Pair of Linear Equations in Two Variables: Systems and Solutions

Introduction to Systems of Linear Equations

Solving Problems with Multiple Conditions

In mathematics, particularly in the context of real-world applications, problems often involve more than one unknown quantity and more than one condition or relationship linking these quantities. While a single linear equation in two variables (like $x+y=5$) can describe one such relationship and has infinitely many solutions, these solutions are restricted only by that single condition.

Consider a problem like: "The sum of two numbers is $10$, and their difference is $2$." We have two unknown numbers and two conditions. Let the numbers be $x$ and $y$. The first condition translates to $x+y=10$, and the second translates to $x-y=2$. We are looking for specific values of $x$ and $y$ that satisfy *both* of these conditions simultaneously.

When we have a set of two or more equations involving the same variables, we call this a system of equations. If all the equations in the system are linear equations, it is called a system of linear equations or, more specifically when there are two equations, a pair of linear equations.

Definition of a Pair of Linear Equations in Two Variables

A pair of linear equations in two variables is a system consisting of exactly two linear equations, both involving the same two variables (which are usually denoted as $x$ and $y$). Each equation in the pair can be written in the standard form of a linear equation in two variables, $ax + by + c = 0$.

A general representation for a pair of linear equations in two variables $x$ and $y$ is:

Here, $a_1, b_1, c_1, a_2, b_2,$ and $c_2$ are real coefficients (constants). For each equation to be a valid linear equation in two variables, the coefficients of the variables in that equation cannot both be zero. That is, $a_1$ and $b_1$ are not both zero (i.e., $a_1^2 + b_1^2 \neq 0$), and $a_2$ and $b_2$ are not both zero (i.e., $a_2^2 + b_2^2 \neq 0$).

Examples of pairs of linear equations in two variables:

• Equation 1: $2x + 3y = 7$ (or $2x + 3y - 7 = 0$)
• Equation 2: $x - y = 1$ (or $x - y - 1 = 0$)

Here, $a_1=2, b_1=3, c_1=-7$ and $a_2=1, b_2=-1, c_2=-1$.

• Equation 1: $p + 2q = 10$ (or $p + 2q - 10 = 0$)
• Equation 2: $3p - q = 5$ (or $3p - q - 5 = 0$)

The variables here are $p$ and $q$. $a_1=1, b_1=2, c_1=-10$ and $a_2=3, b_2=-1, c_2=-5$.

• Equation 1: $x + y = 5$ (or $x + y - 5 = 0$)
• Equation 2: $2x + 2y = 10$ (or $2x + 2y - 10 = 0$)

Here, $a_1=1, b_1=1, c_1=-5$ and $a_2=2, b_2=2, c_2=-10$.

• Equation 1: $x + y = 5$ (or $x + y - 5 = 0$)
• Equation 2: $x + y = 6$ (or $x + y - 6 = 0$)

Here, $a_1=1, b_1=1, c_1=-5$ and $a_2=1, b_2=1, c_2=-6$.

Solution of a Pair of Linear Equations

A solution of a pair of linear equations in two variables is an ordered pair of values $(x, y)$ that satisfies both equations in the system simultaneously. When you substitute the values of $x$ and $y$ from the ordered pair into the first equation, the LHS must equal the RHS, AND when you substitute the same values into the second equation, its LHS must also equal its RHS.

Consider the pair of equations:

Let's test if the ordered pair $(3, 2)$ is a solution:

• Substitute $x=3$ and $y=2$ into Equation (1): LHS $= 3 + 2 = 5$. RHS $= 5$. LHS = RHS. Equation (1) is satisfied.
• Substitute $x=3$ and $y=2$ into Equation (2): LHS $= 3 - 2 = 1$. RHS $= 1$. LHS = RHS. Equation (2) is satisfied.

Since $(3, 2)$ satisfies both equations, it is a solution to this pair of linear equations. As we will see, this particular system has exactly one solution.

