Pair of Linear Equations in Two Variables — Notes

Introduction to Pair of Linear Equations in Two Variables

In everyday life, we encounter situations that involve two or more unknown quantities. Often, these situations can be described using mathematical equations. When these equations involve variables raised to the power of one and have two unknowns, they are called linear equations in two variables. A pair of linear equations in two variables involves two such equations that need to be solved simultaneously to find a common solution.

Consider a scenario where we need to determine two unknown quantities. For example, if Akhila goes to a fair and wants to balance rides on a Giant Wheel and play Hoopla.

Let:

• $x$ be the number of rides on the Giant Wheel.
• $y$ be the number of times she played Hoopla.

The problem states two conditions:

1. "The number of times she played Hoopla is half the number of rides she had on the Giant Wheel." This can be written as an equation:

This equation represents a relationship between the number of rides and the number of Hoopla games.

2. "If each ride costs 3, and a game of Hoopla costs 4, provided she spent ` 20." This gives us another equation based on the total cost:

We now have a pair of linear equations in two variables:

1. $y=\frac{1}{2}x$
2. $3x+4y=20$

The goal of studying pairs of linear equations is to find the values of $x$ and $y$ that satisfy both equations simultaneously. This common solution represents the specific values that meet all the conditions of the problem.

We can rewrite the first equation to have both variables on one side, which is a common form for linear equations: Multiply the first equation by 2:

Rearrange it:

So, the pair of equations is:

1. $x-2y=0$
2. $3x+4y=20$

Solving this system will tell us how many rides Akhila took and how many times she played Hoopla. The methods to solve such systems will be explored in detail in the subsequent sections of this chapter.

Standard Form of a Pair of Linear Equations in Two Variables

A pair of linear equations in two variables, say $x$ and $y$, can be represented in the general standard form as:

where $a_1,b_1,c_1$ and $a_2,b_2,c_2$ are real coefficients, and at least one of $a_1,b_1$ is non-zero, and at least one of $a_2,b_2$ is non-zero.

Let's use the example of Akhila to see how the standard form applies:

1. $x-2y=0$ Here, $a_1=1$, $b_1=-2$, and $c_1=0$.

2. $3x+4y=20$ To fit the standard form $a_{2}x+b_{2}y+c_2=0$, we move 20 to the left side:

Here, $a_2=3$, $b_2=4$, and $c_2=-20$.

The coefficients ($a_1,b_1,a_2,b_2$) determine how the lines represented by these equations behave when graphed, which in turn tells us about the nature of their solution.

Example 1: Representing a Situation as a Pair of Linear Equations

Given: A teacher asks her students to form linear equations based on a word problem. Ritu can row a boat downstream in 20 minutes. She can row the same distance upstream in 25 minutes. If the speed of the boat in still water is 15 km/h, find the speed of the stream.

To find: The speed of the stream.

Solution: Let $x$ km/h be the speed of the boat in still water. Let $y$ km/h be the speed of the stream.

Condition 1: Rowing downstream Speed downstream = Speed of boat + Speed of stream = $(x+y)$ km/h. Time taken = 20 minutes = $\frac{20}{60}$ hours = $\frac{1}{3}$ hours. Distance = Speed $\times$ Time Distance = $(x+y)\times \frac{1}{3}$ km.

Condition 2: Rowing upstream Speed upstream = Speed of boat - Speed of stream = $(x-y)$ km/h. Time taken = 25 minutes = $\frac{25}{60}$ hours = $\frac{5}{12}$ hours. Distance = Speed $\times$ Time Distance = $(x-y)\times \frac{5}{12}$ km.

Since the distance rowed downstream is the same as the distance rowed upstream:

We are given that the speed of the boat in still water is 15 km/h. So, $x=15$. Substitute $x=15$ into the equation:

Now, let's simplify and solve for $y$: Multiply both sides by 12 to clear the denominators:

Distribute:

Collect terms with $y$ on one side and constants on the other:

Solve for $y$:

Answer: The speed of the stream is $\frac{5}{3}$ km/h.

Remember

When dealing with problems involving relative speeds (like boats in rivers or trains moving towards/away from each other), downstream means speeds add up ($v_{boat}+v_{stream}$), and upstream means speeds subtract ($v_{boat}-v_{stream}$). Ensure consistent units (hours for time, km/h for speed).

