Problem 25 - Intermediate Algebra

Question

Solve each system. $$ \left\\{\begin{array}{l} 2 x+z=-2+y \\ 8 x-3 y=-2 \\ 6 x-2 y+3 z=-4 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \((x, y, z) = (\frac{1}{2}, 2, -1)\).

Step 1: Simplify Equations

First, rearrange the terms in each equation to make them easier to work with. Starting with the first equation: \[2x + z = -2 + y\] becomes \[2x - y + z = -2\]The second equation is already simplified:\[8x - 3y = -2\]The third equation is already simplified:\[6x - 2y + 3z = -4\]

Step 2: Eliminate Variables

To make the system easier to solve, try eliminating one of the variables. Let's eliminate \(z\) from the second and third equations.Firstly, from equation 1:\[2x - y + z = -2\]Solve for \(z\):\[z = -2x + y - 2\]Plug \(z\) in equations 2 and 3:For equation 3:\[6x - 2y + 3(-2x + y - 2) = -4\]Simplify:\[6x - 2y - 6x + 3y - 6 = -4\]Combine like terms:\[y - 6 = -4\]This simplifies to:\[y = 2\]

Step 3: Substitute and Solve for x and z

With \(y = 2\), substitute back into equation 2 and solve for \(x\):Equation 2:\[8x - 3(2) = -2\]Simplify:\[8x - 6 = -2\]Add 6 to both sides:\[8x = 4\]Divide by 8:\[x = \frac{1}{2}\]Substitute \(x = \frac{1}{2}\) and \(y = 2\) into the expression for \(z\) from the end of Step 2:\[z = -2(\frac{1}{2}) + 2 - 2\]Simplify:\[z = -1 + 2 - 2 = -1\]

Step 4: Verify the Solution

Check the solution \((x, y, z) = (\frac{1}{2}, 2, -1)\) by plugging these values into the original equations:For equation 1:\[2(\frac{1}{2}) - 2 + (-1) = -2\]Simplifies to:\[1 - 2 - 1 = -2\] Correct!For equation 2:\[8(\frac{1}{2}) - 3(2) = -2\]Simplifies to:\[4 - 6 = -2\]Correct!For equation 3:\[6(\frac{1}{2}) - 2(2) + 3(-1) = -4\]Simplifies to:\[3 - 4 - 3 = -4\]Correct!This verifies that the solution is correct.

Key Concepts

Elimination Method

The elimination method is a powerful tool for solving systems of linear equations. This approach involves manipulating the equations strategically to "eliminate" one of the variables, making it easier to find the solution for the others.

Here's how it works:

• Choose a variable to eliminate from some of the equations.
• Use basic algebraic operations—like addition or subtraction of equations after multiplication by constants—to make the coefficients of that variable opposite in two equations.
• By adding or subtracting those modified equations, one variable disappears, leaving a simpler system to solve.

In our original problem, the task was to eliminate the variable \(z\). By rearranging and substituting the expressions, we could reduce the complexity of the system, eventually leading us to a simplified path for finding \(x\) and \(y\) without \(z\) getting in the way. This step is crucial in making complex systems manageable and is a cornerstone of solving linear algebra problems.

Substitution Method

The substitution method is another technique widely used for tackling systems of equations. This method involves solving one of the equations for a particular variable and then substituting that expression into the other equations.

This strategy simplifies the problem by reducing the number of variables in the equations you're working with.

• Begin by isolating one variable in one of the equations.
• Substitute the expression you derive into the other equations.
• You'll end up with a set of equations with fewer variables, making it easier to solve.

In our step-by-step solution, we first solved for \(z\) in terms of \(x\) and \(y\) from the first equation. This expression was then plugged into the other equations, eliminating the \(z\) variable from them, which made it possible to work on solving just for \(x\) and \(y\). By reducing the number of variables, the substitution method turns a potentially complicated problem into a simpler, more direct solution process.

Solutions of Equations

Finding solutions to a system of equations means determining the values of the variables that satisfy all equations simultaneously. These solutions represent the point at which all the equations intersect if graphed.

In algebra, once you achieve a simplified set of expressions through methods like elimination and substitution, you then solve for each variable methodically. The solution to a system of equations can be:

• A unique solution (a single set of values for the variables, like in this problem).
• No solution (when the equations represent parallel lines that never intersect).
• Infinitely many solutions (when the equations are dependent, representing the same line).

For our specific example, after applying elimination and substitution, the solution set was \((x, y, z) = (\frac{1}{2}, 2, -1)\). This means that these values for \(x\), \(y\), and \(z\) satisfy all the original equations, as verified by substituting these back into the original system. Understanding the nature and types of solutions is vital in contextualizing your algebraic results.