Question 310893: Hi, If you could please show me the steps on how to get the answer to this. I am not sure where to start. i really appreciate any help
4x+y+z=3
-x+y-2z=-11
x+2y+2z=-1
Answer by mollukutti(30)   (Show Source):
You can put this solution on YOUR website! Please remember when you get multiple linear equations like these, start by eliminating the variables.
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1> 4x+y+z=3
2> -x+y-2z=-11
3> x+2y+2z=-1
We will start with equations 1 & 2
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4x + y + z = 3
-x + y - 2z = -11 .... multiply by 4 so as to make the co-efficients of x equal
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4x + y + z = 3
-4x + 4y - 8z = -44
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5y - 7z = -41 (Obtained by adding the above 2 equations)
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Let us consider this as our 4th equation
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4> 5y - 7z = -41
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Now let us take equations 2 & 3
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-x + y - 2z = -11
x + 2y + 2z = -1 .... Here already the co-efficients of x are equal
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3y = -12 (Obtained by adding the above 2 equations)
or, y = -4
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Now let us substitute the value of y in equation (4)
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5y - 7z = -41
or, 5.-4 - 7z = -41
or, -20 -7z = -41
or, -7z = -41 + 20
or, -7z = -21
or, z = 3
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So we have y=-4 and z=3 and we need to fin the value of x hence substitute the values in equation (1) or (2) or (3)
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4x+y+z=3
or, 4x + (-4) + 3 = 3
or, 4x -4 + 3 =3
or, 4x -1 = 3
or, 4x = 3 + 1
or, 4x = 4
or, x = 1
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Hence we have solved for all 3 as follows:
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x=1
y= -4
z=3
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