Equations of two unknowns

Linear simultaneous equations, also known as a system of linear equations, are two or more algebraic equations that share the same variables and are solved together. They are called simultaneous because the solution satisfies all equations in the system at the same time.

General Form:

$$\begin{cases} a_1x + b_1y = c_1 \\ a_2x + b_2y = c_2 \end{cases}$$

Example of one linear equation:

$20x + y = 10$

To solve a system of equations like this, we need at least two equations involving the same variables. For example:

$$\begin{cases} 20x + y = 10 \\ 5x - y = 2 \end{cases}$$

This system can be solved using methods such as:

1. Elimination Method
2. Substitution Method
3. Graphical Method

The solution to a linear simultaneous equation is the pair $(x, y)$ that satisfies both equations at once.

Solve the following simultaneous equations by elimination method.

$6x + y = 15$ $3x + y = 9$

Solution

$$\begin{cases} 6x + y = 15 \quad \text{(1)} \\ 3x + y = 9 \quad \text{(2)} \end{cases}$$

Step 1: Subtract equation (2) from equation (1) to eliminate $y$

$(6x + y) - (3x + y) = 15 - 9$

$6x + y - 3x - y = 6$

$3x = 6$

Step 2: Solve for $x$

$x = \frac{6}{3} = 2$

Step 3: Substitute $x = 2$ into equation (2) to find $y$

$3x + y = 9$

$3(2) + y = 9$

$6 + y = 9$

$y = 9 - 6 = 3$

Final Answer:

$x = 2, \quad y = 3$

Solve the following simultaneous equations by elimination method.

$2x + y = 10$ $3x - 2y = 1$

Solution

$$\begin{cases} 2x + y = 10 \quad \text{(1)} \\ 3x - 2y = 1 \quad \text{(2)} \end{cases}$$

Step 1: Eliminate $y$. To do this, multiply both equations to make the coefficients of $y$ equal. Multiply equation (1) by 2:

$2(2x + y) = 2(10) \Rightarrow 4x + 2y = 20 \quad \text{(3)}$

Equation (2) is:

$3x - 2y = 1 \quad \text{(2)}$

Step 2: Add equations (3) and (2) to eliminate $y$:

$(4x + 2y) + (3x - 2y) = 20 + 1$

$4x + 2y + 3x - 2y = 21$

$7x = 21$

Step 3: Solve for $x$

$x = \frac{21}{7} = 3$

Step 4: Substitute $x = 3$ into equation (1) to find $y$

$2x + y = 10$

$2(3) + y = 10$

$6 + y = 10 \Rightarrow y = 10 - 6 = 4$

Final Answer:

$x = 3, \quad y = 4$

Solve the following linear simultaneous equations by elimination method.

$3x - 2y = 8$ $2x + 5y = -1$

Solution

$$\begin{cases} 3x - 2y = 8 \quad \text{(1)} \\ 2x + 5y = -1 \quad \text{(2)} \end{cases}$$

Step 1: Eliminate one variable. Let's eliminate $x$ by making the coefficients of $x$ equal. Multiply equation (1) by 2:

$2(3x - 2y) = 2(8) \Rightarrow 6x - 4y = 16 \quad \text{(3)}$

Multiply equation (2) by 3:

$3(2x + 5y) = 3(-1) \Rightarrow 6x + 15y = -3 \quad \text{(4)}$

Step 2: Subtract equation (3) from equation (4) to eliminate $x$:

$(6x + 15y) - (6x - 4y) = -3 - 16$

$6x + 15y - 6x + 4y = -19$

$19y = -19$

Step 3: Solve for $y$

$y = \frac{-19}{19} = -1$

Step 4: Substitute $y = -1$ into equation (1) to solve for $x$

$3x - 2y = 8$

$3x - 2(-1) = 8$

$3x + 2 = 8 \Rightarrow 3x = 6 \Rightarrow x = \frac{6}{3} = 2$

Final Answer:

$x = 2, \quad y = -1$

Find the solution of the following simultaneous equations by elimination method.

