Worksheet
Explore the methods to solve a pair of linear equations in two variables, including graphical, substitution, elimination, and cross-multiplication techniques.
Pair of Linear Equations in Two Variables - Practice Worksheet
Strengthen your foundation with key concepts and basic applications.
This worksheet covers essential long-answer questions to help you build confidence in Pair of Linear Equations in Two Variables from Mathematics for Class X (Mathematics).
Basic comprehension exercises
Strengthen your understanding with fundamental questions about the chapter.
Questions
Explain the graphical method of solving a pair of linear equations in two variables. What does the point of intersection represent?
Recall how to plot a line using its equation and interpret the intersection point.
Solution
The graphical method involves plotting two linear equations on the same graph. Each equation represents a straight line. The point where these two lines intersect is the solution to the pair of equations, representing the values of x and y that satisfy both equations simultaneously. If the lines are parallel, there is no solution, indicating the equations are inconsistent. If the lines coincide, there are infinitely many solutions, meaning the equations are dependent. This method is useful for visualizing the relationship between two variables and understanding the nature of their solutions.
Describe the substitution method for solving a pair of linear equations in two variables with an example.
Start by isolating one variable in one equation and then substitute it into the other equation.
Solution
The substitution method involves solving one equation for one variable and then substituting this expression into the other equation. For example, consider the equations 2x + y = 5 and 3x - 2y = 4. Solve the first equation for y: y = 5 - 2x. Substitute this into the second equation: 3x - 2(5 - 2x) = 4. Simplify to find x = 2, then substitute back to find y = 1. This method is straightforward when one equation can easily be solved for one variable.
How does the elimination method work in solving a pair of linear equations? Provide a step-by-step explanation.
Look for or create coefficients that are opposites for one variable to eliminate it.
Solution
The elimination method involves adding or subtracting the equations to eliminate one variable, making it possible to solve for the other. For example, take the equations x + y = 4 and x - y = 2. Adding them eliminates y, giving 2x = 6, so x = 3. Substituting x back into one equation gives y = 1. This method is efficient when the coefficients of one variable are opposites or can be made so by multiplication.
What are the conditions for a pair of linear equations to have a unique solution, no solution, or infinitely many solutions?
Compare the ratios of the coefficients of x, y, and the constants in the two equations.
Solution
A pair of linear equations has a unique solution if the lines intersect at one point, meaning the ratios of the coefficients of x and y are not equal (a1/a2 ≠ b1/b2). No solution exists if the lines are parallel (a1/a2 = b1/b2 ≠ c1/c2), indicating the equations are inconsistent. Infinitely many solutions occur if the lines coincide (a1/a2 = b1/b2 = c1/c2), meaning the equations are dependent. These conditions help determine the nature of the solutions without graphing.
Solve the pair of equations 3x + 2y = 12 and 6x + 4y = 24 using the elimination method. What do you observe?
Notice that the second equation is a multiple of the first, suggesting they represent the same line.
Solution
To solve 3x + 2y = 12 and 6x + 4y = 24, multiply the first equation by 2 to get 6x + 4y = 24. Subtracting this from the second equation gives 0 = 0, which is always true. This indicates that the two equations are equivalent and represent the same line. Therefore, there are infinitely many solutions, as any point on the line satisfies both equations.
A shopkeeper sells two types of pens. The cost of 5 pens of type A and 3 pens of type B is Rs. 79, while the cost of 2 pens of type A and 7 pens of type B is Rs. 89. Find the cost of each type of pen.
Use the elimination method to solve for one variable first.
Solution
Let the cost of type A pen be x and type B pen be y. The equations are 5x + 3y = 79 and 2x + 7y = 89. Multiply the first equation by 2 and the second by 5 to get 10x + 6y = 158 and 10x + 35y = 445. Subtract the first new equation from the second to eliminate x: 29y = 287, so y = 9.89. Substitute y back into the first original equation to find x = 10.26. Thus, type A pen costs approximately Rs. 10.26 and type B pen costs approximately Rs. 9.89.
