Problem 42 - Precalculus with Limits

Question

Solving a System of Linear Equations In Exercises \(25 - 46\) , solve the system of linear equations and check any solutions algebraically. $$ \left\\{ \begin{aligned} 2 x + y - 3 z = & 4 \\ 4 x + 2 z = & 10 \\ - 2 x + 3 y - 13 z = & \- 8 \end{aligned} \right. $$

Step-by-Step Solution

Answer

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

Step 1: Express the Second Equation in terms of x and z

Expresse the 2nd equation in terms of \(x\) to get \(x = \frac{10 - 2z}{4}\) or simply \(x = \frac{5 - z}{2}\). This makes it easier to eliminate \(x\) in the third equation.

Step 2: Substitution

Substitute \(x = \frac{5 - z}{2}\) from the second equation into the third equation to isolate \(y\). This gives us \(-2(\frac{5 - z}{2}) + 3y - 13z = -8\), which simplifies to \(3y - 9z = -3\). Reorganize to isolate \(y\), giving \(y = 3z - 1\).

Step 3: Substitute \(y\)-Value

Substitute \(y = 3z - 1\) and \(x = \frac{5 - z}{2}\) into the first equation. This gives us \(2(\frac{5 - z}{2}) + 3z - 3 - 3z = 4\), which can be simplified to get \(z = 1\).

Step 4: Find x and y

Substitute the value of \(z = 1\) in the equations for \(x\) and \(y\). For \(x\), we have \(x = \frac{5 - 1}{2} = 2\). For \(y\), we have \(y = 3 * 1 - 1 = 2\).

Step 5: Verification

Substitute \(x = 2\), \(y = 2\) and \(z = 1\) into the original equations to verify that they hold true.

Key Concepts

Substitution Method

The substitution method is a powerful technique used to find the solution to systems of linear equations. It involves expressing one variable in terms of another, allowing us to simplify a system by reducing the number of variables. Here’s how you can understand and apply this method:

Generally, you start by isolating one of the variables in one of the equations. This equation is then used to substitute back into the other equations. For instance, in our example, we express the second equation in terms of \(x\) as \(x = \frac{5 - z}{2}\). This is then substituted into the third equation to find expressions for other variables.

By substituting variables strategically, you're able to simplify complex equations and progressively isolate each variable. This step-by-step isolation of variables eventually leads to finding the actual values of all the variables involved in the system.

Algebraic Verification

Algebraic verification is the process of checking whether the solutions found for a system of equations are indeed correct. Once you derive the values of the variables, substitution back into the original equations is crucial. This ensures that the solutions you have obtained are valid.

In the problem above, after determining \(x = 2\), \(y = 2\), and \(z = 1\), these values are substituted back into the original set of equations:

• \(2x + y - 3z = 4\),
• \(4x + 2z = 10\),
• \(-2x + 3y - 13z = -8\).

By substituting the derived values back into these equations, each should hold true. If they do, the solution is verified as correct. This step is essential to confirm that no mistakes were made during calculations.

Linear Equations

Linear equations form the heart of this type of system-solving problem. Each equation in the system represents a line in a multi-dimensional space, and solving the system means finding the point where these lines intersect.

In algebra, a linear equation typically takes the form of \(ax + by + cz = d\), where \(a\), \(b\), \(c\), and \(d\) are constants. The solutions are found where the equations intersect, representing the set of values that satisfies all equations simultaneously.

Understanding linear equations is fundamental so that you can rearrange and manipulate them as needed, whether you're substituting variables or checking solutions. A solid grasp of these equations not only helps in solving systems, but also in visualizing and understanding real-world scenarios where such mathematical models are applied.