Exam-smart

Problems like this often appear as word problems. You'll be given a scenario with two related unknown quantities. The key is to translate the given information into two distinct linear equations by defining variables and setting up relationships based on conditions like speed, distance, time, cost, or quantity. The question might ask for one of the variables or both.

Example 2: Forming Equations from Given Information

Given: Ritu, we encountered in the previous example, had a boat. She can row a boat downstream in 20 minutes and upstream in 25 minutes. Let $x$ km/h be the speed of the boat in still water and $y$ km/h be the speed of the stream.

To find: Form a pair of linear equations in two variables to represent the given information.

Solution: Let $x$ be the speed of the boat in still water (in km/h). Let $y$ be the speed of the stream (in km/h).

Downstream Motion: Speed downstream = $(x+y)$ km/h. Time taken = 20 minutes = $\frac{20}{60}$ hours = $\frac{1}{3}$ hours. Let the distance rowed be $d$ km. Using the formula distance = speed $\times$ time, we get:

This can be written as: $3d=x+y$ (Equation 1)

Upstream Motion: Speed upstream = $(x-y)$ km/h. Time taken = 25 minutes = $\frac{25}{60}$ hours = $\frac{5}{12}$ hours. The distance is the same, $d$ km. Using the formula distance = speed $\times$ time, we get:

This can be written as: $12d=5(x-y)$ (Equation 2)

We have two equations:

1. $3d=x+y$
2. $12d=5(x-y)$

These equations involve three variables ($x$, $y$, and $d$). To get a pair of linear equations in two variables ($x$ and $y$), we need to eliminate $d$.

From Equation 1, $d=\frac{x+y}{3}$. Substitute this into Equation 2:

Expand both sides:

Now, rearrange to get the standard form $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$: Move all terms to one side. Let's move them to the right side to keep the coefficient of $x$ positive:

So, the first equation is $x-9y=0$.

Alternatively, we can also express the relationship by equating the distances directly. Distance downstream = $(x+y)\times \frac{1}{3}$ Distance upstream = $(x-y)\times \frac{5}{12}$ Equating distances:

Multiply by 12:

This equation represents the relationship between the speed of the boat and the speed of the stream based on the given times. If a specific speed of the boat or stream was given, we could form a second equation to solve for the unknowns. For instance, if the speed of the boat in still water was 15 km/h, then $x=15$. Substituting this into $x-9y=0$ would give $15-9y=0$, which leads to $9y=15$, so $y=\frac{15}{9}=\frac{5}{3}$ km/h.

Answer: The pair of linear equations that can be formed from the given information (assuming distance $d$ is the same) are:

1. $x-9y=0$ (This is a single equation relating $x$ and $y$)

If we were given a specific value for one variable, say $x=15$, we would use that to form a system:

1. $x-9y=0$
2. $x=15$

However, the question asks to form a pair of linear equations. If we consider the distance $d$ as one variable and speed $x$ and $y$ as others, we had:

1. $x+y=3d$
2. $x-y=\frac{12d}{5}$

This is a pair of linear equations in $x,y,d$. The problem implies forming equations in terms of the speeds $x$ and $y$. The crucial insight is that for a fixed distance, the relationship derived, $x-9y=0$, is itself the basis for a solution. If a specific value for $x$ or $y$ is given, it forms the second equation. The prompt is slightly ambiguous on what "pair" to form if only one relationship can be directly extracted. However, the standard interpretation is to express all given conditions as equations. Since only one relationship between $x$ and $y$ can be directly formed, and the problem states $x=15$ in the example solution, we use that. The standard form $a_{1}x+b_{1}y+c_1=0$ for the first condition is $x-9y=0$. If we consider the given $x=15$ as the second condition, then the pair is:

1. $x-9y=0$
2. $x=15$

Remember

Word problems require careful translation into algebraic expressions. Identify the unknowns, assign variables, and use the given information to form equations. Pay close attention to units and context (e.g., downstream vs. upstream).

Exam-smart

These questions test your ability to model real-world situations mathematically. You'll be given a narrative and expected to set up the correct system of equations. The "pair of linear equations" may not always be explicitly stated as two separate equations involving only $x$ and $y$ from the start; sometimes, one equation needs to be derived from a relationship, and a specific value given for a variable forms the second equation of the pair.

