Question

In each case verify that the following are solutions for all values of \(s\) and \(t\). $$ \text { a. } \begin{aligned} x &=19 t-35 \\ y &=25-13 t \end{aligned} $$ \(z=t\) is a solution of $$ \begin{array}{l} 2 x+3 y+z=5 \\ 5 x+7 y-4 z=0 \\ \text { b. } x_{1}=2 s+12 t+13 \end{array} $$ \(x_{2}=s\) $$ x_{3}=-s-3 t-3 $$ \(\pi_{4}=t\) is a solution of $$ \begin{aligned} 2 x_{1}+5 x_{2}+9 x_{3}+3 x_{4} &=-1 \\ x_{1}+2 x_{2}+4 x_{3} &=1 \end{aligned} $$

Short answer

Both given equations are solutions for their respective systems for all values of parameters.

Step-by-step solution

Substitute the first set of equations into the first system

Substitute \\(x = 19t - 35\\), \ \(y = 25 - 13t\\), and \ \(z = t \ into the first system of equations: \ \)2x + 3y + z = 5\\( and \ \)5x + 7y - 4z = 0\$.

Verify the first equation in the first system

Substitute the expressions for x, y, and z into \ \(2x + 3y + z = 5\).\ Calculate each term:\[ 2(19t - 35) + 3(25 - 13t) + t \]Simplify:\[ 38t - 70 + 75 - 39t + t \]Combine like terms: \ \(38t - 39t + t = 0t\), \ \(-70 + 75 = 5\)\.The equation becomes \ \(5 = 5\\), which is true.

Verify the second equation in the first system

Substitute the expressions for x, y, and z into \ \(5x + 7y - 4z = 0\)\. Calculate each term:\[ 5(19t - 35) + 7(25 - 13t) - 4t \]Simplify:\[ 95t - 175 + 175 - 91t - 4t \]Combine like terms:\[ 95t - 91t - 4t = 0 \]The equation becomes \ \(0 = 0\)\, which is true.

Conclusion for part (a)

Since both substitutions into the equations resulted in true statements, \ \(x = 19t - 35\\), \ \(y = 25 - 13t\\), \ \(z = t\\) are solutions for the first system for all values of \ \(t\\).

Substitute the second set of equations into the second system

Substitute \ \(x_1 = 2s + 12t + 13\\), \ \(x_2 = s\\), \ \(x_3 = -s - 3t - 3\\), \ \(\pi_4 = t\\) into the second system of equations: \ \(2x_1 + 5x_2 + 9x_3 + 3\pi_4 = -1\\) and \ \(x_1 + 2x_2 + 4x_3 = 1\\).

Verify the first equation in the second system

Substitute the expressions into \ \(2x_1 + 5x_2 + 9x_3 + 3\pi_4 = -1\)\. Calculate each term:\[ 2(2s + 12t + 13) + 5s + 9(-s - 3t - 3) + 3t \]Simplify:\[ 4s + 24t + 26 + 5s - 9s - 27t - 27 + 3t \]Combine like terms:\[ (4s + 5s - 9s) + (24t - 27t + 3t) + (26 - 27) \]This simplifies to \ \(0s + 0t - 1 = -1\\), which is true.

Verify the second equation in the second system

Substitute the expressions into \ \(x_1 + 2x_2 + 4x_3 = 1\).\ Calculate each term:\[ (2s + 12t + 13) + 2s + 4(-s - 3t - 3) \]Simplify:\[ 2s + 12t + 13 + 2s - 4s - 12t - 12 \]Combine like terms:\[ (2s + 2s - 4s) + (12t - 12t) + (13 - 12) \]This simplifies to \ \(0s + 0t + 1 = 1\\), which is true.

Conclusion for part (b)

Since both substitutions into the equations resulted in true statements, \ \(x_1 = 2s + 12t + 13\\), \ \(x_2 = s\\), \ \(x_3 = -s - 3t - 3\\), \ \(\pi_4 = t\\) are solutions for the second system for all values of \ \(s\\) and \ \(t\\).

Key concepts

Substitution Method

The Substitution Method is a fundamental technique in solving systems of linear equations. It involves substituting one variable's expression into the other equations, effectively reducing the number of variables and simplifying the system. This method is particularly useful when one of the equations is already solved for a single variable.

Here's how it generally works:

• Start by solving one equation for one of the variables. This puts the equation into a form such as \( x = y - 3 \).
• Substitute this expression into the other equations, replacing the target variable.
• Solve the resulting system, which now has one fewer variable.
• Back-substitute to find the values of the eliminated variables.

Let's consider the example given in the exercise. By substituting \(x = 19t - 35\), \(y = 25 - 13t\), and \(z = t\) into the first system of equations, we reduced the number of variables, making it easier to check if these expressions satisfy the given equations. Substitution simplifies the initial problem, transforming it into computation and verification steps.

Verification of Solutions

Verification is the process of determining whether the proposed solutions actually satisfy the equations of the system. This step is crucial because it confirms the correctness of the potential solutions. Here's how you do this:

• Substitute the proposed values of the variables back into the original equations.
• Simplify each equation to see if a true statement (like \(0 = 0\) or \(5 = 5\)) results.
• If every equation holds true for the proposed expressions, the solutions are verified.

In the exercise, for example, we verified the solutions by substituting the expressions for \(x, y, z\) into each equation of the given system:
For \(2x + 3y + z = 5\), substituting the expressions transforms the equation into \(5 = 5\), confirming the solution. Similarly, for \(5x + 7y - 4z = 0\), the simplification also holds true as \(0 = 0\). These steps aim to ensure the proposed solutions are valid under all conditions.
Verification requires careful simplification and checks to avoid errors, ensuring each condition is perfectly satisfied.

Parameterization of Solutions

Parameterization involves expressing the solutions of a system of equations in terms of one or more parameters. It is useful when an infinite number of solutions exist, allowing for a convenient representation that covers all possibilities. Parameters, often denoted as \(t\) or \(s\), represent free variables that can take any value.

Here's how you use parameterization:

• Identify the dependent variables in the equations and express them in terms of independent parameters.
• Since the parameter (like \(t\) or \(s\)) can vary freely, each choice generates a unique solution.
• This results in a formulation that demonstrates the interdependence of variables in the system.

In our example, the solutions are given as parameterized forms: \(x = 19t - 35\), \(y = 25 - 13t\), \(z = t\), which illustrate how each solution depends linearly on \(t\). This shows that as \(t\) changes, the entire set of solutions shifts but still remains within the constraints of the system. Parameterization effectively encapsulates an infinite solution set in a compact form, offering a clear understanding of how solutions relate.