Chapter 3 Pair of Linear Equations in Two Variables (Concepts)

Graphs and Solutions of Linear Equations in Two Variables

In earlier classes, we solved equations like $2x + 5 = 0$, which have only one variable and one unique solution. Now, we explore linear equations with two variables. These are incredibly useful for representing relationships between two changing quantities, like the relationship between the number of items bought and the total cost.

What is a Linear Equation in Two Variables?

A linear equation in two variables is any equation that can be written in the standard form:

Here's what each part means:

• $x$ and $y$ are the two variables.
• $a, b,$ and $c$ are real numbers that act as coefficients and a constant.
• The key condition is that $a$ and $b$ cannot both be zero ($a^2 + b^2 \neq 0$). If both were zero, the variables would disappear!

The term "linear" is used because the highest power of each variable is 1, and when you graph the equation, it always forms a perfect straight line.

Examples:

• $3x + 2y - 10 = 0$ (Here $a=3, b=2, c=-10$)
• $x - 4y = 5$ (Can be written as $1x - 4y - 5 = 0$)
• $y = 7$ (Can be written as $0x + 1y - 7 = 0$. Since $b \neq 0$, this is valid).

Solutions of a Linear Equation

A solution to a linear equation in two variables is an ordered pair of numbers $(x, y)$ that makes the equation a true statement. Think of it as a pair of values that perfectly balances the equation.

Unlike equations with one variable, a single linear equation in two variables has infinitely many solutions. For any value you choose for $x$, you can find a corresponding value for $y$ that works, and vice versa.

For the equation $x + y = 5$:

• If $x=1$, then $1+y=5$, so $y=4$. The solution is $(1, 4)$.
• If $x=10$, then $10+y=5$, so $y=-5$. The solution is $(10, -5)$.
• If $y=0$, then $x+0=5$, so $x=5$. The solution is $(5, 0)$.

You can continue finding pairs like this forever. Each pair represents a single point on the graph.

Graph of a Linear Equation

The graph of a linear equation in two variables is a straight line. This line is a visual collection of all the infinite solutions to the equation.

• Every point $(x, y)$ that lies on the line is a solution.
• Every solution $(x, y)$ corresponds to a point on the line.

Steps to Draw the Graph

1. Isolate a variable: It's often easiest to rearrange the equation to solve for $y$ (e.g., write $ax+by+c=0$ as $y = -\frac{a}{b}x - \frac{c}{b}$).
2. Find at least two solutions: Choose simple values for $x$ and calculate the corresponding $y$ values. A great strategy is to find the intercepts:
   • Set $x=0$ to find the y-intercept (where the line crosses the y-axis).
   • Set $y=0$ to find the x-intercept (where the line crosses the x-axis).
   Finding a third point is a good way to check for mistakes.
3. Create a solution table: Organize the $(x, y)$ pairs in a table.
4. Plot the points and draw the line: Plot the points on a graph and connect them with a straight line. Add arrows to both ends to show it continues infinitely.

Worked Examples

Answer:

Solution

Step 1: Isolate y.

$y = 5 - x$

Step 2: Find solutions.

• If $x = 0$, then $y = 5 - 0 = 5$. Solution: $(0, 5)$ (y-intercept).
• If $y = 0$, then $0 = 5 - x \implies x = 5$. Solution: $(5, 0)$ (x-intercept).
• If $x = 2$, then $y = 5 - 2 = 3$. Solution: $(2, 3)$ (check point).

Step 3: Create a solution table.

x	y
0	5
5	0
2	3

Step 4: Plot points and draw the line.

Plot the points $(0, 5)$, $(5, 0)$, and $(2, 3)$ on a graph and connect them with a straight line.

Answer: The graph of $x+y=5$ is a straight line passing through points like $(0,5)$ and $(5,0)$.

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Solution

Step 1: Isolate y.

$2x - 4 = y$, or more commonly, $y = 2x - 4$.

Step 2: Find solutions.

• If $x = 0$, then $y = 2(0) - 4 = -4$. Solution: $(0, -4)$ (y-intercept).
• If $y = 0$, then $0 = 2x - 4 \implies 2x = 4 \implies x = 2$. Solution: $(2, 0)$ (x-intercept).
• If $x = 3$, then $y = 2(3) - 4 = 6 - 4 = 2$. Solution: $(3, 2)$ (check point).

Step 3: Create a solution table.

x	y
0	-4
2	0
3	2

Step 4: Plot points and draw the line.

Plot the points $(0, -4)$, $(2, 0)$, and $(3, 2)$ on a graph and connect them.

