3.3 Linear & Simultaneous Equations

Introduction to Linear Equations

A linear equation is an algebraic equation of the first degree, meaning the highest power of any variable in the equation is $1$. The general form of a linear equation with one variable is:

$$ax + b = 0$$

Where $a$ and $b$ are constants and $x$ is the unknown variable. Solving a linear equation involves finding the value of the variable that makes the equation true.

Solving Simple Linear Equations

To solve a linear equation, you isolate the unknown variable on one side of the equation. This is achieved using the balance method: whatever operation you perform on one side of the equation (addition, subtraction, multiplication, division), you must also perform on the other side.

Transposition Rules

• If a term is added, subtract it from both sides.
• If a term is subtracted, add it to both sides.
• If a variable is multiplied by a coefficient, divide both sides by that coefficient.
• If a variable is divided, multiply both sides.

Worked Example: Rearranging and Solving

Problem: Solve the equation for $x$:

$$3(2x - 4) - 2(x + 1) = 6$$

Step-by-step Solution:

1. Expand the brackets on the left side: $$(6x - 12) - (2x + 2) = 6$$ $$6x - 12 - 2x - 2 = 6$$
2. Combine like terms on the left side: $$4x - 14 = 6$$
3. Add $14$ to both sides to isolate the variable term: $$4x = 6 + 14$$ $$4x = 20$$
4. Divide both sides by $4$ to solve for $x$: $$x = \frac{20}{4} = 5$$

Transposing Formulas with Multiple Variables

In EASA Module 1, you will often need to rearrange a formula to make a different variable the "subject." Let's take the parallel resistance equation:

$$R_t = \frac{R_1 R_2}{R_1 + R_2}$$

Rearrange the equation to make $R_1$ the subject:

1. Multiply both sides by $(R_1 + R_2)$ to clear the fraction: $$R_t(R_1 + R_2) = R_1 R_2$$
2. Expand the bracket: $$R_t R_1 + R_t R_2 = R_1 R_2$$
3. Collect all terms containing $R_1$ on one side of the equation. Subtract $R_t R_1$ from both sides: $$R_t R_2 = R_1 R_2 - R_t R_1$$
4. Factor out $R_1$ on the right side: $$R_t R_2 = R_1(R_2 - R_t)$$
5. Divide both sides by $(R_2 - R_t)$ to isolate $R_1$: $$R_1 = \frac{R_t R_2}{R_2 - R_t}$$

This technique of grouping terms and factoring is vital for resolving EASA formula transposition questions.

Simultaneous Linear Equations with Two Unknowns

A system of simultaneous equations consists of two or more equations containing multiple variables. To find a unique solution, you must have at least as many independent equations as there are variables. For two variables ($x$ and $y$), we need a system of two equations:

$$a_1 x + b_1 y = c_1$$ $$a_2 x + b_2 y = c_2$$

There are two primary algebraic methods to solve these systems: substitution and elimination.

1. The Substitution Method

This method is best when one of the variables has a coefficient of $1$ or $-1$.

1. Rearrange one equation to express one variable in terms of the other (e.g., $y = \text{expression in } x$).
2. Substitute this expression into the other equation.
3. Solve the resulting single-variable equation.
4. Substitute the numerical value back into the first equation to find the second variable.

Worked Example: Solve the system:

$$\begin{cases} 3x + y = 11 \\ 2x - 3y = 0 \end{cases}$$

1. Isolate $y$ in the first equation: $$y = 11 - 3x$$
2. Substitute $(11 - 3x)$ for $y$ in the second equation: $$2x - 3(11 - 3x) = 0$$
3. Solve for $x$: $$2x - 33 + 9x = 0$$ $$11x - 33 = 0$$ $$11x = 33 \implies x = 3$$
4. Substitute $x = 3$ back into the isolated expression for $y$: $$y = 11 - 3(3) = 11 - 9 = 2$$

The solution is $x = 3$, $y = 2$.

2. The Elimination Method

This method is preferred when coefficients do not easily lend themselves to substitution.

1. Multiply one or both equations by appropriate numbers so that the coefficients of one of the variables are equal in magnitude.
2. Add or subtract the equations to eliminate that variable.
3. Solve the resulting single-variable equation.
4. Substitute the value back to find the other variable.

Worked Example: Solve the system:

$$\begin{cases} 4x + 3y = 25 \\ 3x - 2y = 6 \end{cases}$$

1. To eliminate $y$, multiply the first equation by $2$ and the second equation by $3$: $$8x + 6y = 50$$ $$9x - 6y = 18$$
2. Add the two equations to eliminate $y$: $$(8x + 9x) + (6y - 6y) = 50 + 18$$ $$17x = 68$$
3. Solve for $x$: $$x = \frac{68}{17} = 4$$
4. Substitute $x = 4$ into the first equation: $$4(4) + 3y = 25$$ $$16 + 3y = 25$$ $$3y = 9 \implies y = 3$$

The solution is $x = 4$, $y = 3$.

Practical Aviation Applications

1. Static Equilibrium on Landing Gear

During design or weight-and-balance inspections, technicians calculate the landing gear reaction forces. Consider a light aircraft weighing $12,000\text{ N}$ static on the ramp. The nose gear is $2\text{ m}$ forward of the center of gravity (CG), and the main gear is $0.5\text{ m}$ aft of the CG. Let $F_n$ be the nose gear force and $F_m$ be the main gear force. The two physical laws of equilibrium state:

1. Vertical Force Equilibrium: The sum of upward gear forces must equal the aircraft weight: $$F_n + F_m = 12000$$
2. Moment Equilibrium: The sum of moments about the CG must be zero (clockwise equal counter-clockwise): $$2 F_n - 0.5 F_m = 0$$

We have a system of simultaneous equations:

$$\begin{cases} F_n + F_m = 12000 \\ 2 F_n - 0.5 F_m = 0 \end{cases}$$

Let's solve by substitution:

1. From the second equation, isolate $F_m$: $$0.5 F_m = 2 F_n \implies F_m = 4 F_n$$
2. Substitute $F_m = 4 F_n$ into the first equation: $$F_n + 4 F_n = 12000$$ $$5 F_n = 12000 \implies F_n = 2400\text{ N}$$
3. Solve for $F_m$: $$F_m = 4(2400) = 9600\text{ N}$$

Thus, the nose gear carries $2,400\text{ N}$ of load, and the main gear carries $9,600\text{ N}$.

2. Electrical Circuit Analysis (Kirchhoff's Laws)

Simultaneous equations are essential for analyzing multi-loop electrical circuits in aircraft avionics. Using Kirchhoff's laws, we can find unknown branch currents. Consider a two-loop DC circuit that yields the following equations for currents $I_1$ and $I_2$:

$$\begin{cases} 7I_1 - 2I_2 = 12 \\ -2I_1 + 5I_2 = 6 \end{cases}$$

Let's solve using elimination:

1. Multiply the first equation by $2$ and the second equation by $7$ to align the $I_1$ coefficients: $$14I_1 - 4I_2 = 24$$ $$-14I_1 + 35I_2 = 42$$
2. Add the two equations to eliminate $I_1$: $$31I_2 = 66$$ $$I_2 = \frac{66}{31} \approx 2.13\text{ A}$$
3. Substitute $I_2$ back to find $I_1$: $$7I_1 - 2(2.13) = 12$$ $$7I_1 - 4.26 = 12 \implies 7I_1 = 16.26 \implies I_1 \approx 2.32\text{ A}$$

These electrical calculations form the basis for circuit breaker and wire sizing analyses in aircraft modifications.