Problem 13 - Precalculus: Mathematics for Calculus

Question

Perform an operation on the given system that eliminates the indicated variable. Write the new equivalent system. \(\left\\{\begin{array}{rr}3 x+y+z= & 4 \\ -x+y+2 z= & 0 \\ x-2 y-z= & -1\end{array}\right.\) Eliminate the \(x\) -term from the second equation.

Step-by-Step Solution

Answer

Eliminating \(x\) yields: \(4y + 7z = 4\) along with the other equations.

Step 1: Identify the Coefficients of x

In the given system of equations, identify the coefficients of \(x\) in the first and second equations. The coefficients are \(3\) in the first equation and \(-1\) in the second equation.

Step 2: Determine the Multiplying Factor

To eliminate \(x\) from the second equation, we need to multiply the first equation by \(1\) (which is the coefficient of \(-x\) in the second equation) and multiply the second equation by \(3\) (which is the coefficient of \(x\) in the first equation).

Step 3: Multiply and Adjust the Equations

Multiply the entire first equation by \(1\): \(3x + y + z = 4\). Multiply the entire second equation by \(3\): \(-3x + 3y + 6z = 0\).

Step 4: Add the Adjusted Equations

Add the adjusted equations from Step 3 to eliminate the \(x\) term. This gives us: \(3x - 3x + y + 3y + z + 6z = 4 + 0\), which simplifies to \(4y + 7z = 4\).

Step 5: Write the New Equivalent System

The new equivalent system with \(x\) eliminated from the second equation is: \[\begin{array}{rcl} 3x + y + z &=& 4 \ 4y + 7z &=& 4 \ x - 2y - z &=& -1 \end{array}\].

Key Concepts

Elimination Method

The elimination method is a technique used to solve systems of linear equations. This method focuses on removing one variable at a time to simplify the equations. By manipulating the equations through addition or subtraction, one can eliminate a variable and solve for the others more easily. To use this method, follow these steps:

• Identify the variable you want to eliminate (in this case, it's the \(x\) in the second equation).
• Adjust the equations by multiplying them with appropriate factors, so the coefficients of the selected variable are equal.
• Add or subtract the adjusted equations to eliminate the selected variable.

This method reduces the number of equations and variables, bringing you closer to finding a solution.

Linear Equations

Linear equations are fundamental in algebra and describe a straight line when graphed. These equations have variables raised only to the first power and no products of different variables. A general linear equation looks like:\[ ax + by + cz = d \]where:

• \(x, y, z\) are variables.
• \(a, b, c\) are coefficients.
• \(d\) is a constant term.

In the given problem, we have a system of linear equations that we need to solve simultaneously using the elimination method. Understanding linear equations is key to grasping more complex algebraic concepts.

Coefficients

Coefficients are the numerical factors that multiply the variables in an equation. They play a crucial role in solving linear equations, especially when using the elimination method. To eliminate a variable, it's essential to identify these coefficients correctly, as they determine the multiplying factors for each equation. In our example, the coefficients of \(x\) are \(3\) and \(-1\). Steps for using coefficients in the elimination method:

• Identify the coefficients of the variable you want to eliminate (here, \(x\)).
• Choose multiplying factors based on these coefficients to adjust the equations.
• Once the coefficients are adjusted, add or subtract the equations to eliminate the targeted variable.

By using coefficients wisely, solving the system becomes straightforward.

Equivalent System

An equivalent system refers to a new system of equations derived from the original system, sharing the same solutions. The elimination method often results in an equivalent system by modifying the original equations without altering their solutions. For instance, when we eliminate \(x\) from the second equation in our exercise, we form an equivalent system that retains the original system's solutions but in a simplified form. This process:

• Involves operations like multiplication and addition to transform the equations.
• Keeps the number of solutions and their validity unimpaired.

Through this approach, solving systems of equations becomes more manageable, helping us find intersections (solutions) efficiently.