Answer: The graph of $2x-y=4$ is a straight line passing through points like $(0,-4)$ and $(2,0)$.

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Part (i): Graph of $x=3$

The equation $x=3$ can be written in standard form as $1x + 0y - 3 = 0$. This means that for any solution, the value of $x$ must be 3, while the value of $y$ can be anything.

Solutions: Some solutions are $(3, 0), (3, 2), (3, -4)$, etc.

Graph: When we plot these points, we see they all lie on a vertical line that passes through $x=3$ on the x-axis.

Part (ii): Graph of $y=-2$

The equation $y=-2$ can be written as $0x + 1y + 2 = 0$. This means that for any solution, the value of $y$ must be -2, while the value of $x$ can be anything.

Solutions: Some solutions are $(0, -2), (4, -2), (-1, -2)$, etc.

Graph: When we plot these points, we see they all lie on a horizontal line that passes through $y=-2$ on the y-axis.

Answer: The graph of $x=3$ is a vertical line parallel to the y-axis. The graph of $y=-2$ is a horizontal line parallel to the x-axis.

Geometrical Representation of a Pair of Linear Equations

While a single linear equation in two variables represents one straight line with infinite solutions, a pair of linear equations considers two such equations simultaneously. The goal is to find a solution that satisfies both equations at the same time.

The general form for a pair of linear equations in variables $x$ and $y$ is:

where $a_1, b_1, c_1, a_2, b_2, c_2$ are real numbers, and for each equation, the coefficients of $x$ and $y$ are not both zero ($a_1^2 + b_1^2 \neq 0$ and $a_2^2 + b_2^2 \neq 0$).

A solution to this pair of equations is an ordered pair $(x, y)$ that makes both equations true simultaneously. It's the "master key" that unlocks both locks.

Geometrical Interpretation: Two Lines on a Plane

Geometrically, a pair of linear equations represents a pair of straight lines drawn on the same coordinate plane. The solution to the pair of equations is found where the two lines meet, as this intersection point is the only point that lies on both lines.

When you draw two lines on a plane, there are only three possible outcomes for how they can relate to each other. Each outcome corresponds to a different type of solution for the pair of equations.

Case 1: Intersecting Lines (Exactly One Solution)

The two lines cross each other at a single, unique point. This point of intersection is the one and only point that satisfies both equations.

• Number of Solutions: Exactly one.
• System Type: A system with at least one solution is called a consistent system.

Case 2: Parallel Lines (No Solution)

The two lines run in the same direction, like railway tracks, and never meet, no matter how far they are extended. Since they never intersect, there is no point that is common to both lines.

• Number of Solutions: None.
• System Type: A system with no solution is called an inconsistent system.

Case 3: Coincident Lines (Infinitely Many Solutions)

The two equations actually represent the exact same line. One line lies perfectly on top of the other. This usually happens when one equation is just a multiple of the other (e.g., $x+y=2$ and $2x+2y=4$).

• Number of Solutions: Infinitely many. Every point on the line is a solution.
• System Type: This system is also consistent because solutions exist. More specifically, it is called a dependent system because the two equations are not independent of each other.

Summary of Possibilities

Consistent and Inconsistent Systems of Linear Equations

A pair of linear equations is often called a system of linear equations. The primary goal when working with a system is to determine if there's a common solution that satisfies both equations. This leads to the classification of systems as either consistent or inconsistent.

Consistent System of Linear Equations

A system of linear equations is called consistent if it has at least one solution. This means there is at least one ordered pair $(x, y)$ that makes both equations true.

Geometrically, a consistent system is represented by lines that meet or overlap. There are two possibilities:

1. Intersecting Lines (Unique Solution)

The lines cross at exactly one point. This point is the single, unique solution to the system.

2. Coincident Lines (Infinitely Many Solutions)

The two equations represent the exact same line. Since every point on the line is a solution to both, there are infinite solutions. This type of consistent system is also called dependent.

Inconsistent System of Linear Equations

A system of linear equations is called inconsistent if it has no solution. This means there is no ordered pair $(x, y)$ that can satisfy both equations at the same time.

Geometrically, an inconsistent system is represented by two distinct parallel lines. Since parallel lines never intersect, there is no common point, and therefore no solution.

The Algebraic Test: Conditions for Consistency

We can determine if a system is consistent or inconsistent without drawing the graphs. We do this by comparing the ratios of the coefficients of the two equations.

Consider the system:

The nature of the system depends on the relationship between the ratios $\frac{a_1}{a_2}$, $\frac{b_1}{b_2}$, and $\frac{c_1}{c_2}$.