Now consider the pair:

Let's try to find a solution. Suppose $(x_0, y_0)$ is a solution. Then it must satisfy both equations: $x_0 + y_0 = 5$ AND $x_0 + y_0 = 6$. This means $5$ must be equal to $6$, which is impossible ($5 \neq 6$). There is no pair of values $(x, y)$ that can make $x+y$ equal to both $5$ and $6$ at the same time. This pair of equations has no solution.

Finally, consider the pair:

If we examine Equation (6), we can divide both sides by $2$: $\frac{2x}{2} + \frac{2y}{2} = \frac{10}{2}$, which simplifies to $x + y = 5$. This is exactly the same as Equation (5). Any ordered pair $(x, y)$ that satisfies Equation (5) will automatically satisfy Equation (6), and vice-versa. Since a single linear equation in two variables has infinitely many solutions, this pair of equations also has infinitely many solutions (all the solutions that satisfy $x+y=5$).

These examples demonstrate the three possibilities for the number of solutions a pair of linear equations in two variables can have: a unique solution, no solution, or infinitely many solutions. This has a direct and clear geometric interpretation, which we will explore next.

Geometrical Representation of a Pair of Linear Equations

Lines in the Coordinate Plane

As we've established, a single linear equation in two variables, written as $ax + by + c = 0$, corresponds to a straight line when its solutions $(x, y)$ are plotted on a Cartesian coordinate plane. Every point $(x, y)$ that lies on this line is a solution to the equation, and conversely, every solution to the equation corresponds to a point on this line.

When we consider a pair of linear equations in two variables, such as:

each equation represents a distinct straight line in the same coordinate plane (unless they are the same equation). The solution(s) to this system of equations are the ordered pair(s) $(x, y)$ that satisfy both equations simultaneously. Geometrically, the points that satisfy both equations are the points that lie on both lines. These are precisely the points of intersection of the two lines.

Possible Outcomes of Graphing Two Lines

When two distinct straight lines are drawn in the same plane, there are only three possible ways they can be positioned relative to each other. These three geometric possibilities correspond exactly to the three possibilities for the number of solutions a pair of linear equations can have:

1. Case 1: Intersecting Lines

   If the two lines are not parallel, they will intersect at exactly one point. This point is unique and lies on both lines.

   

   The image depicts two lines crossing each other at a single point. The coordinates of this intersection point are the only values of $x$ and $y$ that satisfy both equations.

   Geometrical Interpretation: The lines intersect at a single point.

   Algebraic Interpretation: The pair of linear equations has a unique solution (exactly one solution). This unique solution is the ordered pair corresponding to the coordinates of the intersection point.

   This case occurs when the slopes of the two lines are different.

2. Case 2: Parallel and Distinct Lines

   If the two lines are parallel and do not overlap, they will never intersect, no matter how far they are extended. They maintain a constant distance from each other.

   

   The image shows two lines that are parallel and do not cross. There is no point that lies on both lines simultaneously.

   Geometrical Interpretation: The lines are parallel and distinct.

   Algebraic Interpretation: The pair of linear equations has no solution. There is no ordered pair $(x, y)$ that can satisfy both equations at the same time.

   This case occurs when the slopes of the two lines are the same, but their y-intercepts (or equivalent, the value of $c_1/b_1$ and $c_2/b_2$ assuming $b_1, b_2 \neq 0$) are different.

3. Case 3: Coincident Lines

   If the two lines are not only parallel but also pass through the exact same points, they are called coincident lines. This means the graphs of the two equations are identical; they lie exactly on top of each other.

   

   The image shows two lines that appear as a single line because they overlap perfectly. Every point on one line is also a point on the other line.

   Geometrical Interpretation: The lines are coincident (they are the same line).

   Algebraic Interpretation: The pair of linear equations has infinitely many solutions. Every ordered pair $(x, y)$ that is a solution to one equation is also a solution to the other equation. One equation is simply a multiple of the other (e.g., $a_2 = k \cdot a_1, b_2 = k \cdot b_1, c_2 = k \cdot c_1$ for some non-zero constant $k$).

   This case occurs when the slopes and the y-intercepts (or equivalent, the ratios of corresponding coefficients $a_1/a_2, b_1/b_2, c_1/c_2$) are all the same.

