Example

Solve a System of Three Linear Equations (Example 4.33)

To solve the system of equations:

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

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Updated 2026-06-03

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