Solving Linear Equations with Multiple Variables

Sure, let’s dive into solving linear equations with multiple variables. Linear equations are fundamental in mathematics and are used extensively in various fields such as physics, engineering, economics, and more. Solving these equations involves finding the values of the variables that satisfy the given equations simultaneously.

Here are some examples of solving linear equations with multiple variables:

Two Variables:
Consider the system of equations:

$\begin{align*} 2x + y &= 10 \\ 3x – 2y &= 4 \end{align*}$
To solve this system, you can use the method of substitution or elimination. Let’s use the substitution method here:
From the first equation, solve for $y$ :

$y = 10 – 2x$
Substitute this value of $y$ into the second equation:

$3x – 2(10 – 2x) = 4$
Solve for $x$ :

$3x – 20 + 4x = 4$ $7x = 24$ $x = \frac{24}{7}$
Substitute the value of $x$ back into the first equation to find $y$ :

$y = 10 – 2\left(\frac{24}{7}\right)$ $y = 10 – \frac{48}{7}$ $y = \frac{2}{7}$
So, the solution to the system is $x = \frac{24}{7}$ and $y = \frac{2}{7}$ .
Three Variables:
Consider the system of equations:
$\begin{align*} x + y + z &= 6 \\ 2x – y + 3z &= 8 \\ 3x + 2y – z &= 2 \end{align*}$
To solve this system, you can again use substitution or elimination. Let’s use the elimination method this time:
First, add the first and third equations to eliminate $x$ :
$x + y + z + 3x + 2y – z = 6 + 2$ $4x + 3y = 8$
Next, add twice the first equation and the second equation to eliminate $y$ :
$2x + 2y + 2z + 2x – y + 3z = 12 + 8$ $4x + 2y + 5z = 20$
Now, subtract twice the new first equation from the previous one to eliminate $x$ :
$(4x + 3y) – 2(4x + 2y + 5z) = 8 – 40$ $4x + 3y – 8x – 4y – 10z = -32$ $-4x – y – 10z = -32$
Finally, solve this equation for $z$ :
$z = \frac{-32 + 4x + y}{-10}$
Substitute the values of $x$ and $y$ from the earlier steps into this equation to find $z$ . Let’s assume $x = 2$ and $y = 1$ for simplicity:
$z = \frac{-32 + 4(2) + 1}{-10}$ $z = \frac{-32 + 8 + 1}{-10}$ $z = \frac{-23}{-10}$ $z = \frac{23}{10}$
So, the solution to the system is $x = 2$ , $y = 1$ , and $z = \frac{23}{10}$ .

These examples showcase the process of solving linear equations with multiple variables. Depending on the number of variables and the complexity of the equations, different methods such as substitution, elimination, or matrix operations may be employed to find the solutions.

More Informations

Certainly! Let’s delve deeper into the topic of solving linear equations with multiple variables and explore additional concepts and techniques.

Gaussian Elimination Method

The Gaussian elimination method is a systematic approach to solving systems of linear equations by performing row operations on an augmented matrix. An augmented matrix combines the coefficients of the variables with the constant terms from the equations. The goal is to transform the augmented matrix into reduced row-echelon form, which reveals the solutions to the system.

Example:
Consider the system:

\begin{align*} 2x – y + z &= 1 \\ x + y + z &= 4 \\ 3x – y – 2z &= -3 \end{align*}

Write the augmented matrix:

\begin{pmatrix} 2 & -1 & 1 & | & 1 \\ 1 & 1 & 1 & | & 4 \\ 3 & -1 & -2 & | & -3 \end{pmatrix}

Perform row operations to transform the matrix:
- $R_2 = R_2 – 0.5 \cdot R_1$
- $R_3 = R_3 – 1.5 \cdot R_1$

\begin{pmatrix} 2 & -1 & 1 & | & 1 \\ 0 & 1.5 & 0.5 & | & 3 \\ 0 & -0.5 & -3.5 & | & -4.5 \end{pmatrix}

Further operations:
- $R_3 = R_3 + 0.33 \cdot R_2$

\begin{pmatrix} 2 & -1 & 1 & | & 1 \\ 0 & 1.5 & 0.5 & | & 3 \\ 0 & 0 & -3 & | & -3 \end{pmatrix}

Apply row operations to get the reduced row-echelon form:
- $R_1 = R_1 + 0.5 \cdot R_2$

\begin{pmatrix} 2 & 0 & 1.5 & | & 2.5 \\ 0 & 1.5 & 0.5 & | & 3 \\ 0 & 0 & -3 & | & -3 \end{pmatrix}

Simplify and solve for the variables:
- $z = 1$
- Substitute $z = 1$ into the second equation to solve for $y$
- Substitute $y$ and $z$ into the first equation to solve for $x$

The Gaussian elimination method is powerful for solving systems with many equations and variables. It’s also a precursor to more advanced techniques like LU decomposition and Gauss-Jordan elimination.