The geometrical representation provides a clear and intuitive understanding of the nature of the solutions to a pair of linear equations in two variables. The number of solutions is directly visualized by the number of points where the graphs of the two equations intersect.

Consistent and Inconsistent Pairs of Linear Equations

Classifying Systems by Number of Solutions

As we saw in the previous section, a pair of linear equations in two variables can have one unique solution, no solution, or infinitely many solutions. This difference in the number of solutions leads to a classification of these pairs of equations into two main categories: consistent and inconsistent.

Consistent Pairs of Equations

A pair of linear equations in two variables is defined as consistent if it has at least one solution. This means that there exists at least one ordered pair $(x, y)$ that satisfies both equations simultaneously.

Based on our geometrical interpretation, consistent systems correspond to lines that intersect. This can happen in two ways:

1. Unique Solution:

   A consistent pair has a unique solution if the graphs of the two equations are two distinct lines that intersect at exactly one point.

   Example: Consider the pair of equations $x + y = 5$ and $x - y = 1$. Graphically, these lines intersect at the single point $(3, 2)$. Substituting $(3, 2)$ into both equations: $3+2=5$ (True) and $3-2=1$ (True). This system has a unique solution $(3, 2)$.

   These lines are distinct and intersect.

2. Infinitely Many Solutions:

   A consistent pair has infinitely many solutions if the graphs of the two equations are coincident lines (they are the same line). In this case, every point on the line is a solution to both equations simultaneously, and since a line has infinitely many points, there are infinitely many common solutions.

   Example: Consider the pair of equations $x + y = 5$ and $2x + 2y = 10$. The second equation is obtained by multiplying the first equation by $2$. If you graph both equations, you will draw the exact same line. Any solution to $x+y=5$ (like $(1, 4), (0, 5), (-2, 7)$) will also satisfy $2x+2y=10$. This system has infinitely many solutions.

   These lines are coincident.

   A consistent pair of linear equations signifies that the conditions or constraints described by the equations are compatible and can be satisfied simultaneously by one or more pairs of values for the variables.

   The consistent pair with infinitely many solutions is sometimes specifically referred to as a dependent pair of linear equations, because the two equations are not independent; one equation is a scalar multiple of the other.

   ---

   Inconsistent Pairs of Equations

   A pair of linear equations in two variables is defined as inconsistent if it has no solution. This means there is no ordered pair $(x, y)$ that can satisfy both equations simultaneously.

   Geometrically, inconsistent systems correspond to lines that are parallel and distinct. Because parallel lines never intersect, there is no point that lies on both lines, and thus no common solution.

   Example: Consider the pair of equations $x + y = 5$ and $x + y = 6$. If you try to find a solution $(x, y)$, you would need $x+y$ to be equal to both $5$ and $6$ at the same time, which is impossible. If you graph these equations, you will get two parallel lines (both with a slope of $-1$) but with different y-intercepts ($5$ and $6$). They will never meet.

   An inconsistent pair of linear equations indicates that the conditions or constraints described by the equations are contradictory and cannot be satisfied simultaneously by any pair of values for the variables.

   ---

   Conditions for Consistency and Inconsistency (Based on Coefficients)

   We can determine the nature of a pair of linear equations (whether it is consistent or inconsistent, and how many solutions) without graphing the lines, by simply comparing the ratios of their corresponding coefficients. For a given pair of linear equations in standard form:

   $a_1 x + b_1 y + c_1 = 0$

   ... (1)

   $a_2 x + b_2 y + c_2 = 0$

   ... (2)

   where $a_1, b_1, c_1, a_2, b_2, c_2$ are coefficients and $a_1^2+b_1^2 \neq 0$, $a_2^2+b_2^2 \neq 0$.

   Compare the ratios $\frac{a_1}{a_2}$, $\frac{b_1}{b_2}$, and $\frac{c_1}{c_2}$ (provided the denominators are non-zero. If a denominator is zero, you need to handle the ratios carefully or rearrange the equations).

