Problem 45 - Precalculus Mathematics for Calculus

Question

Solve the system of linear equations. $$\left\\{\begin{array}{l} x-y+6 z=8 \\ x+z=5 \\ x+3 y-14 z=-4 \end{array}\right.$$

Step-by-Step Solution

Answer

The solution is \((x, y, z) = (5, -3, 0)\).

Step 1: Simplify the System of Equations

The given system of equations is: \( \begin{align*} x - y + 6z &= 8, \ x + z &= 5, \ x + 3y - 14z &= -4. \end{align*} \)We have three equations and three variables. We will use the substitution or elimination method to solve it.

Step 2: Solve for x in terms of z

From the second equation, \( x + z = 5 \), we can express \( x \) as:\[ x = 5 - z. \]

Step 3: Substitute x into Other Equations

Substitute \( x = 5 - z \) into the first and third equations:First equation: \((5 - z) - y + 6z = 8 \) simplifies to \( -y + 5z = 3 \). Third equation: \((5 - z) + 3y - 14z = -4 \) simplifies to \( -z + 3y = -9. \)

Step 4: Solve the Simplified System

We have two equations now:\( -y + 5z = 3 \) \( -z + 3y = -9. \)Let's solve for \( y \) in the first equation: \( y = 5z - 3. \) Substitute into the second equation:\[ -z + 3(5z - 3) = -9. \]

Step 5: Solve for z

Expand and simplify the equation:\( -z + 15z - 9 = -9 \)which simplifies to \( 14z = 0. \)Thus, \( z = 0. \)

Step 6: Find y using z = 0

Substitute \( z = 0 \) into \( y = 5z - 3 \):\[ y = 5(0) - 3 = -3. \]

Step 7: Find x using z = 0

Substitute \( z = 0 \) into \( x = 5 - z \):\[ x = 5 - 0 = 5. \]

Step 8: Verify the Solution

Check the solution \( (x, y, z) = (5, -3, 0) \) in each original equation:1. \( 5 - (-3) + 6(0) = 8 \) is true.2. \( 5 + 0 = 5 \) is true.3. \( 5 + 3(-3) - 14(0) = -4 \) is true.All equations are satisfied, confirming our solution is correct.

Key Concepts

System of Linear Equations

A system of linear equations is a collection of two or more linear equations with the same set of variables. In the exercise provided, the system contains three equations involving three variables \(x\), \(y\), and \(z\). An important goal is to find values for these variables that satisfy all the equations simultaneously.

Why do we care about solving these systems? Well, they are foundational in mathematics and are applied in various fields including engineering, economics, and computer science. By finding common solutions, we can tackle problems ranging from financial forecasting to operational research.

Each equation within a system of equations represents a line in mathematical space. Solving the system means finding the point where these lines intersect, which is the solution that satisfies all the equations in the system. In our exercise, solving the system involved finding the values \((x, y, z) = (5, -3, 0)\) where all three lines intersect in a 3-dimensional space.

Substitution Method

The substitution method is a technique used to solve a system of equations by expressing one variable in terms of another, and then substituting this expression into other equations. In our example, we started by expressing \(x\) in terms of \(z\) using the equation \(x + z = 5\). This gave us \(x = 5 - z\).

Once one variable is isolated, we substitute it into the other equations. This technique helps simplify the system by reducing the number of variables. In our exercise, this substitution helped us simplify the first and third original equations, transforming our three-variable system into a two-variable problem.

When using substitution, remember:

• Choose the simplest equation to start with; it's easier to rearrange for a variable.
• Substitute carefully to avoid errors in later steps.
• Check back your derived expressions in the original equations to ensure they are correct.

The substitution method works best for systems where isolating a variable is straightforward, which is nicely demonstrated in this exercise.

Elimination Method

The elimination method is another approach for solving systems of equations. This technique involves adding or subtracting equations to eliminate one or more unknowns. Although, in the original exercise, we didn't employ elimination extensively, understanding it is key to mastering systems of equations.

In another part of solving the example, we used some principles of elimination in simplifying several equations by organizing terms, which eventually allowed straightforward substitutions. The power of the elimination method lies in its ability to help when substitution becomes cumbersome.

Key points for elimination:

• Align terms so the variable you want to eliminate can easily cancel out.
• Multiply equations if necessary so that coefficients of a chosen variable are opposites.
• Add or subtract equations to eliminate the chosen variable.

While elimination may require some upfront adjustments, it effectively simplifies solving when dealing with larger systems or those with less straightforward initial equations.