Example

Solving {3x4z=12y+3z=22x+3y=6\left\{\begin{array}{l} 3x - 4z = -1 \\ 2y + 3z = 2 \\ 2x + 3y = 6 \end{array}\right. by Elimination

To solve the system {3x4z=12y+3z=22x+3y=6\left\{\begin{array}{l} 3x - 4z = -1 \\ 2y + 3z = 2 \\ 2x + 3y = 6 \end{array}\right. by elimination, begin by eliminating zz from the first two equations. Multiplying the first equation by 33 yields 9x12z=39x - 12z = -3, while multiplying the second equation by 44 yields 8y+12z=88y + 12z = 8. Adding these equations cancels the zz term, creating the two-variable equation 9x+8y=59x + 8y = 5. Pair this with the third equation to form a secondary system: {9x+8y=52x+3y=6\left\{\begin{array}{l} 9x + 8y = 5 \\ 2x + 3y = 6 \end{array}\right.. Next, multiply the first equation of this new system by 3-3 to get 27x24y=15-27x - 24y = -15 and the second equation by 88 to get 16x+24y=4816x + 24y = 48. Adding these equations eliminates yy, yielding 11x=33-11x = 33, which simplifies to x=3x = -3. Substitute x=3x = -3 into the original third equation to find y=4y = 4. Then, substitute y=4y = 4 into the original second equation to find z=2z = -2. The solution is the ordered triple (3,4,2)(-3, 4, -2).

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Updated 2026-06-29

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