Question 1151845
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I'll get you started on each problem. 


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Problem 1

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2x+y = 5 ............ equation (1)
-x - 2y + 8z = 8 .... equation (2)
-y + 4z = 6 ......... equation (3)
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Note how there's a 'y' term in the first equation that has a coefficient of 1. This allows us to easily solve for y without involving fractions.
2x+y = 5
2x+y-2x = 5-2x
y = -2x+5


This can be plugged into the other equations to get rid one of the variables


Let's plug it into equation (2)
-x - 2y + 8z = 8
-x - 2( y ) + 8z = 8
-x - 2( -2x+5 ) + 8z = 8 .... replace y with -2x+5, y is eliminated temporarily
-x + 4x - 10 + 8z = 8
3x - 10 + 8z = 8
3x - 10 + 8z + 10 = 8 + 10 ... add 10 to both sides
3x + 8z = 18 .... Call this equation (4)



Plug y = -2x+5 into equation (3) now
-y + 4z = 6
-( y ) + 4z = 6
-( -2x+5 ) + 4z = 6
2x - 5 + 4z = 6
2x - 5 + 4z+5 = 6+5 ... add 5 to both sides
2x + 4z = 11 .... Call this equation (5)



The system of equations we were given reduces to this
3x + 8z = 18 .... Equation (4)
2x + 4z = 11 .... Equation (5)


I'll let you take over from here.


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Problem 2


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3x + 4y + z = 15 .... equation (1)
x - 2y + z = -15 .... equation (2)
2x + y + z = 5 ...... equation (3)
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Solve equation (1) for z
3x + 4y + z = 15
3x + 4y + z-3x = 15-3x .... subtract 3x from both sides
4y+z = 15-3x
4y+z-4y = 15-3x-4y .... subtract 4y from both sides
z = -3x-4y+15


Move onto equation (2) and plug in z = -3x-4y+15
x - 2y + z = -15
x - 2y + ( z ) = -15
x - 2y + ( -3x-4y+15 ) = -15 ... replace z with  -3x-4y+15
x - 2y - 3x-4y+15 = -15
-2x - 6y + 15 = -15
-2x - 6y + 15-15 = -15-15
-2x - 6y = -30 .... call this equation (4)


Do the same for equation (3)
2x + y + z = 5
2x + y + (z) = 5
2x + y + (-3x-4y+15) = 5  ... replace z with  -3x-4y+15
2x + y - 3x-4y+15 = 5
-x -3y + 15 = 5
-x -3y + 15-15 = 5-15
-x -3y = -10 .... call this equation (5)



Equations (4) and (5) form a simpler system
-2x - 6y = -30
-x - 3y = -10


I'll let you solve this system. 
Once you know x and y, use those values to find z.

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