Question

Solve the system of equations. $$\left\\{\begin{aligned} 2 x+3 y-z &=1 \\ x-2 y+z &=7 \\ 3 x+y+2 z &=12 \end{aligned}\right.$$

Short answer

The solution of the system of equations is \( x = 1.4 \), \( y = 3.8 \), \( z = 2.2 \)

Step-by-step solution

Eliminate One of the variables

Add the second equation to the first to eliminate \( z \): \( 2x + 3y - z + x - 2y + z = 1 + 7 \), which simplifies to \( 3x + y = 8 \). This is our fourth equation.

Eliminate the Same Variable from Other Pair

Now, subtract the second equation from the third to also eliminate \( z \): \( 3x + y + 2z - (x - 2y + z) = 12 - 7 \), which simplifies to \( 2x + 3y = 5 \). This is our fifth equation.

Solve the Derived System

We now have a system of two equations (fourth and fifth) with two variables (x and y): \[ \begin{aligned} 3x + y &= 8 \ 2x + 3y &= 5 \end{aligned} \]Subtract the fifth equation from the fourth (three times) to solve for x: \( (3x+y) - 3*(2x+3y) = 8 - 3*5 \), simplifying gives \( -5x = -7 \), thus \( x = (-7)/(-5) = 1.4 \).

Substitute x into Fourth Equation

Substitute the value of \(x = 1.4\) into the fourth equation to solve for \(y\): \(3x + y = 8 \Rightarrow 3*1.4 + y = 8 \Rightarrow y = 8 - 4.2 = 3.8 \).

Substitute x and y into the Original System

Substitute \(x = 1.4\) and \(y = 3.8\) into any of the original equations to find the value for \(z\). Substituting into the first equation gives \(2*1.4 + 3*3.8 - z = 1 \Rightarrow z = 2.8 + 11.4 - 1 = 2.2\).

Key concepts

Algebraic Methods

When dealing with systems of equations, algebraic methods come into play as systematic and reliable approaches to finding the solutions. These methods are built upon basic algebraic principles such as operations with numbers and variables, manipulation of equations, and adhering to equality properties.

One might encounter various algebraic techniques, but two prominent ones are variable elimination and substitution. Each approach has its advantages and can be chosen based on ease of computation and simplicity of equations. A strong grasp of algebraic methods is crucial for solving not just simple linear systems, but also more complex problems encountered in higher mathematics.

Variable Elimination

Variable elimination is a strategy to reduce the number of variables in a system of equations gradually. The goal is to manipulate the equations in such a way that one of the variables cancels itself out.

This often involves adding or subtracting equations from each other. For example, if two equations in a system have the same coefficient for a variable but with opposite signs, adding those equations would result in the cancellation of that variable. Alternatively, if the coefficients are not in such a configuration, we may multiply one or both equations by suitable numbers to create opposing coefficients. This method systematically simplifies the system until one can easily solve the remaining variables.

Substitution Method

The substitution method follows a different path. Here, the focus is on expressing one variable in terms of another and then substituting that expression into another equation.

For instance, if one equation can be rearranged to give, say, \( x = 5 - y \) as an isolated variable expression, then we can use this expression in place of \( x \) in another equation of the system. This substitution turns a multivariable equation into a single-variable one, making it straightforward to solve. The key here is to identify the easiest variable to isolate and substitute across the system.

System of Linear Equations

A system of linear equations is a set of two or more linear equations involving the same set of variables. These equations represent straight lines in two-dimensional space or planes in three-dimensional space.

Their solutions are the points at which these lines or planes intersect. When solving these systems, we're essentially looking for the coordinates of intersection points that satisfy all the equations simultaneously. Linear systems can be consistent (having one or more solutions), inconsistent (having no solution), or dependent (having an infinite number of solutions). Recognizing the nature of the system is an integral part of the solving process.