Problem 51 - Precalculus

Question

Solve the system of linear equations using Gaussian elimination with back- substitution. $$\begin{array}{rr} 2 x+3 y= & 1 \\ x+y= & -2 \end{array}$$

Step-by-Step Solution

Answer

The solution is \(x = -7\) and \(y = 5\).

Step 1: Set up the augmented matrix

Rewrite the system of linear equations as an augmented matrix. Start with the equations:\[\begin{array}{cc|c}2 & 3 & 1 \1 & 1 & -2 \\end{array}\]

Step 2: Use Gaussian elimination to create zeros below the first pivot

To eliminate the first coefficient in the second row, we can subtract the first row multiplied by \(0.5\) from the second row. This gives:\[\begin{array}{cc|c}2 & 3 & 1 \0 & -0.5 & -2.5 \\end{array}\]

Step 3: Simplify the matrix to row-echelon form

Divide the second row by \(-0.5\) to make the leading coefficient 1:\[\begin{array}{cc|c}2 & 3 & 1 \0 & 1 & 5 \\end{array}\]

Step 4: Back-substitution to solve for y

Now that the system is in row-echelon form, we can easily solve for \(y\) from the second equation:\( y = 5 \).

Step 5: Back-substitution to solve for x

Substitute \(y = 5\) back into the first equation to solve for \(x\):\(2x + 3(5) = 1 \)\(2x + 15 = 1 \)\(2x = -14 \)\(x = -7 \).

Key Concepts

Linear Equations

Linear equations are mathematical statements that show the relationship between different variables through constants and coefficients. They typically have the form of elements being added or subtracted, like in the set of equations given in the exercise: \[ 2x + 3y = 1 \] and \[ x + y = -2 \]. Key concepts to keep in mind are:

• Each equation represents a line when plotted on a graph.
• The objective is to find values of \(x\) and \(y\) that satisfy both equations simultaneously.
• Systems of linear equations can have one solution, no solution, or infinitely many solutions depending on the lines' relationship.

When using methods like Gaussian elimination, these equations are manipulated to ease solution finding.

Augmented Matrix

An augmented matrix is a simpler, more condensed representation of a system of linear equations. It turns a set of equations into rows in a matrix, bringing together the coefficients and constants.For the exercise's equations, the augmented matrix looks like:\[ \begin{array}{cc|c} 2 & 3 & 1 \ 1 & 1 & -2 \end{array} \]It denotes:

• The first two columns represent the coefficients of \(x\) and \(y\), respectively.
• The third column after the vertical line (denoted as "|") contains the constants from the right side of each equation.

The simplicity of the augmented matrix format streamlines operations needed for techniques like Gaussian elimination, making calculations efficient and reducing error opportunities.

Row-Echelon Form

Row-echelon form is a specific arrangement for augmented matrices that simplifies solving systems of linear equations. It focuses on having lower triangular zeros (zeros below the leading coefficients or pivots).The necessary steps to achieve this involve Gaussian elimination, where the matrix for the provided exercise became:\[ \begin{array}{cc|c} 2 & 3 & 1 \ 0 & 1 & 5 \end{array} \]This format is desirable because:

• It exhibits a stair-step pattern, easing the back-substitution process.
• Zeros below pivots simplify solving earlier equations without interference from subsequent equations.

Transforming augmented matrices into row-echelon form is vital for efficient problem-solving.

Back-Substitution

Back-substitution is the final, straightforward step after reaching row-echelon form. It involves solving for variables starting from the bottom of the matrix upwards.For instance, in the exercise's matrix:- From the second row, we immediately find:\[ y = 5 \],because it only involves the variable \(y\).- For the first row, substituting back the value of \(y\): \( 2x + 3(5) = 1 \), which simplifies to:\( 2x + 15 = 1 \), leading to:\( 2x = -14 \) and \( x = -7 \).The benefits of back-substitution include:

• It's a direct process where each equation resolves one variable.
• Avoids the complexity of simultaneous equations since earlier solved values are directly used.

This step efficiently and logically completes the solution of the system.