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how to solve and eliminate variables for 4x-3y+6z=18 and -x+5y+4z=48 and 6x-2y+5x = 0

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asked Sep 25, 2017 in ALGEBRA 2 by anonymous
reshown Sep 26, 2017 by bradely

1 Answer

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4x - 3y + 6z  =  18 -------------------> (1)

-x + 5y + 4z  =  48 -------------------> (2)

6x - 2y + 5z  =  0 ---------------------> (3)

Multiply Eq (1) with 2, Eq (3) with 3 and subtract them

Eq (1) X 2 =======>    8x - 6y + 12z  =  36

Eq (3) X 3 =======>  18x - 6y + 15z  =   0

                                  (-)     (+)    (-)        (-)

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

                                    -10x - 3z  =  36 --------------------------> (4)

Multiply Eq (2) with 2, Eq (3) with 5 and add them

Eq (2) X 2 =======>    -2x + 10y + 8z  =  96

Eq (3) X 5 =======>   30x - 10y + 25z  =  0

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

                                     28x + 33z  =  96 ------------------------> (5)

Multiply Eq (4) with 11 and add to Eq (5)

Eq (4) X 11 =======>   -110x - 33z  =  396

Eq (3) X 5 ========>      28x + 33z  =  96

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

                                              -82x  =  492

        ​                                        x  =  - 492/82

        ​                                        x  = - 6

Substitute x = - 6 in Eq (4)

-10(-6) - 3z  =  36

60 - 3z  =  36

- 3z  =  36 - 60

- 3z  =  -24

3z  =  24

z  =  24/3

z  =  8

Substitute x = - 6 and z = 8 in Eq (2)

-(-6) + 5y + 4(8)  =  48
 
6 + 5y + 32  =  48
 
5y + 38  =  48
 
5y  =  48 - 38
 
5y  =  10
 
y  =  10/5
 
y  =  2
 
Answer :
 
Solutions are x = -6, y = 2 and z = 8.
answered Oct 11, 2018 by homeworkhelp Mentor

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