To solve the specific structural system $$\left\{\begin{array}{l} 3x - 4z = -1 \\ 2y + 3z = 2 \\ 2x + 3y = 6 \end{array}ight.$$ by elimination, begin by addressing equations sharing variables to construct a two-variable problem. Multiplying the first equation by $$3$$ yields $$9x - 12z = -3$$, while multiplying the second equation by $$4$$ produces $$8y + 12z = 8$$. Adding these manipulated equations directly cancels out the $$z$$ term, creating the new two-variable equation $$9x + 8y = 5$$.

Form a secondary system by pairing this new structural equation with the original third equation: $$\left\{\begin{array}{l} 9x + 8y = 5 \\ 2x + 3y = 6 \end{array}ight.$$. Multiply the first equation by $$-3$$ to get $$-27x - 24y = -15$$ and the second equation by $$8$$ to deduce $$16x + 24y = 48$$. Adding these together eliminates $$y$$, isolating $$-11x = 33$$, which securely simplifies to $$x = -3$$. Substituting $$x = -3$$ back into the third original equation resolves the second coordinate as $$y = 4$$. Further substituting $$y = 4$$ into the second equation precisely reveals $$z = -2$$. Writing the solution as an ordered triple gives $$(-3, 4, -2)$$, which dependably verifies true across each base equation.

Solving $$\left\{\begin{array}{l} 3x - 4z = -1 \\ 2y + 3z = 2 \\ 2x + 3y = 6 \end{array}ight.$$ by Elimination

To solve the linear system $$\left\{\begin{array}{l} 4x - 3z = -5 \\ 3y + 2z = 7 \\ 3x + 4y = 6 \end{array}ight.$$ using the elimination method, systematically pair equations to reduce the problem's dimensionality. Eliminate $$z$$ from the first two equations by multiplying the first by $$2$$ (yielding $$8x - 6z = -10$$) and the second by $$3$$ (yielding $$9y + 6z = 21$$). Integrating these formulas cancels $$z$$ to reliably state a new equation: $$8x + 9y = 11$$.

Bringing this calculated statement together with the unused third equation outlines a localized two-variable system: $$\left\{\begin{array}{l} 8x + 9y = 11 \\ 3x + 4y = 6 \end{array}ight.$$. Next, choose to eliminate $$y$$ by multiplying the first equation by $$4$$ to obtain $$32x + 36y = 44$$ and the second by $$-9$$ to create $$-27x - 36y = -54$$. Adjoining these equations strictly isolates $$5x = -10$$, simplifying cleanly to $$x = -2$$. Substituting $$x = -2$$ directly back into the third original equation allows deduction of the localized variable $$y$$ as $$3$$. Furthermore, placing $$y = 3$$ into the second equation predictably reveals $$z = -1$$. When written compactly, the functional solution successfully manifests as the tested ordered triple $$(-2, 3, -1)$$.

Solving $$\left\{\begin{array}{l} 4x - 3z = -5 \\ 3y + 2z = 7 \\ 3x + 4y = 6 \end{array}ight.$$ by Elimination

To solve the system $$\left\{\begin{array}{l} x + 2y - 3z = -1 \\ x - 3y + z = 1 \\ 2x - y - 2z = 2 \end{array}\right.$$ by elimination, first eliminate the variable $$z$$. Multiplying the second equation by $$3$$ and adding it to the first equation yields the two-variable equation $$4x - 7y = 2$$. Next, multiply the second equation by $$2$$ and add it to the third equation to generate another two-variable equation, $$4x - 7y = 4$$. This forms a new sub-system: $$\left\{\begin{array}{l} 4x - 7y = 2 \\ 4x - 7y = 4 \end{array}\right.$$. To eliminate a variable from this new sub-system, multiply the second equation by $$-1$$ and add it to the first equation. This completely eliminates the remaining variables, resulting in the mathematically false statement $$0 = -2$$. Because we are left with a false statement, the system is explicitly inconsistent and has no solution.

Solving $$\left\{\begin{array}{l} x + 2y - 3z = -1 \\ x - 3y + z = 1 \\ 2x - y - 2z = 2 \end{array}\right.$$ by Elimination

To solve the system $$\left\{\begin{array}{l} x + 2y + 6z = 5 \\ -x + y - 2z = 3 \\ x - 4y - 2z = 1 \end{array}\right.$$ by elimination, first eliminate the variable $$x$$. Adding the first and second equations together yields the two-variable equation $$3y + 4z = 8$$. Adding the second and third equations together generates another two-variable equation, $$-3y - 4z = 4$$. This forms a new sub-system: $$\left\{\begin{array}{l} 3y + 4z = 8 \\ -3y - 4z = 4 \end{array}\right.$$. Adding these two new equations firmly eliminates both remaining variables, resulting in the mathematically false statement $$0 = 12$$. Because the process leaves a definitively false statement, the system is strictly inconsistent and has no valid solution.

Solving $$\left\{\begin{array}{l} x + 2y + 6z = 5 \\ -x + y - 2z = 3 \\ x - 4y - 2z = 1 \end{array}\right.$$ by Elimination

To solve the system $$\left\{\begin{array}{l} 2x - 2y + 3z = 6 \\ 4x - 3y + 2z = 0 \\ -2x + 3y - 7z = 1 \end{array}\right.$$ by elimination, first eliminate the variable $$x$$. Adding the first and third equations together directly yields the two-variable equation $$y - 4z = 7$$. Next, multiply the third equation by $$2$$ and add it to the second equation to produce another two-variable equation, $$3y - 12z = 2$$. This forms a new sub-system of equations: $$\left\{\begin{array}{l} y - 4z = 7 \\ 3y - 12z = 2 \end{array}\right.$$. To eliminate a variable from this sub-system, multiply the first equation by $$-3$$ and add it to the second equation. This completely removes the variables and results clearly in the false mathematical statement $$0 = -19$$. Because we are left firmly with a false numerical statement, the system is inconsistent and yields no solution.

Solving $$\left\{\begin{array}{l} 2x - 2y + 3z = 6 \\ 4x - 3y + 2z = 0 \\ -2x + 3y - 7z = 1 \end{array}\right.$$ by Elimination

To solve the system $$\left\{\begin{array}{l} x + 2y - z = 1 \\ 2x + 7y + 4z = 11 \\ x + 3y + z = 4 \end{array}\right.$$ by elimination, first eliminate the variable $$x$$. Multiplying the first equation by $$-1$$ and adding it to the third equation yields the two-variable equation $$y + 2z = 3$$. Next, multiply the first equation by $$-2$$ and add it to the second equation to generate another two-variable equation, $$3y + 6z = 9$$. This forms a new sub-system: $$\left\{\begin{array}{l} y + 2z = 3 \\ 3y + 6z = 9 \end{array}\right.$$. To eliminate a variable from this new sub-system, multiply the first equation by $$-3$$ and add it to the second equation. This completely eliminates the variables, resulting precisely in the inherently true mathematical statement $$0 = 0$$. Because the resulting statement is true, the system is explicitly dependent and has infinitely many solutions. To determine the comprehensive solution set, express two variables precisely in terms of the third. Solving $$y + 2z = 3$$ for $$y$$ gives $$y = -2z + 3$$. Substituting this expression for $$y$$ into the first equation ($$x + 2y - z = 1$$) and solving strategically for $$x$$ yields $$x = 5z - 5$$. The complete solution is definitively any ordered triple logically of the form $$(5z - 5, -2z + 3, z)$$, where $$z$$ is uniquely any real number.

Solving $$\left\{\begin{array}{l} x + 2y - z = 1 \\ 2x + 7y + 4z = 11 \\ x + 3y + z = 4 \end{array}\right.$$ by Elimination

To solve the system $$\left\{\begin{array}{l} x + y - z = 0 \\ 2x + 4y - 2z = 6 \\ 3x + 6y - 3z = 9 \end{array}\right.$$ by elimination, one can naturally eliminate the variable $$x$$. Multiplying the first equation by $$-2$$ and adding it to the second equation directly yields the secondary equation $$2y = 6$$. Multiplying the first equation by $$-3$$ and adding it to the third equation yields the secondary equation $$3y = 9$$. Both equations logically simplify firmly to $$y = 3$$. Attempting to eliminate $$y$$ from these two equivalent equations leads to the true mathematical statement $$0 = 0$$, which solidly indicates that the original system is explicitly dependent and has infinitely many solutions. To precisely formulate the general solution, substitute $$y = 3$$ back into the first equation ($$x + y - z = 0$$), which produces $$x + 3 - z = 0$$. Solving strategically for $$x$$ in terms of $$z$$ safely yields $$x = z - 3$$. The comprehensive solution is represented comprehensively by the continuous ordered triple $$(z - 3, 3, z)$$, where $$z$$ firmly denotes any real number.

Solving $$\left\{\begin{array}{l} x + y - z = 0 \\ 2x + 4y - 2z = 6 \\ 3x + 6y - 3z = 9 \end{array}\right.$$ by Elimination

Real-world applications structurally modeled by systems of linear equations involving three distinct variables can be reliably solved by predictably applying the exact same algebraic techniques, such as elimination, used to correctly resolve abstract mathematical systems. Many of these specific application problems act logically as straightforward extensions of the fundamental two-variable scenarios encountered earlier, necessitating the careful translation of the expanded given conditions directly into three separate linear equations to effectively determine the targeted solution.

