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Multivariable linear systems
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The following system is said to be in row-echelon form, which means that it has a “stair-step” pattern with leading coefficients of 1. Using Back-Substitution in Row- Echelon Form
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Back-Substitution Equation 1 Equation 2 Equation 3
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Back-Substitution Equation 1 Equation 2 Equation 3 From equation 3, you know the value of z. To solve for y, substitute z=2 into Equation 2 to obtain
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Back-Substitution Finally, substitute y=-1 and z=2 into equation 1 to obtain. The solution is x=1, y=-1, and z=2, which can be written as the ordered triple (1, -1, 2). Check this in the original system of equations.
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Practice
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Two systems of equations are equivalent if they have the same solution set. To solve a system that is not in row-echelon form, first convert it to an equivalent system that is in row-echelon form by using own or more of the elementary row operations shown in the next example. This process is called Gaussian elimination. Gaussian Elimination
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Interchange two equations. Multiply own of the equations by a nonzero constant. Add a multiple of one equation to another equation. Elementary Row Operations for Systems of Equations
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Example Equation 1 Equation 2 Equation 3
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Example Continued Adding the first equation to the second equation produces a new second equation.
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Example Continued Adding -2 times the first equation to the third equation produces a new third equation.
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Now that all but the first x have been eliminated from the first column, go to work on the second column. (You need to eliminate y from the third equation. Example Continued Adding the second equation to the third equation produces a new third equation.
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Finally, you need a coefficient of 1 for z in the third equation. Example Continued
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You can conclude that the solution is x=1, y=-1, and z=2, written as (1, -1, 2). Example Continued
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Solve the system of linear equations. Try this…
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