Simultaneous equations involve two or more equations with two or more variables that are solved together. The solutions are the values of the variables that satisfy all equations simultaneously. In this section, we will focus on linear equations with two variables, typically represented as $ax + by = c$ and $dx + ey = f$, where $a$, $b$, $c$, $d$, $e$, and $f$ are constants.

**Understanding the Basics**

**Linear Equations**: A linear equation in two variables has the general form $ax + by = c$. These equations graph as straight lines when plotted on a coordinate plane.

**Simultaneous Equations**: When two linear equations are considered together to find a common solution for $x$ and $y$, they are known as simultaneous equations.

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**Solving Methods**: There are three primary methods for solving simultaneous equations:

**Substitution Method****Elimination Method****Graphical Method**

For the purpose of these notes, we will focus on the Substitution and Elimination methods, as they are more algebraically intensive and relevant for exam preparation.

**Substitution Method**

This method involves solving one of the equations for one variable in terms of the other and then substituting this value into the other equation.

**Steps involved:**

1. Solve one equation for one variable in terms of the other.

2. Substitute this expression into the second equation, eliminating the chosen variable.

3. Solve the resulting equation for the remaining variable.

4. Substitute the obtained value back into the original equation (or the equation you used in step 1) to solve for the eliminated variable.

**Example 1**:

Given equations: $x + y = 5$ and $3x - 2y = 4$.

#### Solution:

1. Solve for $y$ from the first equation: $y = 5 - x$.

2. Substitute $y$ into the second equation and solve for $x$:

$3x - 2(5 - x) = 4 \implies x = \frac{14}{5}$3. Substitute $x = \frac{14}{5}$ back into $y = 5 - x$ to find $y$:

$y = 5 - \frac{14}{5} = \frac{11}{5}$**Final Answer:** $x = \dfrac{14}{5}, y = \dfrac{11}{5}$.

### Example 2:

Solve the following system of equations:

- $2x + y = 7$
- $3x - y = 5$

**Solution:**

1. Solve the first equation for y: $y = -2x + 7$

2. Substitute this expression for y in the second equation:

$3x - (-2x + 7) = 5$3. Solve the resulting equation for x:

$5x - 7 = 5$

$5x = 12$

$x = \frac{12}{5}$

4. Substitute $x = \frac{12}{5}$ back into the first equation to solve for y:

$2( \frac{12}{5}) + y = 7$

$\frac{24}{5} + y = 7$

$y = \frac{11}{5}$

Therefore, the solution is: $x = \dfrac{12}{5}, y = \dfrac{11}{5}$

**Elimination Method**

This method involves manipulating the equations algebraically to eliminate one variable by adding or subtracting the equations strategically. Once one variable is eliminated, you can solve the resulting equation for the remaining variable and then substitute this value back into either original equation to solve for the eliminated variable.

**Steps involved:**

1. Manipulate the equations algebraically to eliminate one variable by adding or subtracting the equations strategically. Aim to get the coefficients of one variable to be additive inverses (e.g., one positive and one negative with the same absolute value).

2. Solve the resulting equation for the remaining variable.

3. Substitute the obtained value back into either original equation to solve for the eliminated variable.

### Example 3:

Given equations: $2x + 3y = 7$ and $4x - y = 9$.

#### Solution:

1. Original Equations:

Equation 1: $2x + 3y = 7$

Equation 2: $4x - y = 9$

2. Multiply second equation by 3:

$2x + 3y = 7$

$12x - 3y = 27$

3. Add the equations to eliminate $y$:

$14x = 34$

4. Solve for $x$:

$x = \frac{34}{14} = \frac{17}{7}$

5. Substitute $x$ back to find $y$ (using $4x - y = 9)$:

$y = \frac{5}{7}$

**Final Answer: **$(x = \dfrac{17}{7}, y = \dfrac{5}{7})$

**Example 4:**

Solve the following system of equations:

- $x + 2y = 4$
- $2x + 4y = 8$

**Solution:**

1. Observe that the y coefficients have a 2:1 ratio.

We can eliminate y by subtracting the first equation from twice the second equation.

$(2) \cdot (x + 2y = 4) - (x + 2y = 7)$$x = 0$2. Substitute x = 0 into the first equation to solve for y:

$0 + 2y = 4$

$2y = 4$

$y = 2$

Therefore, the solution is: $x = 0, \quad y = 2$

**Real-World Application**

**Example 5:**

Scenario: Selling tickets where adult tickets cost £5 each, child tickets cost £3 each, 20 tickets sold in total, and £70 collected.

#### Solution:

1. Equations based on the scenario: $x + y = 20$ and $5x + 3y = 70$.

2. Solve the system of equations: $x = 5, y = 15$

**Conclusion:** 5 adult tickets and 15 child tickets were sold.