Problem 29 - College Algebra and Calculus: An Applied Approach

Question

Solve the system of equations. $$\left\\{\begin{array}{l}12 x+5 y+z=0 \\ 12 x+4 y-z=0\end{array}\right.$$

Step-by-Step Solution

Answer

The solution set for the system of equations is \( y = -8/3*x \) and \( z = -4x \) where x is any real number.

Step 1: Setup

We have given a system of two linear equations:\(12x + 5y + z = 0 \)and\(12x + 4y - z = 0 \).

Step 2: Apply the Elimination Method

We start by adding these two equations to eliminate the variable z:\( (12x + 5y + z) + (12x + 4y - z) = 0 + 0. This simplifies to \(24x + 9y = 0 \).

Step 3: Converting to y-intercept form

Rearrange the equation in terms of y to find its value in relation to x:\(9y = -24x \) Divide by 9 on both sides:\(y = -24x / 9 \)Simplify further to:\( y = -8/3*x \).

Step 4: Substitute y's value in one of the original equations.

We choose the first equation and substitute y's value:\(12x + 5*(-8/3*x) + z = 0\)This simplifies to \(4x + z = 0\)Therefore, \(z = -4x\) .

Step 5: Conclusion

We've found the values of y and z that make the given system of equations true for all values of x. To represent the entire solution set, we can combine these into a parameterized solution set: \(y = -8/3*x , z = -4x\).

Key Concepts

Elimination Method

The elimination method is a strategy used to solve systems of linear equations. It's particularly useful when you want to eliminate one of the variables by adding or subtracting the equations. In our exercise, the objective was to eliminate the variable \(z\) from the system. To carry this out:

• We took the two given equations: \(12x + 5y + z = 0\) and \(12x + 4y - z = 0\).
• We added them together, which effectively canceled out \(z\):

By doing this, we arrive at a simpler equation, \(24x + 9y = 0\), with only two variables, \(x\) and \(y\). The elimination method is a powerful tool because it simplifies complex systems into more manageable problems.

Linear Equations

Linear equations form the backbone of the system presented in the exercise. These equations define a straight line when plotted on a graph. They follow the general format \(ax + by + cz = d\), where \(a\), \(b\), \(c\), and \(d\) are constants. In our problem:

• Equation 1 is: \(12x + 5y + z = 0\)
• Equation 2 is: \(12x + 4y - z = 0\)

Since both these equations are linear, a solution will correspond to the point where the lines intersect, or in this parameterized solution, a line of intersection. Understanding linear equations is important because they are foundational for many aspects of mathematics and are used frequently in various applications, such as calculating budget constraints or predicting trends.

Parameterized Solution

A parameterized solution describes an entire set of solutions to a system of equations by expressing one or more of the variables in terms of a parameter, often denoted as \(t\). In the given exercise:

• We expressed \(y\) as \(y = -\frac{8}{3}x\)
• We expressed \(z\) as \(z = -4x\)

By doing so, \(x\) becomes the free variable or parameter for this system. This is because we can choose any real number for \(x\), and there will be corresponding values for \(y\) and \(z\) that satisfy both original equations. Parameterized solutions are powerful in capturing the entire range of possible solutions, allowing one to understand and manipulate the results flexibly and comprehensively.