   1. Condition for Unique Solution (Consistent - Independent):

      If the ratio of the coefficient of $x$ in the first equation to the coefficient of $x$ in the second equation is NOT equal to the ratio of the coefficient of $y$ in the first equation to the coefficient of $y$ in the second equation. The lines intersect at exactly one point.

      $ \frac{a_1}{a_2} \neq \frac{b_1}{b_2} $

      [Condition for Unique Solution]

      Example: $2x + 3y - 7 = 0$ and $x - y - 1 = 0$. Here $a_1=2, b_1=3, c_1=-7$ and $a_2=1, b_2=-1, c_2=-1$. The ratio $\frac{a_1}{a_2} = \frac{2}{1} = 2$. The ratio $\frac{b_1}{b_2} = \frac{3}{-1} = -3$. Since $2 \neq -3$, the condition $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ is satisfied. The lines intersect at one point, and the system has a unique solution.

   2. Condition for No Solution (Inconsistent):

      If the ratio of the coefficient of $x$ is equal to the ratio of the coefficient of $y$, but this ratio is NOT equal to the ratio of the constant terms. The lines are parallel and distinct.

      $ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} $

      [Condition for No Solution]

      Example: $x + y - 5 = 0$ and $x + y - 6 = 0$. Here $a_1=1, b_1=1, c_1=-5$ and $a_2=1, b_2=1, c_2=-6$. The ratio $\frac{a_1}{a_2} = \frac{1}{1} = 1$. The ratio $\frac{b_1}{b_2} = \frac{1}{1} = 1$. The ratio $\frac{c_1}{c_2} = \frac{-5}{-6} = \frac{5}{6}$. We see that $\frac{a_1}{a_2} = \frac{b_1}{b_2} = 1$, but $1 \neq \frac{c_1}{c_2}$. The condition $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ is satisfied. The lines are parallel and distinct, and the system has no solution.

   3. Condition for Infinitely Many Solutions (Consistent - Dependent):

      If the ratio of the coefficient of $x$ is equal to the ratio of the coefficient of $y$, and this ratio is also equal to the ratio of the constant terms. The lines are coincident.

      $ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} $

      [Condition for Infinitely Many Solutions]

      Example: $x + y - 5 = 0$ and $2x + 2y - 10 = 0$. Here $a_1=1, b_1=1, c_1=-5$ and $a_2=2, b_2=2, c_2=-10$. The ratio $\frac{a_1}{a_2} = \frac{1}{2}$. The ratio $\frac{b_1}{b_2} = \frac{1}{2}$. The ratio $\frac{c_1}{c_2} = \frac{-5}{-10} = \frac{1}{2}$. We see that $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \frac{1}{2}$. The condition is satisfied. The lines are coincident, and the system has infinitely many solutions.

      Understanding the classification of linear equation pairs into consistent (having at least one solution) and inconsistent (having no solution) categories, and knowing the conditions on the coefficients that determine these classifications, is crucial for predicting the nature and number of solutions before attempting to solve a system.

Graphical Method of Solving a Pair of Linear Equations

Finding the Intersection Point

The graphical method for solving a pair of linear equations in two variables is based on the understanding that the graph of each linear equation is a straight line. The solutions to the system of equations are the points $(x, y)$ that satisfy both equations simultaneously. Geometrically, these are the points that lie on both lines, which means they are the points where the lines intersect.

By plotting the graphs of the two linear equations on the same coordinate plane, we can visually determine the number of solutions (based on whether the lines intersect, are parallel, or are coincident) and, if there is a unique solution, find its approximate coordinates by reading the point of intersection from the graph.

Steps for the Graphical Method

To solve a pair of linear equations graphically:

1. Graph the First Equation:

   • Find at least two distinct solutions (ordered pairs $(x, y)$) for the first linear equation. Choosing the x-intercept (by setting $y=0$ and solving for $x$) and the y-intercept (by setting $x=0$ and solving for $y$) is often convenient, provided these points are distinct and easy to plot.
   • Plot these solutions as points on a coordinate plane.
   • Draw a straight line accurately through these plotted points. Extend the line and label it with the equation.

