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Answer:
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Answer:
5x + 4y – z = 0
10y – 3z = 11
z = 3
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It's fairly easy to see how to proceed in this case. I'll just back-substitute the z-value from the third equation into the second equation, solve the result for y, and then plug z and y into the first equation and solve the result for x.
10y – 3(3) = 11
10y – 9 = 11
10y = 20
y = 2
5x + 4(2) – (3) = 0
5x + 8 – 3 = 0
5x + 5 = 0
5x = –5
x = –1
Then the solution is (x, y, z) = (–1, 2, 3).
The reason this system was easy to solve is that the system was "triangular"; this refers to the equations having the form of a triangle, because of the lower equations containing only the later variables.
system of equations with "triangle" of terms highlighted
The point is that, in this format, the system is simple to solve. And Gaussian elimination is the method we'll use to convert systems to this upper triangular form, using the row operations we learned when we did the addition method.
Solve the following system of equations using Gaussian elimination.
–3x + 2y – 6z = 6
5x + 7y – 5z = 6
x + 4y – 2z = 8
No equation is solved for a variable, so I'll have to do the multiplication-and-addition thing to simplify this system. In order to keep track of my work, I'll write down each step as I go. But I'll do my computations on scratch paper. Here is how I did it:
The first thing to do is to get rid of the leading x-terms in two of the rows. For now, I'll just look at which rows will be easy to clear out; I can switch rows later to get the system into "upper triangular" form. There is no rule that says I have to use the x-term from the first row, and, in this case, I think it will be simpler to use the x-term from the third row, since its coefficient is simply "1". So I'll multiply the third row by 3, and add it to the first row. I do the computations on scratch paper:
[-3x + 2y - 6z = 6] + [3x + 12y - 6z = 24] = [14y - 12z = 30]
...and then I write down the results:
system of equations with updated first row
(When we were solving two-variable systems, we could multiply a row, rewriting the system off to the side, and then add down. There is no space for this in a three-variable system, which is why we need the scratch paper.)
Warning: Since I didn't actually do anything to the third row, I copied it down, unchanged, into the new matrix of equations. I used the third row, but I didn't actually change it. Don't confuse "using" with "changing".
To get smaller numbers for coefficients, I'll multiply the first row by one-half: