Question

Solve the system of equations. $$\left\\{\begin{aligned} 3 x+2 z &=13 \\ x+2 y+z &=-5 \\\\-3 y-z &=10 \end{aligned}\right.$$

Short answer

The solution set of the system of equations is \(\{x = -3, y = -6, z = 8\}\).

Step-by-step solution

Identify The Variables

The variables in the given equations are \(x\), \(y\), and \(z\). Each equation features a different configuration of variables.

Rearrange The Equations

Rearrange the 2nd equation as \(x = -2y - z - 5\) and use this equation to substitute \(x\) in the 1st equation. The substituted equation is \(3(-2y - z - 5) + 2z = 13\). Simplifying this one gets \(-6y - 3z -15 + 2z = 13\) and by further simplification we reach at \(-6y - z = 28\).

Solve For The Variables

Now we have two new equations with variables \(y\) and \(z\), \(-6y - z = 28\) and \(-3y - z = 10\). Subtract the second equation from the first and the result is \(-3y = 18\). So \(y = -6\). Substitute \(y = -6\) into the second equation and solve for \(z\). \(-3*(-6) - z = -10\), which gives \(z = 8\). Finally, Substitute \(y = -6\) and \(z = 8\) into the rearranged second equation to solve for \(x\). The equation is \(x = -2*(-6) - 8 - 5\) which gives \(x = -3\).

Check The Solution Set

Verify the solution set \(\{x = -3, y = -6, z = 8\}\) by substituting these values into the original equations to ensure the left-hand side equals to the right-hand side

Key concepts

Linear Algebra

Linear algebra is a vital branch of mathematics that deals with vectors, vector spaces, linear mappings, and the systems of linear equations. In the context of linear equations, it primarily involves finding the values of variables that satisfy multiple equations at once, a scenario commonly referred to as a 'system of equations'.

When looking at the system of equations in our exercise, we discern three equations with three unknowns. Linear algebra provides methods to solve such systems, one of which is the 'substitution method' employed in the given solution. The system is built in such a way that solutions for the variables must work for all equations simultaneously, embodying the beauty of linear algebra's structure and applications.

Substitution Method

The substitution method is a powerful technique used in linear algebra for solving systems of equations. It involves expressing one variable in terms of others from one equation and then 'substituting' this expression into another equation. This step-by-step reduction can simplify a system to the point where one variable can be directly determined, after which back-substitution will progressively reveal the values of the remaining variables.

In our exercise, we first solve for variable 'x' from the second equation and then substitute it into the first and third equations, allowing us to resolve the system step by step. Substitution is especially useful for systems with multiple equations and is often more straightforward than other methods when one of the equations can be conveniently solved for one of the variables.

Variable Isolation

Variable isolation is a fundamental technique within the substitution method, wherein we manipulate an equation to 'isolate' one variable on one side of the equation. This essentially translates to having the subject variable by itself on one side and all other terms on the opposite side.

In the given exercise, this was accomplished by rearranging the second equation to solve for 'x', effectively isolating it. Variable isolation is crucial as it paves the way to substitute the isolated variable into the other equations, reducing the system to fewer variables and thereby inching closer to finding the solution.

Equation Simplification

Equation simplification is essential in solving algebraic equations efficiently. It involves combining like terms, reducing equations to their simplest form, and performing basic arithmetic operations to make equations more manageable. Simplification might include distributing multiplication over addition, collecting like terms, and dividing both sides of an equation to isolate a variable.

In our step by step solution, after substituting 'x' into the first equation, we distributed the multiplication and combined like terms to simplify the equation to \( -6y - z = 28 \). This simplification assists in making the procedure of solving for variables much clearer and allows us to use another method, like addition or subtraction, to solve the simplified system of equations.