Example

Solving {3x−4z=03y+2z=−32x+3y=−5\left\{\begin{array}{l} 3x - 4z = 0 \\ 3y + 2z = -3 \\ 2x + 3y = -5 \end{array}\right. by Elimination

To solve the system {3x−4z=03y+2z=−32x+3y=−5\left\{\begin{array}{l} 3x - 4z = 0 \\ 3y + 2z = -3 \\ 2x + 3y = -5 \end{array}\right. by elimination, notice that each given equation represents a relationship between only two of the three variables. First, select a pair of equations and manipulate them to eliminate a shared variable. For example, multiplying the second equation by 22 yields 6y+4z=−66y + 4z = -6. Adding this resulting statement to the first equation algebraically eliminates the variable zz, generating 3x+6y=−63x + 6y = -6. Pairing this derived statement with the third original equation creates a simplified two-variable system: {3x+6y=−62x+3y=−5\left\{\begin{array}{l} 3x + 6y = -6 \\ 2x + 3y = -5 \end{array}\right.. To solve for xx, multiply the second equation by −2-2 to obtain −4x−6y=10-4x - 6y = 10 and add it to the first equation to cleanly eliminate yy, yielding −x=4-x = 4, which simplifies to x=−4x = -4. Sequentially substituting x=−4x = -4 back into the third equation calculates y=1y = 1. Finally, inserting the initially found value x=−4x = -4 into the first equation allows for solving the remaining variable, resulting in z=−3z = -3. After verifying these substituted parameters independently against all three original statements, the absolute solution reliably forms the ordered triple (−4,1,−3)(-4, 1, -3).

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Updated 2026-06-18

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