Worked Examples

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Graphical Method of Solving a Pair of Linear Equations

The graphical method is a visual way to solve a system of linear equations. Since each equation represents a straight line, finding the "solution" to the system is the same as finding the coordinates of the point(s) where these two lines meet. This point is the only one that lies on both lines, and therefore, its coordinates are the only ones that satisfy both equations.

Steps for the Graphical Method

To find the solution to a pair of linear equations graphically, follow these systematic steps:

1. Prepare Each Equation: For each equation, find at least two solution pairs $(x, y)$. The easiest way is often to find the x- and y-intercepts:
   • Find the y-intercept by setting $x=0$ and solving for $y$. This gives a point $(0, y)$.
   • Find the x-intercept by setting $y=0$ and solving for $x$. This gives a point $(x, 0)$.
   Finding a third point is a good way to verify that your line is correct.
2. Create Solution Tables: Organize the solution pairs for each equation into separate tables. This keeps your work neat and easy to read.
3. Plot the Lines: On a single sheet of graph paper, draw a Cartesian coordinate system. Plot the points from the first table and connect them with a straight line. Label this line with its equation. Do the same for the second equation.
4. Find the Solution from the Graph: Look at the two lines you have drawn and determine their relationship:
   • If the lines intersect: They meet at one point. The coordinates of this point $(x, y)$ are the unique solution to the system.
   • If the lines are parallel: They never meet. There is no solution to the system.
   • If the lines are coincident: The two lines are identical. There are infinitely many solutions, as every point on the line is a solution.
5. Verify the Solution (Highly Recommended): If you found a unique solution point, substitute its x and y values back into both original equations to make sure they are both true. This confirms your graphical result is accurate.

Worked Examples

Answer:

Solution

Step 1: Find solutions for each equation.

For Equation 1: $x + 2y = 3$ (or $y = \frac{3-x}{2}$)

• If $x = 1$, $y = \frac{3-1}{2} = 1$. Point: $(1, 1)$.
• If $x = 3$, $y = \frac{3-3}{2} = 0$. Point: $(3, 0)$.

For Equation 2: $4x + 3y = 2$ (or $y = \frac{2-4x}{3}$)

• If $x = -1$, $y = \frac{2-4(-1)}{3} = \frac{6}{3} = 2$. Point: $(-1, 2)$.
• If $x = 2$, $y = \frac{2-4(2)}{3} = \frac{-6}{3} = -2$. Point: $(2, -2)$.

Step 2: Create Solution Tables.

$x + 2y = 3$

x	y
1	1
3	0

$4x + 3y = 2$

x	y
-1	2
2	-2

Step 3 & 4: Plot the lines and find the intersection.

Plot the points for each line on the same graph. You will see that the lines intersect at a single point.

From the graph, the point of intersection is clearly $(-1, 2)$.

Step 5: Verify the solution.

Check $(-1, 2)$ in both equations:

Eq 1: $(-1) + 2(2) = -1 + 4 = 3$. (True)

Eq 2: $4(-1) + 3(2) = -4 + 6 = 2$. (True)

Answer: The solution is $x = -1, y = 2$.

Answer:

Solution

Step 1 & 2: Find solutions and create tables.

For Equation 1: $x + 2y = 4$ (or $y = \frac{4-x}{2}$)

• If $x = 0$, $y = \frac{4-0}{2} = 2$. Point: $(0, 2)$.
• If $x = 4$, $y = \frac{4-4}{2} = 0$. Point: $(4, 0)$.

For Equation 2: $2x + 4y = 12$ (or $y = \frac{12-2x}{4}$)

• If $x = 0$, $y = \frac{12-0}{4} = 3$. Point: $(0, 3)$.
• If $x = 6$, $y = \frac{12-12}{4} = 0$. Point: $(6, 0)$.

$x + 2y = 4$

x	y
0	2
4	0

$2x + 4y = 12$

x	y
0	3
6	0

Step 3 & 4: Plot the lines and observe.

When we plot both lines on the same graph, we see that they are parallel and never intersect.

Answer: The lines are parallel, so there is no solution. The system is inconsistent.

Answer:

Solution

Step 1 & 2: Find solutions and create tables.

For Equation 1: $2x + 3y = 9$ (or $y = \frac{9-2x}{3}$)

• If $x = 0$, $y = \frac{9-0}{3} = 3$. Point: $(0, 3)$.
• If $x = 3$, $y = \frac{9-6}{3} = 1$. Point: $(3, 1)$.

