Problem 38 - Algebra and Trigonometry Real Mathematics, Real People

Question

Solve the system of linear equations and check any solution algebraically. $$\left\\{\begin{array}{rr} x-2 y+z= & 2 \\ 2 x+2 y-3 z= & -4 \\ 5 x\quad+z= & 1 \end{array}\right.$$

Step-by-Step Solution

Answer

The solution of the system of equations is \(x = -1/6\), \(y = -1/2\), and \(z = 1/6\).

Step 1: Simplify the Equations

The second equation can be simplified. Divide the whole equation by 2 gives: \(x + y - 1.5z = -2\) . Now, the system of equations to work with becomes:(1) \( x - 2y + z = 2 \) (2) \( x + y - 1.5z = -2 \) (3) \( 5x + z = 1 \)

Step 2: Eliminate Variables

Now one can eliminate the same variable from two equations. Subtract equation (2) from equation (1):(1 - 2: ) \(-3y + 2.5z = 4\)Store this as equation (4). Now add equation (2) and equation (3):(2 + 3: ) \(6x = -1\)From this, one gets the solution for \(x\) directly:\(x = -1/6\)

Step 3: Solve for Remaining Variables

Substitute x = -1/6 back into equation (2) and equation (3) to get two new equations in terms of \(y\) and \(z\). Substitute into equation (2) gives:\(y = 1.5z - 4/3\)Substitute into equation (3) gives:\(1 - 5/6 = z\)Solving for \(z\) gives \(z = 1/6\). Then, substituting \(z = 1/6\) back into \(y = 1.5z - 4/3\) gives \(y = -1/2\)

Step 4: Check the Solution

Finally, substitute \(x = -1/6\), \(y = -1/2\), and \(z = 1/6\) back into the original three equations to verify that they are indeed solutions.

Key Concepts

System of Equations

A system of equations is a collection of two or more equations with the same set of variables. In our exercise, we have three equations with variables \( x \), \( y \), and \( z \).
These equations are solved simultaneously to find a set of values that satisfy each one of them.
The solutions to the system, if they exist, provide values for the variables that make all the given equations true at the same time.

• Understanding how to set up and interpret a system of equations is foundational in algebra.
• Visualization: Each equation can represent a plane in 3-dimensional space, and the solution corresponds to the intersection points of these planes.

Understanding a system of equations lays the groundwork for various mathematical and real-world applications where multiple conditions must be satisfied simultaneously.

Algebraic Solution

An algebraic solution involves using algebraic manipulations to solve a problem, which in this case is applying algebra techniques to find the solutions of the system of equations.
This involves operations such as addition, subtraction, multiplication, or division applied to equations to simplify and solve them.

• The first step in our exercise simplified the second equation by dividing it, which made it easier to eliminate variables later on.
• Solving algebraically ensures accuracy since it relies on mathematical principles rather than estimation or graphical solutions.

It’s like a puzzle where you're rearranging pieces to reveal the final picture. Each manipulation moves you closer to the solution without changing the original "puzzle" conditions.

Variable Elimination

Variable elimination is a technique used to solve systems of equations where you strategically remove variables to simplify the equations until you can directly solve for one of the variables.
In this approach, you manipulate the equations to make one variable disappear from two of the equations.
In our exercise:

• Equation (2) was subtracted from equation (1) to eliminate \(x\), simplifying it to solve for \(y\) and \(z\).
• The combination of equations (2) and (3) allowed for solving \(x\) directly.

By eliminating a variable, you reduce the number of variables you need to solve for, effectively narrowing down your search for the solution. Working systematically like this helps to manage complex systems and reach a solution efficiently.