Question
Asked Oct 30, 2019

I am stuck on this equation and unsure how to solve. Here is the equation

Solve by system of equations: 

 
5x+3y+z
=
−8
x−3y+2z
=
20
14x−2y+3z
=
20

 

check_circleExpert Solution
Step 1

Given,

5x 3y z-8
(1)
x-3y 2z 20 ->(2)
14x-2y 3z 20 ->(3)
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5x 3y z-8 (1) x-3y 2z 20 ->(2) 14x-2y 3z 20 ->(3)

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Step 2
Equation (1) Equation (2), we get
(5x+3y z)+(x-3y +2z)8+20
6x+3z 12 -»(4)
2x Equation (2)-3x Equation (3), we get
2(x-3y 2z-3(14x -2y +3z) (2x 20) -(3 x 20)
40x-5z= -20 ->(5)
help_outline

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Equation (1) Equation (2), we get (5x+3y z)+(x-3y +2z)8+20 6x+3z 12 -»(4) 2x Equation (2)-3x Equation (3), we get 2(x-3y 2z-3(14x -2y +3z) (2x 20) -(3 x 20) 40x-5z= -20 ->(5)

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Step 3

Now solving equations...

5x Equation (4)+3x Equation (5), we get
5(6x+3z)+3(-40x - 5z) = (5x12)+(3x(-20))
-90x-30
1
3
1
Now substituting x=-in equation (4), we get
3
1
+3z 12 2+3z = 12 3z =10z
10
61
10
Now substituting x =-and z =in equation (1), we get
3
3
10
+3y+
3
15
+3y 85+3y -8= 3y13 y-
3
13
5
_
3
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5x Equation (4)+3x Equation (5), we get 5(6x+3z)+3(-40x - 5z) = (5x12)+(3x(-20)) -90x-30 1 3 1 Now substituting x=-in equation (4), we get 3 1 +3z 12 2+3z = 12 3z =10z 10 61 10 Now substituting x =-and z =in equation (1), we get 3 3 10 +3y+ 3 15 +3y 85+3y -8= 3y13 y- 3 13 5 _ 3

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