2. Graph the Second Equation:

   • Find at least two distinct solutions (ordered pairs $(x, y)$) for the second linear equation. Again, the intercepts are good choices if suitable.
   • Plot these solutions as points on the *same* coordinate plane used for the first equation.
   • Draw a straight line accurately through these new plotted points. Extend the line and label it with the equation.

3. Observe the Intersection:

   Examine the relationship between the two lines you have drawn:
   • If the two lines intersect at a single point, the pair of equations has a unique solution. This is a consistent and independent system.
   • If the two lines appear parallel and distinct (they do not intersect), the pair of equations has no solution. This is an inconsistent system.
   • If the two lines coincide (they are the same line, one lying exactly on top of the other), the pair of equations has infinitely many solutions. This is a consistent and dependent system.

4. Determine the Solution (if unique):

   If the lines intersect at a single point, carefully read the x-coordinate and the y-coordinate of this intersection point from the graph. This ordered pair $(x, y)$ is the unique solution to the system of equations.

5. Check the Solution:

   Substitute the coordinates of the intersection point into both of the original equations. If both equations are satisfied (LHS = RHS), then the solution is correct. This check is important because reading coordinates from a graph can sometimes be imprecise, especially if the intersection point is not at integer coordinates.

Answer:

Step 1: Graph the first equation: $x + y = 5$.

Let's find the intercepts:

• Set $x=0$: $0 + y = 5 \implies y = 5$. Point: $(0, 5)$.
• Set $y=0$: $x + 0 = 5 \implies x = 5$. Point: $(5, 0)$.
• Let's find one more point for accuracy. Set $x=2$: $2 + y = 5 \implies y = 5 - 2 = 3$. Point: $(2, 3)$.

Table of solutions for $x+y=5$:

$x$	$y$	$(x, y)$
0	5	$(0, 5)$
5	0	$(5, 0)$
2	3	$(2, 3)$

Plot these points and draw the line for $x+y=5$.

Step 2: Graph the second equation: $x - y = 1$.

Let's find the intercepts:

• Set $x=0$: $0 - y = 1 \implies -y = 1 \implies y = -1$. Point: $(0, -1)$.
• Set $y=0$: $x - 0 = 1 \implies x = 1$. Point: $(1, 0)$.
• Let's find one more point. Set $x=3$: $3 - y = 1 \implies -y = 1 - 3 \implies -y = -2 \implies y = 2$. Point: $(3, 2)$.

Table of solutions for $x-y=1$:

$x$	$y$	$(x, y)$
0	-1	$(0, -1)$
1	0	$(1, 0)$
3	2	$(3, 2)$

Plot these points on the same graph and draw the line for $x-y=1$.

Step 3 & 4: Observe the intersection and determine the solution.

Looking at the graph where both lines are plotted:

The two lines intersect at a single point. By observing the coordinates of this point, we see it is $(3, 2)$.

The unique solution is $(3, 2)$.

Step 5: Check the solution.

Substitute $x=3$ and $y=2$ into the original equations:

• Equation 1: $x + y = 5 \implies 3 + 2 = 5$. $5 = 5$. (True).
• Equation 2: $x - y = 1 \implies 3 - 2 = 1$. $1 = 1$. (True).

Since the ordered pair $(3, 2)$ satisfies both equations, it is the correct unique solution.

Answer: The unique solution to the pair of linear equations is $\textbf{(3, 2)}$.

Answer:

Step 1: Graph the first equation: $2x + 4y = 10$. (Note: This equation can be simplified by dividing all terms by 2 to get $x + 2y = 5$. Graphing $x+2y=5$ is equivalent and easier). Let's use $x+2y=5$.

Find solutions for $x+2y=5$:

• Set $x=0$: $0 + 2y = 5 \implies 2y = 5 \implies y = 2.5$. Point: $(0, 2.5)$.
• Set $y=0$: $x + 2(0) = 5 \implies x = 5$. Point: $(5, 0)$.
• Set $x=1$: $1 + 2y = 5 \implies 2y = 4 \implies y = 2$. Point: $(1, 2)$.

Table of solutions for $2x+4y=10$ (or $x+2y=5$):

$x$	$y$	$(x, y)$
0	2.5	$(0, 2.5)$
5	0	$(5, 0)$
1	2	$(1, 2)$

Plot these points and draw the line for $2x+4y=10$.

