Question 1203611: x+4y-6z=-1
2x-y+2z=-7
-x+2y-4z=5
Found 6 solutions by josgarithmetic, MathLover1, math_tutor2020, greenestamps, ikleyn, MathTherapy:
Answer by josgarithmetic(39617)   (Show Source):
You can put this solution on YOUR website! You could try a site like matrixcalc.org and enter the matrix for that system. You can check the steps included there if interested or need. Solution for your given system was indicated to be this:
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-3=x
-1=y
-1=z
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Answer by MathLover1(20850)  (Show Source):
Answer by math_tutor2020(3817) (Show Source):
You can put this solution on YOUR website!
I'll show how to solve this system using matrix row reduction.
The goal is to get the matrix into reduced row echelon form (RREF).
We have this given system
x+4y-6z=-1
2x-y+2z=-7
-x+2y-4z=5
which is the same as
1x+4y-6z=-1
2x-1y+2z=-7
-1x+2y-4z=5
It forms this matrix
The last column represents the right hand values -1, -7 and 5.
The rest of the matrix represents the coefficients.
Normally a matrix does not have separating grid lines, which is unfortunate. But I'll use grid lines to help separate the values. It should hopefully make things look a bit cleaner.
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Here are the steps for RREF.
| 1 | 4 | -6 | -1 | | | 0 | -9 | 14 | -5 | R2 - 2*R1 --> R2 | | -1 | 2 | -4 | 5 | |
Notation like R2 - 2*R1 --> R2 means "subtract off twice of row 1 from row 2. Then store the results in row 2 (we overwrite row 2)".
| 1 | 4 | -6 | -1 | | | 0 | -9 | 14 | -5 | | | 0 | 6 | -10 | 4 | R3+R1 --> R3 |
| 1 | 4 | -6 | -1 | | | 0 | 1 | -14/9 | 5/9 | (-1/9)*R2 -> R2 | | 0 | 6 | -10 | 4 | |
| 1 | 4 | -6 | -1 | | | 0 | 1 | -14/9 | 5/9 | | | 0 | 0 | -2/3 | 2/3 | R3 - 6*R2 -> R3 |
| 1 | 4 | -6 | -1 | | | 0 | 1 | -14/9 | 5/9 | | | 0 | 0 | 1 | -1 | (-3/2)*R3 -> R3 |
The matrix is in row echelon form (REF) but not RREF. This is because we have a lower triangular region of zeros below the main diagonal pivot entries.
| 1 | 4 | -6 | -1 | | | 0 | 1 | 0 | -1 | R2 + (14/9)*R3 -> R2 | | 0 | 0 | 1 | -1 | |
| 1 | 0 | -6 | 3 | R1 - 4*R2 -> R1 | | 0 | 1 | 0 | -1 | | | 0 | 0 | 1 | -1 | |
| 1 | 0 | 0 | -3 | R1 + 6*R3 -> R1 | | 0 | 1 | 0 | -1 | | | 0 | 0 | 1 | -1 | |
For more practice with RREF, here is a very useful tool
http://www.math.odu.edu/~bogacki/lat/
It is called "linear algebra toolkit". It is a collection of matrix solvers that show step by step solutions.
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To summarize:
We started with this 3x4 matrix
and ended up with this reduced row echelon form (RREF)
The original 3x3 sub-block has morphed into the 3x3 identity matrix which has the main diagonal of all "1"s, and the rest are "0"s.
The right hand side of this RREF matrix are the solution values.
Answer:
x = -3
y = -1
z = -1
I'll let the student check each equation.
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
The solution from tutor @mathtutor_2020 shows a solution using what is commonly called Gauss-Jordan elimination. That is one good method for solving a system of 3 linear equations.
Tutor MathLover uses a purely algebraic approach; but her propensity to use substitution instead of elimination leads down an unnecessarily difficult path to the solution.
And her solution only shows a path to the solution -- it doesn't do anything to teach YOU how to get the solution.
Here is an algebraic solution using only elimination.
