Example

Solve a System of Three Linear Equations (Example 4.33)

To solve the system of equations:

{3x4z=03y+2z=32x+3y=5\begin{cases} 3x - 4z = 0 \\\\ 3y + 2z = -3 \\\\ 2x + 3y = -5 \end{cases}

First, eliminate zz from the first two equations by multiplying the second equation by 2 and adding it to the first equation: 3x4z=03x - 4z = 0 6y+4z=66y + 4z = -6 Adding these equations yields a new equation with variables xx and yy: 3x+6y=63x + 6y = -6 Next, solve the new system formed by this equation and the third equation, which also contains variables xx and yy. Multiply the third equation by -2 and add it to the new equation to eliminate yy: 3x+6y=63x + 6y = -6 4x6y=10-4x - 6y = 10 Adding these equations yields x=4-x = 4, which simplifies to x=4x = -4. Substitute x=4x = -4 into the third equation to solve for yy: 2(4)+3y=52(-4) + 3y = -5 8+3y=5-8 + 3y = -5 3y=33y = 3 y=1y = 1 Substitute x=4x = -4 into the first equation to solve for zz: 3(4)4z=03(-4) - 4z = 0 124z=0-12 - 4z = 0 4z=12-4z = 12 z=3z = -3 Checking the solution (4,1,3)(-4, 1, -3) in all three original equations confirms that it makes all three equations true.

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Updated 2026-05-25

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