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Find x,y,z 2x+3y+5z=30, x+y+z=10, 2x-5y+4z=20

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Reduction method
asked Nov 2 in BASIC MATH by anonymous
reshown Nov 2 by bradely

1 Answer

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The system of equations are
 
2x + 3y + 5z  =  30 ---------------------> (1)
 
x + y + z  =  10 ----------------------> (2)
 
2x - 5y + 4z  =  20 ------------------------> (3)
 
Multiply Eq (2) with 5 and add to Eq (3)
 
Eq (2) X 5 ======>  5x + 5y + 5z  =  50
 
Eq (3) =========>  2x -  5y + 4z  =  20
                            ----------------------------------
 
                                        7x + 9z  =  70 --------------------> (4)
 
Multiply Eq (2) with 5 and subtract it from Eq (1)
 
 
Eq (3) =========>  2x + 3y + 5z  =  30
 
Eq (2) X 3 ======>  3x + 3y + 3z  =  30
 
                                (-)   (-)    (-)         (-)
 
                             ----------------------------------
 
                                     - x + 2z  =  0
 
                                       x  =  2z --------------------------> (5)
 
Substitute x = 2z in Eq (4)
 
7(2z) + 9z  =  70
 
14z + 9z  =  70
 
23z  =  70
 
z  =  70/23
 
Substitute z = 70/23 in Eq (5)
 
x  =  2(70/23)
 
x  =  140/23
 
Substitute z = 70/23 and z = 140/23 in Eq (2)
 
70/23 + y + 140/23  =  10
 
(70 + 140)/23 + y  =  10
 
(210)/23 + y  =  10
 
y  =  10 - (210)/23
 
y  =  (230 - 210)/23
 
y  =  20/23
 
Answer :
 
The Solutions are z = 70/23,  y = 20/23  and  z = 140/23.
answered Nov 3 by homeworkhelp Mentor

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