Try this: A boat travels 16 km in 2 hours downstream and 4 hours upstream. Find the speed of the boat in still water and the speed of the stream. Answer: Speed of boat in still water = 6 km/h, Speed of stream = 2 km/h.

Graphical Method of Solution of a Pair of Linear Equations

A pair of linear equations in two variables can be represented graphically as two straight lines on a Cartesian plane. The solution to such a system is the point (or points) where these lines intersect. The nature of the intersection determines the type of solution.

Types of Solutions and Corresponding Line Pair Behaviour

There are three possibilities for the graphical representation of a pair of linear equations:

1. Intersecting Lines: The lines cross at exactly one point. This means the pair of equations has a unique solution. Such a pair is called a consistent pair of linear equations.
2. Parallel Lines: The lines never meet. This means the pair of equations has no solution. Such a pair is called an inconsistent pair of linear equations.
3. Coincident Lines: The two lines lie exactly on top of each other. This means every point on the line is a solution, leading to infinitely many solutions. Such a pair is called a dependent pair of linear equations, and it is always consistent.

We can determine the nature of the solution by comparing the ratios of the coefficients of the variables and the constant terms for a pair of linear equations in the general form:

To solve a pair of linear equations graphically:

1. Represent each equation as a line on a graph paper.
2. To draw a line, find at least two points that satisfy the equation. These points can be found by choosing values for one variable (e.g., $x=0$ or $x=1$) and calculating the corresponding value for the other variable ($y$).
3. Plot these points and draw a straight line through them.
4. Repeat for the second equation.
5. Observe the intersection point(s). If they intersect at one point $(x_0,y_0)$, then $(x_0,y_0)$ is the unique solution. If the lines are parallel, there is no solution. If the lines are coincident, there are infinitely many solutions.

Example 1: Intersecting Lines

Given: The pair of equations:

1. $x+3y=6$
2. $2x-3y=12$

To find: Whether the pair is consistent and to solve them graphically.

Solution: First, we rewrite the equations in the general form to find the ratios of coefficients: Equation 1: $x+3y-6=0⟹a_1=1,b_1=3,c_1=-6$ Equation 2: $2x-3y-12=0⟹a_2=2,b_2=-3,c_2=-12$

Compare the ratios:

Since $\frac{a_1}{a_2}\mathrm{\ne }\frac{b_1}{b_2}$, the lines will intersect at a single point, meaning the pair is consistent and has a unique solution.

Now, let's find points to plot:

For Equation 1 ($x+3y=6$): If $x=0$, then $0+3y=6⟹y=2$. Point A(0, 2). If $y=0$, then $x+3(0)=6⟹x=6$. Point B(6, 0).

For Equation 2 ($2x-3y=12$): If $x=0$, then $2(0)-3y=12⟹-3y=12⟹y=-4$. Point P(0, -4). If $y=-2$, then $2x-3(-2)=12⟹2x+6=12⟹2x=6⟹x=3$. Point Q(3, -2).

Plotting points A(0, 2), B(6, 0), P(0, -4), and Q(3, -2) and drawing lines AB and PQ, we observe they intersect at point B(6, 0).

Answer: The solution is $x=6$ and $y=0$. The pair of equations is consistent.

Example 2: Coincident Lines

Given: The pair of equations:

1. $5x-8y+1=0$
2. $3x-\frac{24}{5}y+\frac{3}{5}=0$ (Note: The textbook text seems to have a typo here, it refers to multiplying Eq (2) by 5/3 to get Eq (1). Assuming the correct form of Eq (2) would lead to coincident lines with Eq (1) when scaled.)

Let's assume a corrected form for clarity that leads to coincident lines. If we take:

1. $5x-8y+1=0$
2. $10x-16y+2=0$

To find: Whether the pair has no solution, unique solution, or infinitely many solutions.

Solution: Equation 1: $a_1=5,b_1=-8,c_1=1$ Equation 2: $a_2=10,b_2=-16,c_2=2$

Compare the ratios:

Since $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$, the lines represented by these equations are coincident. This means there are infinitely many solutions.

Example 3: Intersecting Lines (Word Problem)

Given: Champa bought pants and skirts.

1. The number of skirts ($y$) is two less than twice the number of pants ($x$): $y=2x-2$.
2. The number of skirts ($y$) is four less than four times the number of pants ($x$): $y=4x-4$.

To find: The number of pants and skirts Champa bought.

