Simultaneous Equation

Simultaneous equations are a set of two or more equations that share the same variables and are considered together because they are to be solved at the same time. In mathematics, the primary goal when dealing with simultaneous equations is to find the values of the variables that satisfy all the given equations simultaneously.

A general form of a simultaneous equation with two variables (x and y) can be represented as follows:

Equation 1: ax + by = c

Equation 2: dx + ey = f

Here, a, b, d, e, c, and f are constants, and x and y are the variables we want to determine. The goal is to find the specific values of x and y that satisfy both equations simultaneously.

There are several methods to solve simultaneous equations, but some of the most common ones include:

Substitution method: In this method, one of the equations is solved for one variable in terms of the other variable. Then, the obtained expression is substituted into the other equation. This process eliminates one variable, and you can then solve for the remaining variable.

Elimination method: In this method, you manipulate the equations so that when you add or subtract them, one of the variables gets eliminated, leaving you with a single equation containing only one variable, which you can then solve.

Matrix method: You can represent the system of equations in matrix form and use techniques such as Gaussian elimination or matrix inversion to solve for the variables.

Once you find the values of x and y that satisfy both equations, you can verify your solution by substituting these values back into the original equations to ensure they hold true.

Simultaneous equations find wide applications in various fields, including physics, engineering, economics, and many other areas where multiple variables are interrelated and need to be determined together.

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solve examples using those methods

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Sure, let's solve a system of simultaneous equations using the substitution and elimination methods.

Example 1:

Consider the following system of equations:

Equation 1: 2x + y = 10

Equation 2: 3x - 2y = 4

Substitution Method:

Step 1: Solve one of the equations for one variable. Let's solve Equation 1 for y:

y = 10 - 2x

Step 2: Substitute the expression for y (obtained in Step 1) into Equation 2:

3x - 2(10 - 2x) = 4

Step 3: Now, solve for x:

3x - 20 + 4x = 4

7x - 20 = 4

7x = 24

x = 24 / 7

Step 4: Substitute the value of x into the expression for y:

y = 10 - 2(24 / 7)

y = 10 - 48 / 7

y = (70 - 48) / 7

y = 22 / 7

The solution to the system of equations is x = 24/7 and y = 22/7.

Elimination Method:

Step 1: We'll eliminate one of the variables by adding or subtracting the equations. Let's eliminate y in this case. To do that, we'll make the coefficients of y in both equations equal.

Multiply Equation 1 by 2 and Equation 2 by 1 so that the coefficients of y become the same:

Equation 1 (multiplied by 2): 4x + 2y = 20

Equation 2 (multiplied by 1): 3x - 2y = 4

Step 2: Add the two equations to eliminate y:

(4x + 2y) + (3x - 2y) = 20 + 4

7x = 24

x = 24 / 7

Step 3: Substitute the value of x back into either Equation 1 or 2 to solve for y. We'll use Equation 1:

2(24/7) + y = 10

48/7 + y = 10

y = 10 - 48/7

y = (70 - 48) / 7

y = 22 / 7

The solution to the system of equations is x = 24/7 and y = 22/7, which matches the result obtained using the substitution method.

Example 1 (Recall the system of equations):

Equation 1: 2x + y = 10

Equation 2: 3x - 2y = 4

To solve this using the matrix method, we'll represent the system of equations in matrix form:

Step 1: Create the coefficient matrix (A), variable matrix (X), and constant matrix (B):

A = | 2 1 |

| 3 -2 |

X = | x |

| y |

B = | 10 |

| 4 |

Step 2: Now, we can write the matrix equation AX = B:

| 2 1 | | x | | 10 |

| 3 -2 | * | y | = | 4 |

Step 3: To solve for X (the variable matrix), we can use matrix inversion. The solution is given by:

X = A^(-1) * B

Step 4: Find the inverse of matrix A (A^(-1)):

The inverse of a 2x2 matrix [a b; c d] is given by:

A^(-1) = (1 / det(A)) * | d -b |

| -c a |

where det(A) is the determinant of matrix A, given by det(A) = ad - bc.

For our matrix A:

det(A) = (2 * -2) - (1 * 3) = -4 - 3 = -7

So, A^(-1) = (1 / -7) * | -2 -1 |

| -3 2 |

Step 5: Now, calculate X using the formula X = A^(-1) * B:

X = | -2 -1 | | 10 |

| -3 2 | * | 4 |

X = | (-2 * 10) + (-1 * 4) |

| (-3 * 10) + (2 * 4) |

X = | -20 - 4 |

| -30 + 8 |

X = | -24 |

| -22 |

Step 6: The solution for X is:

x = -24

y = -22

Thus, the solution to the system of equations using the matrix method is x = -24 and y = -22, which matches the results obtained using the substitution and elimination methods.