Explain how to form a pair of linear equations from a word problem. Use the example of two numbers whose sum is 50 and difference is 10.
Define variables for the unknowns and translate the given conditions into equations.
Solution
To form a pair of linear equations from a word problem, identify the variables and the relationships between them. For example, let the two numbers be x and y. Their sum is 50: x + y = 50. Their difference is 10: x - y = 10. These two equations form a system that can be solved using substitution or elimination. Solving them gives x = 30 and y = 20. This approach can be applied to various real-life situations by carefully defining variables and translating words into mathematical equations.
What is the significance of the graphical representation of a pair of linear equations in understanding their solutions?
Consider how the position and slope of the lines affect their intersection.
Solution
The graphical representation of a pair of linear equations provides a visual understanding of their solutions. The point of intersection represents the unique solution if the lines intersect. Parallel lines indicate no solution, while coinciding lines represent infinitely many solutions. This method helps in comprehending the nature of the solutions, whether they are consistent, inconsistent, or dependent, without algebraic manipulation. It also aids in verifying the solutions obtained by other methods.
Solve the pair of equations 0.4x + 0.3y = 1.7 and 0.7x - 0.2y = 0.8 using the substitution method.
Start by expressing y in terms of x from the first equation and then substitute into the second.
Solution
First, solve the first equation for y: 0.3y = 1.7 - 0.4x, so y = (1.7 - 0.4x)/0.3. Substitute this into the second equation: 0.7x - 0.2[(1.7 - 0.4x)/0.3] = 0.8. Multiply through by 0.3 to eliminate denominators: 0.21x - 0.2(1.7 - 0.4x) = 0.24. Expand and simplify: 0.21x - 0.34 + 0.08x = 0.24, leading to 0.29x = 0.58, so x = 2. Substitute x back into the expression for y: y = (1.7 - 0.8)/0.3 = 3. Thus, the solution is x = 2, y = 3.
A fraction becomes 1/2 when 1 is subtracted from the numerator and 1 is added to the denominator. It becomes 1/3 when 1 is added to the numerator and 1 is subtracted from the denominator. Find the original fraction.
Set up equations based on the given conditions and solve the system.
Solution
Let the numerator be x and the denominator be y. The first condition gives (x - 1)/(y + 1) = 1/2, leading to 2x - 2 = y + 1 or 2x - y = 3. The second condition gives (x + 1)/(y - 1) = 1/3, leading to 3x + 3 = y - 1 or 3x - y = -4. Subtract the first equation from the second: x = -7. Substitute x into the first equation: -14 - y = 3, so y = -17. Thus, the original fraction is -7/-17, which simplifies to 7/17.
Pair of Linear Equations in Two Variables - Mastery Worksheet
Advance your understanding through integrative and tricky questions.
This worksheet challenges you with deeper, multi-concept long-answer questions from Pair of Linear Equations in Two Variables to prepare for higher-weightage questions in Class X Mathematics.
Intermediate analysis exercises
Deepen your understanding with analytical questions about themes and characters.
Questions
Solve the pair of linear equations graphically: 2x + 3y = 8 and 4x + 6y = 7. What do you observe about their solutions?
Compare the ratios of the coefficients to determine the nature of the lines.
Solution
The equations 2x + 3y = 8 and 4x + 6y = 7 are parallel lines since their coefficients are proportional (a1/a2 = b1/b2 ≠ c1/c2). Therefore, they have no solution, indicating an inconsistent pair of equations.
A fraction becomes 9/11 when 2 is added to both numerator and denominator. If 3 is added to both, it becomes 5/6. Find the fraction.
Set up equations based on the given conditions and solve them using substitution or elimination.
Solution
Let the fraction be x/y. The equations formed are (x+2)/(y+2) = 9/11 and (x+3)/(y+3) = 5/6. Solving these, we find x = 7 and y = 9. Thus, the fraction is 7/9.