Step 2: Graph the second equation: $x + 2y = 3$.

Find solutions for $x+2y=3$:

• Set $x=0$: $0 + 2y = 3 \implies 2y = 3 \implies y = 1.5$. Point: $(0, 1.5)$.
• Set $y=0$: $x + 2(0) = 3 \implies x = 3$. Point: $(3, 0)$.
• Set $x=1$: $1 + 2y = 3 \implies 2y = 2 \implies y = 1$. Point: $(1, 1)$.

Table of solutions for $x+2y=3$:

$x$	$y$	$(x, y)$
0	1.5	$(0, 1.5)$
3	0	$(3, 0)$
1	1	$(1, 1)$

Plot these points on the same graph and draw the line for $x+2y=3$.

Step 3 & 4: Observe the intersection and determine the solution.

Looking at the graph where both lines are plotted:

The two lines appear to be parallel and do not intersect. This indicates there is no point common to both lines.

Step 5: Check (optional, but good practice).

We can check the coefficients ratios: For $2x+4y-10=0$ and $x+2y-3=0$, $\frac{a_1}{a_2} = \frac{2}{1} = 2$, $\frac{b_1}{b_2} = \frac{4}{2} = 2$, $\frac{c_1}{c_2} = \frac{-10}{-3} = \frac{10}{3}$. Since $\frac{a_1}{a_2} = \frac{b_1}{b_2}$ but $\neq \frac{c_1}{c_2}$, the lines are parallel and distinct. There is no solution.

Answer: The pair of linear equations has $\textbf{no solution}$.

Limitations of the Graphical Method:

While the graphical method provides a great visual understanding of the solution, its accuracy is limited by the precision of drawing and reading the graph. If the intersection point has fractional or irrational coordinates, it can be difficult to determine the exact solution from the graph. For precise solutions, algebraic methods are preferred.

Algebraic Methods of Solving a Pair of Linear Equations (Substitution, Elimination, Cross-Multiplication)

While the graphical method provides a visual representation of the solution(s) to a pair of linear equations, its accuracy is limited, especially when the intersection point does not have integer coordinates. Algebraic methods provide exact solutions. There are three primary algebraic methods used to solve a system of two linear equations in two variables: the Substitution Method, the Elimination Method, and the Cross-Multiplication Method.

Let the given pair of linear equations in standard form be:

Method 1: Substitution Method

The substitution method involves solving one of the equations for one variable in terms of the other, and then substituting that expression into the other equation. This transforms the system of two equations in two variables into a single equation in one variable, which we know how to solve.

Steps:

1. Express One Variable:

   Choose one of the two equations (often, pick the equation that looks simpler or where a variable has a coefficient of 1 or -1) and solve it for one variable in terms of the other. For example, if you choose Equation (1), express $y$ in terms of $x$ to get $y = \text{an expression involving } x$, or express $x$ in terms of $y$ to get $x = \text{an expression involving } y$.

2. Substitute:

   Substitute the expression obtained in Step 1 into the *other* equation. For instance, if you solved Equation (1) for $y$, substitute that expression for $y$ into Equation (2). If you solved for $x$, substitute that expression for $x$ into Equation (2). This will result in a linear equation containing only one variable.

3. Solve for the First Variable:

   Solve the linear equation obtained in Step 2 for the remaining variable. This will give you the numerical value of one of the variables (e.g., the value of $x$).

4. Find the Second Variable:

   Substitute the numerical value found in Step 3 back into the expression derived in Step 1 (where one variable was expressed in terms of the other). Solve for the numerical value of the second variable (e.g., the value of $y$).

5. State and Check the Solution:

   The solution is the ordered pair $(x, y)$. Verify your solution by substituting both values into *both* of the original equations to ensure that both equations are satisfied.

Answer:

Method 2: Elimination Method

The elimination method aims to eliminate one of the variables by making its coefficients in both equations either equal or additive inverses (opposite in sign) and then adding or subtracting the equations. This results in a single equation with one variable.

