Example

Solve a Consistent System of Three Linear Equations with Dependent Equations (Example 4.35)

An example demonstrating how to solve a consistent system of three linear equations with dependent equations, which has infinitely many solutions. Consider the system: {x+2yz=1 2x+7y+4z=11 x+3y+z=4\begin{cases} x + 2y - z = 1 \ 2x + 7y + 4z = 11 \ x + 3y + z = 4 \end{cases} Step 1: Eliminate xx using the first and third equations by multiplying the first by 1-1 and adding it to the third, yielding y+2z=3y + 2z = 3. Step 2: Eliminate xx again using the first and second equations by multiplying the first by 2-2 and adding it to the second, yielding 3y+6z=93y + 6z = 9. Step 3: Eliminate yy using the two new equations by multiplying the first by 3-3 and adding it to the second, which results in the true statement 0=00 = 0. This indicates a dependent system with infinitely many solutions. Solving the first new equation for yy gives y=2z+3y = -2z + 3. Substituting this into the original first equation and solving for xx gives x=5z5x = 5z - 5. The solutions are of the form (5z5,2z+3,z)(5z - 5, -2z + 3, z) where zz is any real number.

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Updated 2026-05-26

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