Question

Match each system of equations with its solution. [The solutions are labeled (a), (b), (c), and (d).] (a) \((-1,0,3)\) (b) \((6,2,-2)\) (c) \((2,1,-3)\) (d) \((4,-1,5)\) $$ \left\\{\begin{aligned} -x-2 y+5 z &=23 \\ -3 x+y+6 z &=17 \\ 9 x+2 y-7 z &=-1 \end{aligned}\right. $$

Short answer

After checking each solution, it can be concluded that Solution (b) \((6,2,-2)\) is the correct match for the system of equations, as it is the only one that satisfies all three equations.

Step-by-step solution

Check Solution (a)

Substitute the coordinates of Solution (a) \((-1,0,3)\) into the system of equations. If each equation holds true, then this is the correct match. Otherwise, proceed to the next solution.

Check Solution (b)

Substitute the coordinates of Solution (b) \((6,2,-2)\) into the system of equations. If each equation holds true, then this is the correct match. Otherwise, proceed to the next solution.

Check Solution (c)

Substitute the coordinates of Solution (c) \((2,1,-3)\) into the system of equations. If each equation holds true, then this is the correct match. Otherwise, proceed to the next solution.

Check Solution (d)

Substitute the coordinates of Solution (d) \((4,-1,5)\) into the system of equations. If each equation holds true, then this is the correct match.

Key concepts

System of Linear Equations

A system of linear equations consists of multiple linear equations that have common variables. In algebra, solving such a system means finding the values for the variables that satisfy all equations simultaneously. In the exercise provided, we are dealing with a system of three equations with three variables:

\[ \begin{aligned} -x-2y+5z &= 23, \ -3x+y+6z &= 17, \ 9x+2y-7z &= -1. \end{aligned} \]
To solve such a system, one must find a set of numbers for the variables \(x, y, z\) that will make all three equations correct when substituted into them. These types of problems are fundamental in algebra and have applications across science, engineering, economics, and more.

Substitution Method

The substitution method is a technique used to solve systems of equations by finding an expression for one variable in terms of another and substituting this expression into the other equations. This process can lead to an easier-to-solve equation and eventually to the values of all the variables. In the context of our textbook exercise, this method would involve solving one of the equations for one variable and plugging this value in to the other equations.

For instance, if we could solve the first equation for \(x\), we would have \(x = \text{expression in } y \text{ and } z\), which we could then substitute in the other equations. This could potentially lead to simpler equations with fewer variables, making it easier to find the solution.

Algebraic Solutions

When we talk about algebraic solutions, we refer to the process of solving equations using algebraic manipulations such as addition, subtraction, multiplication, division, substitution, and elimination. These techniques allow us to systematically reduce systems of equations to simpler forms that can be solved more easily. In our exercise, we seek an algebraic solution by testing each option \(a, b, c, d\) by substituting the variables \(x, y, z\) with the provided values and checking if all equations hold true.

Algebraic solutions are based on properties of equality and require logical and systematic application of algebraic principles. In educational practices, teaching students to solve systems of equations algebraically prepares them for higher-level math as well as developing critical thinking and problem-solving skills.

Matrix Solution Method

The matrix solution method, also known as the matrix inversion method, uses matrix algebra to solve systems of linear equations. In this method, the system of equations is represented in matrix form. This form allows us to utilize powerful matrix operations to find solutions. For the given exercise, the matrix representation of the coefficients of variables would look like this:

\[ \begin{bmatrix} -1 & -2 & 5 \ -3 & 1 & 6 \ 9 & 2 & -7 \end{bmatrix} \]
With the right side of the equations represented as another matrix and using matrix operations like the inverse of a matrix, we can solve for the variables. Though not used explicitly in our textbook exercise, this method is integral for larger and more complex systems of equations, especially when dealing with many variables and equations, where traditional algebraic methods may not be practical.