Steps:

1. Make Coefficients Equal:

   Multiply one or both equations by appropriate non-zero constants so that the absolute values of the coefficients of one of the variables (either $x$ or $y$) become equal in both equations. The goal is to make the coefficients of the chosen variable $\pm k$ and $\mp k$ or $k$ and $k$ in the two equations.

2. Eliminate a Variable:

   • If the variables you want to eliminate have coefficients with the same sign (e.g., both $+6y$ or both $-5x$), subtract one equation from the other.
   • If the variables you want to eliminate have coefficients with opposite signs (e.g., $+6y$ and $-6y$ or $-5x$ and $+5x$), add the two equations.

   This operation will eliminate the chosen variable, leaving a single linear equation in the other variable.

3. Solve for the First Variable:

   Solve the resulting linear equation for the remaining variable (e.g., solve for $x$). This gives the numerical value of one variable.

4. Find the Second Variable:

   Substitute the numerical value found in Step 3 back into either of the original equations and solve for the value of the second variable.

5. State and Check the Solution:

   The solution is the ordered pair $(x, y)$. Verify your solution by substituting both values into *both* of the original equations.

Answer:

Method 3: Cross-Multiplication Method

The cross-multiplication method is a formulaic approach for solving a pair of linear equations that have a unique solution. The formulas for $x$ and $y$ are derived from the general solution of the system $a_1 x + b_1 y + c_1 = 0$ and $a_2 x + b_2 y + c_2 = 0$.

Steps:

1. Write in Standard Form:

   Ensure both given equations are written in the standard form $a_1 x + b_1 y + c_1 = 0$ and $a_2 x + b_2 y + c_2 = 0$. Identify the coefficients $a_1, b_1, c_1$ and $a_2, b_2, c_2$.

2. Arrange Coefficients:

   Write down the coefficients in a specific arrangement, ignoring the variables and the equals signs. A helpful pattern is to list the coefficients in the order $b, c, a, b$ for both equations, one row below the other: $$ \begin{array}{ccccc} b_1 & c_1 & a_1 & b_1 \\ b_2 & c_2 & a_2 & b_2 \end{array} $$

3. Apply the Cross-Multiplication Formula:

   Use the pattern of cross-multiplication shown below the coefficients to set up the formula: $$ \frac{x}{b_1 c_2 - b_2 c_1} = \frac{y}{c_1 a_2 - c_2 a_1} = \frac{1}{a_1 b_2 - a_2 b_1} $$

   Visualization of Cross-Multiplication:

   Consider the arrangement of coefficients:

   $$ \begin{array}{c} \text{for } x: \begin{array}{cc} b_1 & c_1 \\ \downarrow \times \uparrow \\ b_2 & c_2 \end{array} \\ \text{Denominator for } x = b_1 c_2 - b_2 c_1 \end{array} \quad \quad \begin{array}{c} \text{for } y: \begin{array}{cc} c_1 & a_1 \\ \downarrow \times \uparrow \\ c_2 & a_2 \end{array} \\ \text{Denominator for } y = c_1 a_2 - c_2 a_1 \end{array} \quad \quad \begin{array}{c} \text{for } 1: \begin{array}{cc} a_1 & b_1 \\ \downarrow \times \uparrow \\ a_2 & b_2 \end{array} \\ \text{Denominator for } 1 = a_1 b_2 - a_2 b_1 \end{array} $$

   Notice the order of subscripts and coefficients in the denominators: Start with $b_1c_2 - b_2c_1$ under $x$, then cycle the pairs $(c_1, c_2)$ and $(a_1, a_2)$ to get $c_1a_2 - c_2a_1$ under $y$, and finally $(a_1, a_2)$ and $(b_1, b_2)$ to get $a_1b_2 - a_2b_1$ under $1$.