Solving Applications Using Systems of Linear Equations with Three Variables

To solve the system $$ \left\{\begin{array}{l} 3x - 4z = 0 \\ 3y + 2z = -3 \\ 2x + 3y = -5 \end{array}\right. $$ by elimination, notice that each given algebraic 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 $$ 2 $$ yields $$ 6y + 4z = -6 $$. Adding this resulting statement to the initial first equation algebraically eliminates the variable $$ z $$, generating $$ 3x + 6y = -6 $$. Pairing this derived statement with the third original equation creates a simplified two-variable system: $$ \left\{\begin{array}{l} 3x + 6y = -6 \\ 2x + 3y = -5 \end{array}\right. $$. To solve for $$ x $$, multiply the second equation by $$ -2 $$ to obtain $$ -4x - 6y = 10 $$ and add it to the first to cleanly eliminate $$ y $$, yielding $$ -x = 4 $$, which means $$ x = -4 $$. Sequentially substituting $$ x = -4 $$ back into the third equation calculates $$ y = 1 $$. Finally, inserting the initial found value $$ x = -4 $$ into the first equation allows for solving the remaining variable, resulting in $$ z = -3 $$. After verifying these substituted parameters independently against all three original statements, the absolute solution reliably forms the ordered triple $$ (-4, 1, -3) $$.

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

To solve a system of linear equations with three variables using the elimination method, follow a systematic seven-step procedure. First, write all equations in standard form, clearing any fractional coefficients. Second, choose one variable to eliminate and use a pair of equations to do so by multiplying them to create opposite coefficients and adding them together. Third, select a different pair of equations and eliminate the exact same variable, which produces a second new equation. Fourth, take these two newly formed equations, which now represent a system of two equations with two variables, and solve them. Fifth, substitute the two found values back into one of the original three-variable equations to calculate the third variable. Sixth, express the final answer as an ordered triple. Seventh, verify the solution by checking that the ordered triple satisfies all three of the original equations.

Claude

The absolute solutions verifying a functional system composed broadly of three linear equations simply represent the respective substituted values corresponding securely to the variables that ensure each one of the available equations calculates consistently true. Crucially, a confirmed solution for this structural configuration is mapped algebraically using a proper ordered triple $$(x, y, z)$$ designating a real common juncture uniting all the separate incorporated planes.

Solutions of a System of Linear Equations with Three Variables

To solve a system of two linear equations by elimination, follow this seven-step procedure:

1. Write both equations in standard form. If any coefficients are fractions, clear them first.
2. Make the coefficients of one variable opposites. Decide which variable to eliminate, and multiply one or both equations by carefully chosen constants so that the coefficients of that variable become opposites.
3. Add the two resulting equations to eliminate that variable.
4. Solve the new equation for the remaining variable.
5. Substitute this solution back into one of the original equations and solve for the other variable.
6. Write the final solution as an ordered pair $$(x, y)$$.
7. Check the ordered pair by substituting its coordinates into both original equations to verify that each yields a true statement.

Solving a System of Linear Equations by Elimination

A functioning system of three linear equations possessing three identical variables structurally consists of three explicitly distinct linear equations of the standard form $$ax+by+cz=d$$, considered completely simultaneously. Geometrically speaking, successfully finding the holistic solution to such a clustered system literally demands observing directly what specific regions the three associated three-dimensional planes happen to occupy in common.

System of Three Linear Equations with Three Variables

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OpenStax Intermediate Algebra

Solving a System of Linear Equations with Three Variables by Elimination

An inconsistent system composed of three linear equations with three variables is defined as a system that possesses no mathematical solution. Graphically, when the three equations are mapped as individual planes in three-dimensional space, there are no mutual intersection points connecting all three planes simultaneously, so no single ordered triple $$(x, y, z)$$ can satisfy every equation in the system. When algebraically resolving an inconsistent system of three variables through elimination, the process consistently eliminates all variables entirely, leaving a definitively false numerical statement, such as $$0 = -2$$.

Inconsistent System of Three Linear Equations

When algebraically solving a system consisting of three linear equations and three variables, if the process completely eliminates all variables but results in a mathematically true numerical statement, such as $$0 = 0$$, the system has infinitely many solutions. This signifies that the system is consistent with dependent equations. In such cases, indicating the comprehensive solution set involves displaying explicitly how two of the variables logically depend upon the third variable.

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