Questions on Algebra: Systems of Linear Equations answered by real tutors!

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Question 152808: solve the system:
x+3y+5z=20
y-4z=-16
3x-2y+9z=36
: solve the system:
x+3y+5z=20
y-4z=-16
3x-2y+9z=36

Answer by Electrified_Levi(89) About Me  (Show Source):
You can put this solution on YOUR website!
Hi, Hope I can help,
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solve the system:
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x+3y+5z=20
y-4z=(-16)
3x-2y+9z=36
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This is the way I usually solve these problems(pretty easy once you know how to do it) ( There is no fast way to solve these problems)
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First, solve for a letter in all the equations( since we already have a 2 system/variable equation for equation 2, we don't have to solve for any letter in equation 2)( Since  y - 4z = (-16) that means we will need to solve for "x" in our other equations)
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We will rewrite the 3 equations so it makes more sense( the second equation has no "x's" in the equation so it has "0x")
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x+3y+5z=20
0x + y - 4z =( -16)
3x-2y+9z=36
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We will switch the equations around
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x+3y+5z=20
3x-2y+9z=36
0x + y - 4z = (-16)
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We will solve "x" in our first two equations(can't solve "x" in our third equation, since it is 0x)
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First equation
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 x+3y+5z=20
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We will move "3y" over to the right side
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 x+3y+5z=20 =  x+3y - 3y +5z=20 -3y
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 x+3y - 3y +5z=20 -3y =  x + 5z= 20 - 3y
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We will move "5z" to the right side
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 x + 5z= 20 - 3y =  x + 5z - 5z = 20 - 3y - 5z
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 x + 5z - 5z = 20 - 3y - 5z =  x = 20 - 3y - 5z
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We will switch the letters around
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 x = 20 - 3y - 5z =  x = (-3y) - 5z + 20
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This is our First Answer  (-3y) - 5z + 20
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We will now solve "x" in our second equation
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 3x-2y+9z=36
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We will move (-2y) to the right side
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 3x-2y+9z=36 =  3x-2y + 2y +9z=36 + 2y
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 3x-2y + 2y +9z=36 + 2y =  3x + 9z = 36 + 2y
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We will move "9z" to the right side
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 3x + 9z = 36 + 2y =  3x + 9z - 9z = 36 + 2y - 9z
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 3x + 9z - 9z = 36 + 2y - 9z =  3x = 36 + 2y - 9z
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We will switch the letters around
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 3x = 36 + 2y - 9z =  3x = 2y - 9z + 36
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We will now divide each side by "3"
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 3x = 2y - 9z + 36 =  3x/3 = (2y - 9z + 36)/3
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 3x/3 = (2y - 9z + 36)/3 =  cross (3)x/cross(3) = (2y - 9z + 36)/3
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 x = (2y - 9z + 36)/3
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Our second answer =  (2y - 9z + 36)/3
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We can now solve to get a 2 system(variable) equation
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We will put our two answers together in an equation, since "x" = both of the answers, our answers equal each other
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First answer =  (-3y) - 5z + 20
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Second answer =  (2y - 9z + 36)/3
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Our equation will equal  (-3y) - 5z + 20 = (2y - 9z + 36)/3
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 (-3y) - 5z + 20 = (2y - 9z + 36)/3 =  ((-3y) - 5z + 20)/1 = (2y - 9z + 36)/3
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We will use cross multiplication to get rid of the fractions
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 highlight((-3y) - 5z + 20)/1 = (2y - 9z + 36)/ highlight(3)
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 (-9y) - 15z + 60 = 2y - 9z + 36
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We will move (-9y) to the right side
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 (-9y) + 9y - 15z + 60 = 2y + 9y - 9z + 36
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 (-9y) + 9y - 15z + 60 = 2y + 9y - 9z + 36 =  (-15z) + 60 = 11y - 9z + 36
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We will move the (-15z) to the right side
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 (-15z) + 60 = 11y - 9z + 36 =  (-15z) + 15z + 60 = 11y - 9z + 15z + 36
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 (-15z) + 15z + 60 = 11y - 9z + 15z + 36 =  60 = 11y + 6z + 36
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We will move the "36" to the left side
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 60 = 11y + 6z + 36 =  60 - 36 = 11y + 6z + 36 - 36
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 60 - 36 = 11y + 6z + 36 - 36 =  24 = 11y + 6z
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We will rearrange,  24 = 11y + 6z =  11y + 6z = 24
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Our second 2 system/variable equation =  11y + 6z = 24
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We will put our two 2 system/variable equations side by side
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First equation =  y-4z=(-16)