4. Solve for $x$ and $y$:

   Equate the first fraction with the third fraction to solve for $x$. Equate the second fraction with the third fraction to solve for $y$. $$ \frac{x}{b_1 c_2 - b_2 c_1} = \frac{1}{a_1 b_2 - a_2 b_1} \implies x = \frac{b_1 c_2 - b_2 c_1}{a_1 b_2 - a_2 b_1} $$ $$ \frac{y}{c_1 a_2 - c_2 a_1} = \frac{1}{a_1 b_2 - a_2 b_1} \implies y = \frac{c_1 a_2 - c_2 a_1}{a_1 b_2 - a_2 b_1} $$

5. Condition for Unique Solution:

   This method yields a unique solution only if the denominator $a_1 b_2 - a_2 b_1 \neq 0$. If $a_1 b_2 - a_2 b_1 = 0$, it means $\frac{a_1}{a_2} = \frac{b_1}{b_2}$ (assuming $a_2, b_2 \neq 0$). This is the condition for the lines to be parallel or coincident. In such cases, the system has either no solution or infinitely many solutions, and the cross-multiplication method cannot be directly applied to find a unique solution (it will result in division by zero or $0/0$ forms).

6. Check the Solution:

   Substitute the found values of $x$ and $y$ into both of the original equations to verify the solution.

Answer:

Algebraic methods are preferred for their precision. The substitution method is generally easiest when one variable has a coefficient of 1 or -1. The elimination method is often efficient when coefficients can be easily made equal or opposite. The cross-multiplication method is a formula that can be applied directly after writing equations in standard form, provided a unique solution exists.

Equations Reducible to a Pair of Linear Equations

Non-Linear Equations that can be Transformed

Not all pairs of equations involving two variables are linear. However, some pairs of non-linear equations have a special structure that allows them to be converted or "reduced" into a pair of linear equations by making appropriate substitutions for the variables or expressions involving the variables. Once transformed into a linear system, these equations can then be solved using the standard algebraic methods (Substitution, Elimination, or Cross-Multiplication) that we have already discussed.

Identifying Reducible Equations

Pairs of equations that are reducible to linear form often contain terms where the variables appear in the denominator, or where certain expressions involving variables repeat across different terms or equations.

Common patterns indicating that a pair of equations is reducible to linear form include the presence of terms like:

• $\frac{1}{x}, \frac{1}{y}$
• $\frac{1}{ax+b}, \frac{1}{cy+d}$
• $\frac{1}{x+y}, \frac{1}{x-y}$
• Expressions like $(x+y), (x-y), (2x+3y)$, etc., appearing in a way that a substitution would linearize the equations.

The key is to identify the repeating non-linear expressions and replace them with new variables.

Method for Solving Reducible Equations

To solve a pair of equations that is reducible to a pair of linear equations, follow these steps:

1. Identify and Substitute:

   Carefully examine the given equations and identify the non-linear expressions involving variables that, if replaced by new variables (say, $u$ and $v$), would transform the equations into a linear system in $u$ and $v$. Make these substitutions. For example, if terms like $\frac{1}{x}$ and $\frac{1}{y}$ are present, let $u = \frac{1}{x}$ and $v = \frac{1}{y}$.

2. Form the Linear System:

   Substitute the new variables identified in Step 1 into the original equations. This should result in a pair of linear equations in terms of the new variables (e.g., $u$ and $v$).

3. Solve the Linear System:

   Solve the resulting pair of linear equations (in $u$ and $v$) using any of the algebraic methods (Substitution, Elimination, or Cross-Multiplication). This will give you the numerical values of the new variables (i.e., find the values of $u$ and $v$).

4. Substitute Back and Solve for Original Variables:

   Substitute the numerical values found in Step 3 back into the original substitutions you made in Step 1 (e.g., if $u = \frac{1}{x}$ and $v = \frac{1}{y}$, replace $u$ and $v$ with their values). Solve these equations for the original variables ($x$ and $y$).

5. Check the Solution:

   Verify the ordered pair solution $(x, y)$ you found by substituting these values back into *both* of the original (non-linear) equations to ensure they are satisfied. Remember that if any variable appeared in a denominator in the original equations, the solution must not make that denominator zero.

Answer:

Answer:

Recognizing equations that are reducible to linear form and applying the appropriate substitution is a valuable problem-solving technique. It allows us to leverage the methods for solving linear systems to solve a wider variety of equations.