Solution: We have the pair of equations:

1. $y=2x-2$
2. $y=4x-4$

Let's rewrite them in the general form: Equation 1: $2x-y-2=0⟹a_1=2,b_1=-1,c_1=-2$ Equation 2: $4x-y-4=0⟹a_2=4,b_2=-1,c_2=-4$

Compare ratios:

Since $\frac{a_1}{a_2}\mathrm{\ne }\frac{b_1}{b_2}$, the lines intersect at a unique point, meaning a unique solution exists.

Now, find points to plot:

For Equation 1 ($y=2x-2$): If $x=1$, then $y=2(1)-2=0$. Point A(1, 0). If $x=2$, then $y=2(2)-2=2$. Point B(2, 2).

For Equation 2 ($y=4x-4$): If $x=1$, then $y=4(1)-4=0$. Point A(1, 0). If $x=2$, then $y=4(2)-4=4$. Point C(2, 4).

Plotting points A(1, 0), B(2, 2), and C(2, 4). The lines representing the equations intersect at point A(1, 0).

Answer: Champa bought 1 pant and 0 skirts.

Try this:

Given the pair of equations: $x-y=1$ and $2x+y=5$. Check graphically if they have a unique solution, and if so, find it. Answer: The lines intersect at (2, 1).

Remember

When drawing graphs for linear equations, always find at least two points. For word problems, carefully translate the statements into algebraic equations before graphing.

Exam-smart

You might be given a graph showing two lines and asked to find the solution, or asked to determine if a system is consistent/inconsistent/dependent based on the ratio of coefficients without graphing. You could also be given a word problem and asked to represent it graphically.

Algebraic Methods of Solving a Pair of Linear Equations

While graphical methods provide a visual understanding of solutions, algebraic methods offer precise numerical solutions, especially useful for non-integral coordinates. These methods involve manipulating the equations directly to isolate variables.

Substitution Method

The substitution method involves solving one equation for one variable and then substituting that expression into the other equation. This reduces the system to a single equation with a single variable.

Steps:

1. Choose an equation and express one variable in terms of the other.
2. Substitute this expression into the other equation.
3. Solve the resulting equation for the single variable.
4. Substitute the value found back into the expression from Step 1 to find the value of the second variable.

Example 4:

Given: The pair of equations: $7x\text{–}15y=2$ (1) $x+2y=3$ (2)

To find: The values of $x$ and $y$ that satisfy both equations.

Solution: Step 1: From Equation (2), express $x$ in terms of $y$: $x=3\text{–}2y$ (3) Step 2: Substitute this expression for $x$ into Equation (1):

Step 3: Substitute the value of $y$ back into Equation (3):

Answer: $x=\frac{49}{29}$, $y=\frac{19}{29}$

Example 5: Age Problem

Given: Aftab's statement to his daughter: "Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be."

To find: The present ages of Aftab and his daughter.

Solution: Let $s$ be Aftab's current age and $t$ be his daughter's current age. Seven years ago: Aftab's age was $s-7$, daughter's age was $t-7$.

$s-7t+42=0$ (1)

Three years from now: Aftab's age will be $s+3$, daughter's age will be $t+3$.

$s-3t-6=0$ (2)

Using the substitution method: From Equation (2), express $s$ in terms of $t$: $s=3t+6$ (3)

Substitute this into Equation (1):

Substitute the value of $t$ back into Equation (3):

Answer: Aftab is 42 years old, and his daughter is 12 years old.

Example 6: Cost Problem (Infinite Solutions)

Given: Cost of 2 pencils and 3 erasers is 9. Cost of 4 pencils and 6 erasers is 18.

To find: The cost of each pencil and eraser.

Solution: Let $x$ be the cost of a pencil and $y$ be the cost of an eraser. $2x+3y=9$ (1) $4x+6y=18$ (2)

From Equation (1), express $x$ in terms of $y$:

$x=\frac{9-3y}{2}$ (3)

Substitute this into Equation (2):

This is a true statement with no variables. This indicates that the two original equations are dependent and represent the same line. There are infinitely many solutions, meaning we cannot find a unique cost for each pencil and eraser.

Example 7: Parallel Lines (No Solution)

Given: Two rails represented by the equations: $x+2y\text{–}4=0$ (1) $2x+4y\text{–}12=0$ (2)

To find: Whether the rails will cross each other (i.e., if there is a common solution).

