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Question 347565: solve the system using eliminaton or sustitution
-3x+y-z=-2
2x-y-2z=-12
4x+2y+z=1
Answer by haileytucki(390)   (Show Source):
You can put this solution on YOUR website! -3x+y-z=-2_2x-y-2z=-12_4x+2y+z=1
Move all terms not containing y to the right-hand side of the equation.
y=3x+z-2_2x-y-2z=-12_4x+2y+z=1
Replace all occurrences of y with the solution found by solving the last equation for y. In this case, the value substituted is 3x+z-2.
y=3x+z-2_2x-(3x+z-2)-2z=-12_4x+2y+z=1
Replace all occurrences of y with the solution found by solving the last equation for y. In this case, the value substituted is 3x+z-2.
y=3x+z-2_2x-(3x+z-2)-2z=-12_4x+2(3x+z-2)+z=1
Multiply -1 by each term inside the parentheses.
y=3x+z-2_2x-3x-z+2-2z=-12_4x+2(3x+z-2)+z=1
Since 2x and -3x are like terms, add -3x to 2x to get -x.
y=3x+z-2_-x-z+2-2z=-12_4x+2(3x+z-2)+z=1
Since -z and -2z are like terms, subtract 2z from -z to get -3z.
y=3x+z-2_-x-3z+2=-12_4x+2(3x+z-2)+z=1
Multiply 2 by each term inside the parentheses.
y=3x+z-2_-x-3z+2=-12_4x+6x+2z-4+z=1
Since 4x and 6x are like terms, add 6x to 4x to get 10x.
y=3x+z-2_-x-3z+2=-12_10x+2z-4+z=1
Since 2z and z are like terms, add z to 2z to get 3z.
y=3x+z-2_-x-3z+2=-12_10x+3z-4=1
Move all terms not containing x to the right-hand side of the equation.
y=3x+z-2_-x=3z-2-12_10x+3z-4=1
Subtract 12 from -2 to get -14.
y=3x+z-2_-x=3z-14_10x+3z-4=1
Multiply each term in the equation by -1.
y=3x+z-2_-x*-1=3z*-1-14*-1_10x+3z-4=1
Multiply -x by -1 to get x.
y=3x+z-2_x=3z*-1-14*-1_10x+3z-4=1
Simplify the right-hand side of the equation by multiplying out all the terms.
y=3x+z-2_x=-3z+14_10x+3z-4=1
Replace all occurrences of x with the solution found by solving the last equation for x. In this case, the value substituted is -3z+14.
y=3x+z-2_x=-3z+14_10(-3z+14)+3z-4=1
Multiply 10 by each term inside the parentheses.
y=3x+z-2_x=-3z+14_-30z+140+3z-4=1
Since -30z and 3z are like terms, subtract 3z from -30z to get -27z.
y=3x+z-2_x=-3z+14_-27z+140-4=1
Subtract 4 from 140 to get 136.
y=3x+z-2_x=-3z+14_-27z+136=1
Since 136 does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 136 from both sides.
y=3x+z-2_x=-3z+14_-27z=-136+1
Add 1 to -136 to get -135.
y=3x+z-2_x=-3z+14_-27z=-135
Divide each term in the equation by -27.
y=3x+z-2_x=-3z+14_-(27z)/(-27)=-(135)/(-27)
Simplify the left-hand side of the equation by canceling the common factors.
y=3x+z-2_x=-3z+14_z=-(135)/(-27)
Simplify the right-hand side of the equation by simplifying each term.
y=3x+z-2_x=-3z+14_z=5
Replace all occurrences of z with the solution found by solving the last equation for z. In this case, the value substituted is 5.
y=3x+(5)-2_x=-3z+14_z=5
Replace all occurrences of z with the solution found by solving the last equation for z. In this case, the value substituted is 5.
y=3x+(5)-2_x=-3(5)+14_z=5
Remove the parentheses around the expression 5.
y=3x+5-2_x=-3(5)+14_z=5
Subtract 2 from 5 to get 3.
y=3x+3_x=-3(5)+14_z=5
Multiply -3 by each term inside the parentheses.
y=3x+3_x=-15+14_z=5
Add 14 to -15 to get -1.
y=3x+3_x=-1_z=5
Replace all occurrences of x with the solution found by solving the last equation for x. In this case, the value substituted is -1.
y=3(-1)+3_x=-1_z=5
Multiply 3 by each term inside the parentheses.
y=-3+3_x=-1_z=5
Add 3 to -3 to get 0.
y=0_x=-1_z=5
This is the solution to the system of equations.
y=0_x=-1_z=5
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