

Result:

              The values of (x,y,z) are (3,-1,-2) , if the given are equations 2x+2y+z=- 2,-x-2y+2z=-5 and 2x+4y+z=0.

8 0
4 years ago
What is the value of X How do u do it
Nutka1998 [239]
Hello here is a
solution :

5 0
4 years ago
I NEED HELP PLEASE IM BEGGING YOU
horrorfan [7]

Answer:

$2

$1.26

Step-by-step explanation:

pls give brainliest if you can

8 0
3 years ago
Read 2 more answers
6x + 12 = 10x − 4 1. Solve for x using two different methods
solong [7]

Answer:

x=53/4 or 4x=53

Step-by-step explanation:

First move the variable to the left-hand side and change its sign.

6x+12-10x=-41 to 6x-10x=-41-12

Then collect the like terms

6x-10x=-41-12 to -4x=-41-12 which equals to -4x=-53

Now you divide both sides of the equation by -4

-4x=-53 to x=53 over 4 then if you want you divide which gives you x=13.25

Second way:

Kind of the same thing as the first one

Move the variable to the left-hand side and change its sign

6x+12=10x-41 to 6x=12-10x=-41

Then you collect the like terms

6x-10x=-41-12 to -4x=-41-12

Now you multiply both sides of the equation by -1

-4x=-53 to 4x=53

5 0
3 years ago
What is the sum of 2 10/12 + 3 3/12
aliina [53]
6 1/12 am I right? I'm pretty sure I'm right
7 0
3 years ago
Read 2 more answers
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