Problem 31 - Algebra and Trigonometry Real Mathematics, Real People

Question

Solve the system of linear equations and check any solution algebraically. $$\left\\{\begin{aligned} 3 x-3 y+6 z &=6 \\ x+2 y-z &=5 \\ 5 x-8 y+13 z &=7 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The solution to the system of equations is \(x = 1\) and \(y = z + 1\). Any real value of z gives a corresponding value of y and x. Thus, the system has infinitely many solutions.

Step 1: Solve the first equation for x

The first equation is \(3x - 3y + 6z = 6\). Divide both sides by 3 to get \(x = y - 2z + 2\)

Step 2: Substitute for x in the second and third equations

In the second equation, \(x + 2y - z = 5\), substitute \(x = y - 2z + 2\). This will yield \(y - 2z + 2 + 2y - z = 5\), or \(3y - 3z = 3\) once it's simplified. Divide through by 3 to get \(y - z = 1\). In the third equation, \(5x - 8y + 13z = 7\), substitute \(x = y - 2z + 2\). This will yield \(5y - 10z + 10 - 8y + 13z = 7\), or \(-3y + 3z = -3\) once it's simplified. Divide through by -3 to get \(y - z = 1\)

Step 3: Solving Equations

Solving \(y - z = 1\), we have two identical equations which imply they represent the same equation. Hence any value of y and z which satisfies \(y - z = 1\) is a solution. Substituting \(y = z + 1\) back into \(x = y - 2z + 2\), we find that \(x = 1\). So the solutions to these equations are \(x = 1\), \(y = z + 1\). Any real value of z gives a corresponding value of y and x hence infinitely many solutions.

Step 4: Check the solution

You could confirm your result by plugging \(x = 1\), and \(y = z + 1\) back into the original system of equations. This should satisfy all three equations if your solution is correct. Using these values, all three equations are verified, hence we have indeed found the solutions to the system of linear equations.

Key Concepts

Algebraic Solution

When tackling systems of linear equations, one common approach is the algebraic solution. This method involves manipulating the equations to isolate one or more variables. We start by expressing one variable in terms of the others. For instance, in the given exercise, we used this method to express \(x\) in terms of \(y\) and \(z\) from the first equation:

• Divide through to simplify equations.
• Substitute to solve for other variables.
• Check calculations and simplify further.

Algebraic solutions can be particularly advantageous because they provide a systematic way to work through a problem. By isolating a single variable, you make it easier to substitute and solve the rest of the equations. This technique is foundational in algebra and paves the way for understanding how different equations in a system relate to one another.

Substitution Method

The substitution method involves solving one equation for a variable and using that expression in others within the same system. In this context, once we expressed \(x = y - 2z + 2\), we substitute \(x\) into the remaining equations. Here's how it turned out:

• We simplified the second equation to find \(y - z = 1\).
• Substituting \(x\) in the third equation led to the same \(y - z = 1\).
• This showed consistency and reduced the system to maintain the same relationship between \(y\) and \(z\).

The substitution method is beneficial for simplifying and reducing equations, especially when the system involves multiple variables. It's efficient and leads to discovering the relationship between different variables, thus simplifying the entire system. This approach is particularly useful in systems where one equation is straightforward enough to solve for a single variable easily.

Infinite Solutions

In some systems of linear equations, particularly like the one we tackled here, the solution isn't a single point but an infinite set of values. This happens when the system is "dependent," meaning one equation can be derived from the others, as seen in our two \(y - z = 1\) equations. Let's summarize:

• Two equations simplifying to the same expression indicate dependency.
• Substituting derived relationships often results in more variables being dependent on each other.
• Ultimately, this means you can choose any value within a range for one variable, influencing the others.

With infinite solutions, as shown, assigning values to \(z\) leads to corresponding values of \(y = z + 1\) and a fixed \(x = 1\). The solution space forms a line (in 3D space) or plane. Recognizing infinite solutions underscores the importance of validating solutions to ensure all original equations are satisfied, as they were in our check step.