In a system of 3 linear equations, the objective is to first reduce the system to 2 equations by combining the given equations in a way that one of the variables is eliminated. Then that system of 2 equations can be solved by again eliminating one of the variables by combining the two equations in an appropriate way.
(1) x+4y-6z=-1
(2) 2x-y+2z=-7
(3) -x+2y-4z=5
Seeing the "x" in (1) and the "-x" in (3), add those two equations to get an equation in only y and z:
(4) 6y-10z=4
(4) 3y-5z=2
Now find another way to eliminate x using a different pair of the original equations. One way to do that is to double (3) and add to (2):
2x-y+2z=-7
-2x+4y-8z=10
(5) 3y-6z=3
Equations (4) and (5) are now a system of 2 equations in y and z. We could simplify (5); however, in their current forms (4) and (5) both contain the term 3y, so subtracting (4) from (5) will give us an equation in only z:
3y-6z=3
-3y+5z=-2
-z=1
(6) z=-1
Now substitute z=-1 in either (4) or (5) and solve for y:
3y+6=3
3y=-3
(7) y=-1
And last substitute y=-1 and z=-1 in any of the original equations to solve for x:
x-4+6=-1
(8) x=-3
ANSWER: (x,y,z) = (-3,-1,-1)
Answer by ikleyn(52787)  (Show Source):
You can put this solution on YOUR website! .
You are given this system of three linear equations
x + 4y - 6z = -1 (1)
2x - y + 2z = -7 (2)
-x + 2y - 4z = 5 (3)
Quick inspection shows that the part "-y + 2z" in equation (2) and the part "2y - 4z" in equation (3)
fit very well for mutual destruction. So, you multiply equation (2) by 2 and add it with equation (3).
The modified equations are
4x - 2y + 4z = -14 (2')
-x + 2y - 4z = 5 (3')
After adding these equations, you get
3x = -14 + 5 = 9, which implies x = -9/3 = -3, so one unknown is just found.
Now you plug in x= -3 into equations (1) and (3). You get a system of two equations
-3 + 4y - 6z = -1 (1'')
-(-3) + 2y - 4z = 5 (3'')
You simplify these equations further
4y - 6z = 2 (1''')
2y - 4z = 2 (3''')
Next you divide all the term in (1''') by 2, keeping (3''') as is. You get
2y - 3z = 1 (1'''')
2y - 4z = 2 (3'''')
Next you subtract equation (3'''') from equation (1''''). You get then
z = 1 - 2 = -1. Thus, one more unknown is found.
To find the last unknown, y, use equation (1''''). Substitute there z= -1 to get
2y - 3*(-1) = 1, and obtain then 2y + 3 = 1, 2y = 1 - 3 = -2, y= -2/2 = -1.
ANSWER. x= -3; y= -1; z= -1.
Solved.
Answer by MathTherapy(10552)  (Show Source):
You can put this solution on YOUR website!
x+4y-6z=-1
2x-y+2z=-7
-x+2y-4z=5
I've said it many times before and I will continue to say it!! Something is seriously wrong with that
MATHLOVER "character." I don't think this person is capable of learning to do math the way it should be
done, without taking the student through "hell," especially when some math problems are so easy to solve.
x + 4y - 6z = - 1 ----- eq (i)
2x - y + 2z = - 7 ----- eq (ii)
- x + 2y - 4z = 5 ------- eq (iii)
Now, looking at the system of equations, since she's so "bent" on using substitution, then why not
solve eq (i) for x, since its coefficient is 1, or eq (ii) for y, since its coefficient is - 1, or even
eq (iii) for x, since it too has a coefficient of - 1? She instead chose to solve the one equation, # 1,
for y - with a coefficient of 4 - that produces a fractional expression for y, which in most cases
can cause errors and sheer torment to most students!
Why does this person keep doing these things? Is this how a tutor - a mere quasi-tutor, in her case - helps others?
One doesn't have to even bother applying substitution as it's quite obvious/clear as daylight that when eqs
(i) & (iii) are combined/added, x is immediately eliminated, which means that one or another pair of equations would
need to be manipulated in order to eliminate x too. A pair of equations in "y" and "z" would then remain to be solved!
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