For Equation 2: $4x + 6y = 18$ (or $y = \frac{18-4x}{6}$)

• If $x = 0$, $y = \frac{18-0}{6} = 3$. Point: $(0, 3)$.
• If $x = 3$, $y = \frac{18-12}{6} = 1$. Point: $(3, 1)$.

Notice that both equations yield the exact same solution points.

$2x + 3y = 9$

x	y
0	3
3	1

$4x + 6y = 18$

x	y
0	3
3	1

Step 3 & 4: Plot the lines and observe.

When we plot the points, we find that the points for both equations lie on the very same line. The lines are coincident.

Answer: The lines are coincident, which means there are infinitely many solutions. Any point on the line, such as $(0, 3)$ or $(3, 1)$, is a solution.

Algebraic Methods of Solving a Pair of Linear Equations

While the graphical method provides a visual understanding of solutions, it can be imprecise, especially with non-integer answers. Algebraic methods provide exact solutions by manipulating the equations directly. We will explore the three main algebraic methods.

1. The Substitution Method

The core idea of the Substitution Method is to solve one equation for one variable and then "substitute" that expression into the other equation. This reduces the system to a single equation with a single variable, which is easy to solve.

Steps for the Substitution Method:

1. Isolate a Variable: Choose one equation and solve it for one variable (either $x$ or $y$). It's easiest to pick an equation where a variable has a coefficient of 1 or -1.
2. Substitute: Plug the expression from Step 1 into the other equation. This will create a new equation with only one variable.
3. Solve: Solve this new equation to find the numerical value of that variable.
4. Back-Substitute: Substitute the value you just found back into the expression from Step 1 to find the value of the other variable.
5. State the Solution: Write the solution as an ordered pair $(x, y)$ and verify it in both original equations.

Answer:

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2. The Elimination Method

The Elimination Method works by adding or subtracting the two equations in order to eliminate one of the variables. This is done by first making the coefficients of one variable opposites (like +5 and -5) or identical.

Steps for the Elimination Method:

1. Align Equations: Write the equations so that like terms are aligned in columns.
2. Match Coefficients: Multiply one or both equations by suitable numbers so that the coefficients of one variable are either the same or opposites.
3. Eliminate: Add the equations if the coefficients are opposites. Subtract the equations if the coefficients are the same. This will result in a new equation with only one variable.
4. Solve: Solve this new equation for the remaining variable.
5. Back-Substitute: Substitute the value you just found into either of the original equations to find the value of the other variable.
6. State the Solution: Write the solution as an ordered pair $(x, y)$.

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3. The Cross-Multiplication Method

The Cross-Multiplication Method is a formula-based approach that directly calculates the solution for a system with a unique solution. It is derived from the elimination method.

Steps for the Cross-Multiplication Method:

1. Standard Form: First, write both equations in the standard form $ax + by + c = 0$.
2. Identify Coefficients: Identify the coefficients $a_1, b_1, c_1$ from the first equation and $a_2, b_2, c_2$ from the second.
3. Apply the Formula: Use the following formula to set up the relationships:

   $\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$

   A diagram can help remember the order of multiplication:
4. Solve for x and y: Solve for $x$ by equating the first and last parts of the formula. Solve for $y$ by equating the second and last parts.

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Equations Reducible to a Pair of Linear Equations

Sometimes, we encounter pairs of equations that are not linear at first glance because the variables appear in the denominator, under a radical, or in some other non-linear form. However, many of these systems can be transformed or 'reduced' into a standard pair of linear equations by making a suitable substitution. Once the system is linear, we can solve it using any familiar algebraic method (elimination, substitution, etc.) and then convert the solution back to the original variables.

The General Strategy

1. Identify the Pattern: Look for the non-linear expressions that are common in both equations (e.g., terms like $\frac{1}{x}$, $\frac{1}{y-2}$, or $\frac{1}{x+y}$).
2. Make a Substitution: Introduce new variables (commonly $u$ and $v$) to represent these common non-linear expressions.
3. Form a Linear System: Rewrite the original equations using these new variables. The result should be a standard pair of linear equations in terms of $u$ and $v$.
4. Solve the Linear System: Solve this new system for $u$ and $v$ using the elimination or substitution method.
5. Back-Substitute: Replace $u$ and $v$ with their original expressions in terms of $x$ and $y$. This will give you one or more simple equations.
6. Solve for Original Variables: Solve these final equations to find the values of the original variables, $x$ and $y$.
7. Verify: Check your final solution in the original, non-linear equations to ensure it is correct.

Worked Examples

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