Linear Equations in Two Variables

A linear equation in two variables is a mathematical statement that equates a linear polynomial to a constant. This type of equation can be expressed as:

ax + by = c

where a, b, and c are real numbers, and x and y are variables. For example:

2x + 3y = 6
5x – y = 4
-3x + 2y = -12

Understanding linear equations

Linear equations in two variables represent straight lines when graphed on the Cartesian plane. The general form of the equation can be rearranged into slope-intercept form:

y = mx + b

where m is the slope of the line and b is the y-intercept. Here's how you can convert the normal form:

ax + by = c => by = -ax + c => y = (-a/b)x + (c/b)

Example 1: Transforming an equation

Let's convert the equation 2x + 3y = 6 to slope-intercept form.

2x + 3y = 6
3y = -2x + 6
y = (-2/3)x + 2

The slope is -2/3 and the y-intercept is 2.

Graphical representation

To graph a linear equation in two variables, plot the points that satisfy the equation and draw a line through them. Let's consider the equation y = 2x + 1.

In this example, we plotted the points (0, 1), (1, 3) and (-1, -1) and drew a line through them. This line is the graphical representation of the equation y = 2x + 1.

Solution of linear equations

The solution of a linear equation in two variables is the set of all points (x, y) that satisfy the equation. These points lie on the line represented by the equation.

Example 2: Finding a solution

Find two solutions of the equation x - y = 2.

1. Choose x = 0: 
   0 - y = 2 
   => y = -2 
   Solution: (0, -2)

2. Choose x = 4: 
   4 – y = 2 
   => y = 2 
   Solution: (4, 2)

Therefore, the two solutions are (0, -2) and (4, 2).

Obstructions

Intercepts are the points where the line crosses the axes. To find the x-intercept, set y = 0 and solve for x. To find the y-intercept, set x = 0 and solve for y.

Example 3: Finding the intercept

Find the intercepts of the equation 3x + 4y = 12.

1. X-intercept: set y = 0
   3x + 4(0) = 12 
   => 3x = 12 
   => x = 4 
   X-intercept: (4, 0)

2. Y-intercept: set x = 0
   3(0) + 4y = 12 
   => 4y = 12 
   => y = 3 
   Y-intercept: (0, 3)

Hence the x-intercept is (4, 0) and the y-intercept is (0, 3).

Parallel and perpendicular lines

Two lines are parallel if their slopes are the same. They never cross each other. Two lines are perpendicular if the product of their slopes is -1.

Example 4: Checking parallel lines

Determine whether the lines represented by 2x + 3y = 6 and 4x + 6y = 9 are parallel.

Equation 1 Slope: 2x + 3y = 6 => y = (-2/3)x + 2 => slope = -2/3
Equation 2 Slope: 4x + 6y = 9 => y = (-4/6)x + 3/2 => slope = -2/3

Since the slopes are equal, the lines are parallel.

Example 5: Checking perpendicular lines

Determine whether the lines represented by y = (1/2)x + 3 and 2y = -x + 5 are perpendicular.

Equation 1 Slope: y = (1/2)x + 3 => slope = 1/2
Equation 2 slope: 2y = -x + 5 => y = (-1/2)x + 5/2 => slope = -1/2

The product of the slopes is (1/2) * (-2) = -1, so the lines are perpendicular.

Systems of linear equations

A system of linear equations is a set of two or more equations that have the same variables. Solving a system involves finding the intersection point(s) on the graph.

Example 6: Solving a system by substitution

Solve the system:

x + 2y = 3
3x – y = -2

Substitute from the first equation:

x = 3 – 2y

Substitute into the second equation:

3(3 - 2y) - y = -2
9 - 6y - y = -2
9 – 7y = -2
7y = 11
y = 11/7

Re-substitute to find x:

x = 3 - 2(11/7)
x = 3 – 22/7
x = (21/7) - (22/7)
x = -1/7

Its solution is (-1/7, 11/7).

Example 7: Solving a system by elimination

Solve the system:

2x + 3y = 7
4x – 6y = 8

Multiply the first equation by 2:

4x + 6y = 14

Now, subtract the second equation:

(4x + 6y) – (4x – 6y) = 14 – 8
12y = 6
y = 1/2

Re-substitute to find x:

2x + 3(1/2) = 7
2x + 3/2 = 7
2x = 7 - 3/2
2x = 14/2 - 3/2
2x = 11/2
x = 11/4

The solution is (11/4, 1/2).

Applications of linear equations

Linear equations are widely used to solve real-world problems. Examples include calculating distance, cost, profit, and many other situations.

Example 8: Cost calculation

John buys apples and bananas. If apples cost $2 per banana and bananas cost $3 per banana, and he spends a total of $18, express this situation as a linear equation.

Let x = number of apples, y = number of bananas
Equation: 2x + 3y = 18

Example 9: Break-even analysis

An entrepreneur sells handmade cards for $5. The materials cost $50 and each card costs $1 to make. How many cards must be sold to break even?

Let x = number of cards
Revenue = 5x
Cost = 50 + 1x
Equation for Break-Even:
5x = 50 + x
4x = 50
x = 12.5

Therefore, 13 cards need to be sold to break even.

Understanding linear equations in two variables involves recognizing their forms, graphing them, finding solutions, and applying them to practical problems. They form the basis for understanding complex algebra and are helpful in a variety of fields. By mastering these concepts, students can develop important problem-solving skills applicable to many real-world scenarios.

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