Matrix Inversion Method

Another method for solving linear equations involves matrix inversion. Given a system $Ax = b$ , where $A$ is the coefficient matrix, $x$ is the vector of variables, and $b$ is the constant vector, we can find the solution using the formula $x = A^{-1}b$ , where $A^{-1}$ is the inverse of matrix $A$ .

Example:
Consider the system:

\begin{align*} x + 2y &= 5 \\ 3x + 4y &= 11 \end{align*}

Write the coefficient matrix $A$ and the constant vector $b$ :

A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} , \quad b = \begin{pmatrix} 5 \\ 11 \end{pmatrix}

Find the inverse of matrix $A$ :

A^{-1} = \frac{1}{ad – bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}

where $a = 1$ , $b = 2$ , $c = 3$ , and $d = 4$ .

Calculate $A^{-1}b$ to get the solution vector $x$ :

x = A^{-1}b = \frac{1}{1 \cdot 4 – 2 \cdot 3} \begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix} \begin{pmatrix} 5 \\ 11 \end{pmatrix}

x = \frac{1}{-2} \begin{pmatrix} 4(5) – 2(11) \\ -3(5) + 1(11) \end{pmatrix} = \frac{1}{-2} \begin{pmatrix} -2 \\ -4 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}

So, the solution to the system is $x = 1$ and $y = 2$ .

Cramer’s Rule

Cramer’s Rule is a method used to solve systems of linear equations using determinants. For a system $Ax = b$ , where $A$ is the coefficient matrix and $b$ is the constant vector, the solution for each variable $x_i$ is given by $x_i = \frac{{\text{det}(A_i)}}{{\text{det}(A)}}$ , where $A_i$ is the matrix obtained by replacing the $i$ -th column of $A$ with vector $b$ .

Example:
Consider the system:

\begin{align*} x – y + 2z &= 5 \\ 2x + y + z &= 4 \\ 3x – y – 3z &= 2 \end{align*}

Write the coefficient matrix $A$ and the constant vector $b$ :

A = \begin{pmatrix} 1 & -1 & 2 \\ 2 & 1 & 1 \\ 3 & -1 & -3 \end{pmatrix} , \quad b = \begin{pmatrix} 5 \\ 4 \\ 2 \end{pmatrix}

Calculate the determinant of $A$ ( $\text{det}(A)$ ) and the determinants $\text{det}(A_1)$ , $\text{det}(A_2)$ , $\text{det}(A_3)$ :

\text{det}(A) = \begin{vmatrix} 1 & -1 & 2 \\ 2 & 1 & 1 \\ 3 & -1 & -3 \end{vmatrix} = -4

\text{det}(A_1) = \begin{vmatrix} 5 & -1 & 2 \\ 4 & 1 & 1 \\ 2 & -1 & -3 \end{vmatrix} = 13

\text{det}(A_2) = \begin{vmatrix} 1 & 5 & 2 \\ 2 & 4 & 1 \\ 3 & 2 & -3 \end{vmatrix} = -39

\text{det}(A_3) = \begin{vmatrix} 1 & -1 & 5 \\ 2 & 1 & 4 \\ 3 & -1 & 2 \end{vmatrix} = -13

Calculate the solutions using Cramer’s Rule:

x = \frac{{\text{det}(A_1)}}{{\text{det}(A)}} = \frac{13}{-4} = -\frac{13}{4}

y = \frac{{\text{det}(A_2)}}{{\text{det}(A)}} = \frac{-39}{-4} = \frac{39}{4}

z = \frac{{\text{det}(A_3)}}{{\text{det}(A)}} = \frac{-13}{-4} = \frac{13}{4}

The solutions are $x = -\frac{13}{4}$ , $y = \frac{39}{4}$ , and $z = \frac{13}{4}$ .

These methods—Gaussian elimination, matrix inversion, and Cramer’s Rule—offer different approaches to solving systems of linear equations with multiple variables. Each method has its strengths and weaknesses depending on the size and complexity of the system, making them valuable tools in mathematical problem-solving.