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Second equation =  11y + 6z = 24
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 y-4z=(-16)
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 11y + 6z = 24
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We will now need to solve for a letter again( we will solve "y" since it is the easiest to solve)
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First equation  y-4z=(-16)
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We will move (-4z) to the right side
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 y-4z=(-16) =  y-4z + 4z =(-16) + 4z
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 y-4z + 4z =(-16) + 4z =  y = (-16) + 4z
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 y = (-16) + 4z =  y = 4z - 16 ( rearranging the numbers)
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Our first answer =  4z - 16
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Our second equation  11y + 6z = 24
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We will move "6z" to the right side
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 11y + 6z = 24 =  11y + 6z - 6z = 24 - 6z
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 11y + 6z - 6z = 24 - 6z =  11y = 24 - 6z
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Rearranging  11y = 24 - 6z =  11y =  (-6z) + 24
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We will divide each side by "11"
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 11y =  (-6z) + 24 =  11y/11 =  ((-6z) + 24)/11
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 11y/11 =  ((-6z) + 24)/11 =  cross (11)y/cross (11) =  ((-6z) + 24)/11
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 cross (11)y/cross (11) =  ((-6z) + 24)/11 =  y =  ((-6z) + 24)/11
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Our second answer is  ((-6z) + 24)/11
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We can now put both of our answers into an equation( since "y" equals both of our answers)
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Answer 1 =  4z - 16
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Answer 2 =  ((-6z) + 24)/11
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We can make our equation, it equals  4z - 16 = ((-6z) + 24)/11
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 4z - 16 = ((-6z) + 24)/11 =  (4z - 16)/1 = ((-6z) + 24)/11
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We will cross multiply to get rid of the fractions
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 highlight(4z - 16)/1 = ((-6z) + 24)/highlight(11)
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It becomes,  44z - 176 = (-6z) + 24
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We will move (-6z) to the left side
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 44z - 176 = (-6z) + 24 =  44z + 6z - 176 = (-6z) + 6z + 24
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 44z + 6z - 176 = (-6z) + 6z + 24 =  50z - 176 = 24
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We will move (-176) to the right side
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 50z - 176 = 24 =  50z - 176 + 176 = 24 + 176
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 50z - 176 + 176 = 24 + 176 =  50z = 200
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We will divide each side by "50"
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 50z = 200 =  50z/50 = 200/50
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 50z/50 = 200/50 =  cross(50)z/cross(50) = 4
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 cross(50)z/cross(50) = 4 =  z = 4
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We found that "z" = 4, we can replace "z" with "4" in one of our 2 system/variable equations
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 y-4z=(-16)
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 11y + 6z = 24
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We will use the first equation
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 y-4z=(-16) =  y-4(4)=(-16)
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 y-4(4)=(-16) =  y - 16 = (-16)
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We will move (-16) to the right
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 y - 16 = (-16) =  y - 16 + 16 = (-16) + 16
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 y - 16 + 16 = (-16) + 16 =  y = 0
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We found that "y" = "0"
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We can now find "x", we need to replace "y" and "z" in one of the three original equations
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y = 0
z = 4
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x+3y+5z=20
y-4z=(-16)
3x-2y+9z=36
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We will use the first equation, since it will be the easiest( we can't use the second equation)
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 x+3y+5z=20 =  x+3(0)+5(4)=20
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 x+3(0)+5(4)=20 =  x+0+20=20
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 x+0+20=20 =  x+20=20
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We will move the "20" over to the right side
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 x+20=20 =  x+20-20=20 - 20
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 x=0
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We found that "x" = "0", we now have all 3 variables, "x","y", and "z"
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x = 0
y = 0
z = 4
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We can check by replacing the letters with numbers in our third equation
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 3x-2y+9z=36 =  3(0)- 2(0) + 9(4) = 36
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 3(0)- 2(0) + 9(4) = 36 =  0 - 0 + 36 = 36
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 0 - 0 + 36 = 36 =  36 = 36 ( True)
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x = 0
y = 0
z = 4
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The solution set = (x,y,z), our solution set = (0,0,4)
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Hope I helped, Levi