Solving Systems of Linear Equations in Two Variables

Solving Systems of Equations with Linear Equations in Two Variables

Solving systems of equations with linear equations in two variables involves finding the values of the variables that satisfy all the equations in the system. This process can be done through several methods, including substitution, elimination, and matrix operations.

Substitution Method

The substitution method involves solving one of the equations for one of the variables and then substituting the resulting expression into the other equation. This process can be repeated until the variables are expressed in terms of one another and only one variable remains to be solved for.

For example, consider the following system of equations:

$$ \begin{aligned} x + 2y &= 6 \ 3x - 4y &= 5 \end{aligned} $$

To solve this system using the substitution method, we first solve the first equation for x:

$$ x = 6 - 2y $$

Now, substitute this expression for x into the second equation:

$$ 3(6 - 2y) - 4y = 5 $$

Expanding and simplifying this equation, we get:

$$ 18 - 6y - 4y = 5 $$

$$ -10y = -13 $$

$$ y = 1.3 $$

Now that we have a value for y, we can substitute it back into the expression we found for x:

$$ x = 6 - 2(1.3) $$

$$ x = 6 - 2.6 $$

$$ x = 3.4 $$

Thus, the solution to this system of equations is x = 3.4 and y = 1.3.

Elimination Method

The elimination method involves adding or subtracting the equations in the system to eliminate one of the variables. This process can be repeated until the variables are expressed in terms of one another and only one variable remains to be solved for.

For example, consider the following system of equations:

$$ \begin{aligned} 2x + 3y &= 8 \ -x - 4y &= -2 \end{aligned} $$

To solve this system using the elimination method, we first add the two equations:

$$ 2x + 3y + (-x - 4y) = 8 + (-2) $$

$$ x - y = 4 $$

Now, we can solve this equation for x:

$$ x = 4 + y $$

Substitute this expression for x into the first equation:

$$ 2(4 + y) + 3y = 8 $$

Expanding and simplifying this equation, we get:

$$ 8 + 2y + 3y = 8 $$

$$ 5y = 0 $$

$$ y = 0 $$

Now that we have a value for y, we can substitute it back into the expression we found for x:

$$ x = 4 + 0 $$

$$ x = 4 $$

Thus, the solution to this system of equations is x = 4 and y = 0.

Matrix Operations

Solving systems of equations can also be done using matrix operations. A system of equations can be represented as an augmented matrix. Performing operations on this matrix, such as row operations, can lead to the matrix being in row echelon form or reduced row echelon form, which can be used to solve for the variables.

For example, consider the same system of equations from the previous examples:

$$ \begin{aligned} 2x + 3y &= 8 \ -x - 4y &= -2 \end{aligned} $$

Representing this system as an augmented matrix:

$$ \begin{bmatrix} 2 & 3 \ -1 & -4 \end{bmatrix} $$

Performing row operations on this matrix, we can obtain the following matrix in row echelon form:

$$ \begin{bmatrix} 1 & 1.5 \ -1 & -2 \end{bmatrix} $$

From this matrix, we can see that the variables x and y are both expressed in terms of y. To solve for y, we can set the second row equal to zero and solve for y:

$$ -y = 2 $$

$$ y = -2 $$

Now that we have a value for y, we can substitute it back into the expression we found for x:

$$ x = 1.5(-2) $$

$$ x = -3 $$

Thus, the solution to this system of equations is x = -3 and y = -2.

In conclusion, solving systems of equations with linear equations in two variables can be done using various methods, including substitution, elimination, and matrix operations. Each method has its own set of steps and processes to follow, but ultimately leads to finding the values of the variables that satisfy all the equations in the system.