Solution: From Equation (1), express $x$ in terms of $y$: $x=4\text{–}2y$ (3)

Substitute this into Equation (2):

This is a false statement. This indicates that the equations are inconsistent and represent parallel lines. Therefore, the rails will not cross.

Remember

When solving by substitution, if you arrive at a true statement (like $18=18$), the system has infinitely many solutions. If you arrive at a false statement (like $-4=0$), the system has no solution.

Exam-smart

In word problems solved by substitution, look for phrases like "exceeds by", "times as old as", or statements comparing quantities. The substitution method is useful for translating these relationships into equations and solving them. Pay attention to whether the final step yields a specific value or a statement of identity/contradiction.

Elimination Method

The elimination method involves multiplying one or both equations by suitable constants so that the coefficients of one variable become equal or opposite. Then, by adding or subtracting the equations, that variable is eliminated, leaving a single equation with one variable.

Steps:

1. Multiply both equations by non-zero constants to make the coefficients of one variable numerically equal.
2. Add or subtract one equation from the other to eliminate the variable.

• If coefficients are same (e.g., $3x$ and $3x$), subtract.
• If coefficients are opposite (e.g., $3x$ and $-3x$), add.

3. Solve the resulting equation for the remaining variable.
4. Substitute this value back into either original equation to find the value of the other variable.

Example 8: Income and Expenditure

Given: Ratio of incomes of two persons is 9:7. Ratio of their expenditures is 4:3. Each saves `2000 per month.

To find: Their monthly incomes.

Solution: Let incomes be $9x$ and $7x$. Let expenditures be $4y$ and $3y$. Savings = Income - Expenditure. Equation (1): $9x-4y=2000$ Equation (2): $7x-3y=2000$

To eliminate $y$: Multiply Equation (1) by 3: $3\times (9x-4y)=3\times 2000\to 27x-12y=6000$ (3) Multiply Equation (2) by 4: $4\times (7x-3y)=4\times 2000\to 28x-12y=8000$ (4)

The coefficients of $y$ are now equal (-12y). Subtract Equation (3) from Equation (4):

Substitute $x=2000$ into Equation (1):

The monthly incomes are $9x=9\times 2000=18000$ and $7x=7\times 2000=14000$.

Answer: Their monthly incomes are 18,000 and 14,000.

Example 9: Inconsistent System (No Solution)

Given: $2x+3y=8$ (1) $4x+6y=7$ (2)

To find: All possible solutions.

Solution: To eliminate $x$, multiply Equation (1) by 2: $2\times (2x+3y)=2\times 8\to 4x+6y=16$ (3)

Now compare Equation (3) and Equation (2):

Subtracting Equation (2) from Equation (3):

This is a false statement. Therefore, the pair of equations has no solution. The lines are parallel.

Example 10: Two-Digit Number

Given: The sum of a two-digit number and the number formed by reversing its digits is 66. The digits of the number differ by 2.

To find: The number(s).

Solution: Let the ten's digit be $x$ and the unit's digit be $y$. The number is $10x+y$. The reversed number is $10y+x$.

Condition 1: Sum of the number and its reverse is 66.

Divide by 11: $x+y=6$ (1)

Condition 2: The digits differ by 2. This can be either $x-y=2$ or $y-x=2$.

Case 1: $x-y=2$ (2) Solve equations (1) and (2) by elimination: Add (1) and (2):

Substitute $x=4$ into (1):

The number is $10x+y=10(4)+2=42$.

Case 2: $y-x=2$ (3) (or $-x+y=2$) Solve equations (1) and (3) by elimination: Add (1) and (3):

Substitute $y=4$ into (1):

The number is $10x+y=10(2)+4=24$.

Answer: The numbers are 42 and 24.

Remember

When using the elimination method, if you arrive at a statement like $0=0$ (a true statement), it means the lines are coincident, and there are infinitely many solutions. If you arrive at a statement like $0=k$ where $k\mathrm{\ne }0$ (a false statement), the lines are parallel, and there is no solution.

Exam-smart

Elimination is particularly efficient when coefficients are easy to make equal or opposite. In problems involving digit manipulation (like two-digit numbers), setting up the initial equations using place value ($10x+y$) is crucial. Also, remember that "differ by 2" implies two possible scenarios ($x-y=2$ or $y-x=2$).