$\frac{2}{x} - \frac{3}{y} = -3$ $\frac{1}{x} + \frac{6}{y} = 1$

Solution

$$\begin{cases} \frac{2}{x} - \frac{3}{y} = -3 \quad \text{(1)} \\ \frac{1}{x} + \frac{6}{y} = 1 \quad \text{(2)} \end{cases}$$

Step 1: Let $u = \frac{1}{x}$ and $v = \frac{1}{y}$, so the equations become:

$$\begin{cases} 2u - 3v = -3 \quad \text{(3)} \\ u + 6v = 1 \quad \text{(4)} \end{cases}$$

Step 2: Eliminate one variable. Multiply equation (4) by 2 so that we can eliminate $u$:

$2(u + 6v) = 2(1) \Rightarrow 2u + 12v = 2 \quad \text{(5)}$

Now subtract equation (3) from equation (5):

$(2u + 12v) - (2u - 3v) = 2 - (-3)$

$2u + 12v - 2u + 3v = 5 \Rightarrow 15v = 5 \Rightarrow v = \frac{5}{15} = \frac{1}{3}$

Step 3: Substitute $v = \frac{1}{3}$ into equation (4) to find $u$:

$u + 6v = 1 \Rightarrow u + 6\left(\frac{1}{3}\right) = 1 \Rightarrow u + 2 = 1 \Rightarrow u = 1 - 2 = -1$

Step 4: Recall $u = \frac{1}{x}$, $v = \frac{1}{y}$, so:

$\frac{1}{x} = -1 \Rightarrow x = -1, \quad \frac{1}{y} = \frac{1}{3} \Rightarrow y = 3$

Final Answer:

$x = -1, \quad y = 3$

Definition: The substitution method involves selecting one equation and expressing one variable in terms of the other. This expression is then substituted into the second equation to eliminate one variable, allowing us to solve for the other.

Steps:

1. Choose one of the two equations.
2. Make one variable the subject of the equation.
3. Substitute this expression into the second equation.
4. Solve the resulting equation (with one variable).
5. Substitute the found value back into one of the original equations to find the other variable.
6. Write the final solution as an ordered pair $(x, y)$.

Example 1

$$\begin{cases} y = 2x + 1 \\ x + y = 10 \end{cases}$$

Substitute into second equation:

$x + (2x + 1) = 10$ $3x + 1 = 10$ $3x = 9$ $x = 3$ $y = 2(3) + 1 = 7$

Final Answer: $x = 3,\ y = 7$

Example 2

$$\begin{cases} x = y - 2 \\ 2x + y = 16 \end{cases}$$

Substitute into second equation:

$2(y - 2) + y = 16$ $2y - 4 + y = 16$ $3y = 20$ $y = \frac{20}{3}$ $x = \frac{20}{3} - 2 = \frac{14}{3}$

Final Answer: $x = \frac{14}{3},\ y = \frac{20}{3}$

Example 3

$$\begin{cases} y = 5x \\ 2x + y = 18 \end{cases}$$

Substitute into second equation:

$2x + 5x = 18$ $7x = 18$ $x = \frac{18}{7}$ $y = 5 \cdot \frac{18}{7} = \frac{90}{7}$

Final Answer: $x = \frac{18}{7},\ y = \frac{90}{7}$

Example 4

$$\begin{cases} y = 4 - x \\ 2x + 3y = 12 \end{cases}$$

Substitute:

$2x + 3(4 - x) = 12$ $2x + 12 - 3x = 12$ $-x = 0$ $x = 0$ $y = 4 - 0 = 4$

Final Answer: $x = 0,\ y = 4$

Example 5

$$\begin{cases} x = 2y + 1 \\ 3x + y = 13 \end{cases}$$

Substitute:

$3(2y + 1) + y = 13$ $6y + 3 + y = 13$ $7y = 10$ $y = \frac{10}{7}$ $x = 2 \cdot \frac{10}{7} + 1 = \frac{20}{7} + \frac{7}{7} = \frac{27}{7}$

Final Answer: $x = \frac{27}{7},\ y = \frac{10}{7}$

Example 6

$$\begin{cases} y = 3x - 4 \\ x + 2y = 17 \end{cases}$$

Substitute:

$x + 2(3x - 4) = 17$ $x + 6x - 8 = 17$ $7x = 25$ $x = \frac{25}{7}$ $y = 3 \cdot \frac{25}{7} - 4 = \frac{75}{7} - \frac{28}{7} = \frac{47}{7}$

Final Answer: $x = \frac{25}{7},\ y = \frac{47}{7}$