The sum of a two-digit number and the number obtained by reversing its digits is 66. If the digits differ by 2, find the number.
Express the numbers in terms of their digits and set up equations based on the given conditions.
Solution
Let the digits be x and y. The equations are 10x + y + 10y + x = 66 and |x - y| = 2. Solving gives two possible numbers: 42 and 24.
Compare the graphical and algebraic methods of solving a pair of linear equations. Which method is more efficient and why?
Consider the accuracy and applicability of each method in different scenarios.
Solution
Graphical method provides a visual representation and is useful for understanding the nature of solutions, but it's less precise for non-integral solutions. Algebraic methods (substitution, elimination) are more efficient for exact solutions, especially with non-integral values.
A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹27 for seven days, while Susy paid ₹21 for five days. Find the fixed and additional charges.
Set up equations based on the total charges for different durations.
Solution
Let fixed charge be ₹x and additional charge per day be ₹y. The equations are x + 4y = 27 and x + 2y = 21. Solving gives x = 15 and y = 3.
Explain the conditions under which a pair of linear equations has no solution, a unique solution, or infinitely many solutions.
Analyze the ratios of the coefficients of the equations.
Solution
No solution: Lines are parallel (a1/a2 = b1/b2 ≠ c1/c2). Unique solution: Lines intersect (a1/a2 ≠ b1/b2). Infinitely many solutions: Lines coincide (a1/a2 = b1/b2 = c1/c2).
The coach of a cricket team buys 7 bats and 6 balls for ₹3800. Later, she buys 3 bats and 5 balls for ₹1750. Find the cost of each bat and ball.
Use the elimination method to solve the system of equations.
Solution
Let cost of a bat be ₹x and a ball be ₹y. The equations are 7x + 6y = 3800 and 3x + 5y = 1750. Solving gives x = 500 and y = 50.
Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?
Set up equations based on the age conditions given.
Solution
Let Jacob's age be x and son's age be y. The equations are x + 5 = 3(y + 5) and x - 5 = 7(y - 5). Solving gives x = 40 and y = 10.
A rectangular garden's length is 4m more than its width. Half the perimeter is 36m. Find the dimensions of the garden.
Express the perimeter in terms of the width and set up the equation.
Solution
Let width be x and length be x + 4. The equation is (2x + 2(x + 4))/2 = 36. Solving gives x = 16. Thus, dimensions are 16m and 20m.
The taxi charges in a city consist of a fixed charge plus a charge per km. For 10km, the charge is ₹105, and for 15km, it's ₹155. Find the fixed charge and the rate per km.
Set up linear equations based on the total charges for different distances.
Solution
Let fixed charge be ₹x and rate per km be ₹y. The equations are x + 10y = 105 and x + 15y = 155. Solving gives x = 5 and y = 10.
Pair of Linear Equations in Two Variables - Challenge Worksheet
Push your limits with complex, exam-level long-form questions.
The final worksheet presents challenging long-answer questions that test your depth of understanding and exam-readiness for Pair of Linear Equations in Two Variables in Class X.
Advanced critical thinking
Test your mastery with complex questions that require critical analysis and reflection.
Questions
Akhila went to a fair and spent ` 20 on rides and games. The number of times she played Hoopla is half the number of rides she had on the Giant Wheel. Each ride costs ` 3, and a game of Hoopla costs ` 4. Formulate the situation as a pair of linear equations and solve it graphically.
Consider expressing y in terms of x for the first equation and then substitute into the second equation for graphical representation.
Solution
Let the number of rides be x and the number of Hoopla games be y. The equations are y = (1/2)x and 3x + 4y = 20. Plotting these equations on a graph will give the solution where they intersect.
Explain the conditions under which a pair of linear equations in two variables has no solution, a unique solution, or infinitely many solutions. Provide examples for each case.
Think about the graphical representation of each scenario to understand the conditions.
Solution
A pair of linear equations has no solution if the lines are parallel (a1/a2 = b1/b2 ≠ c1/c2), a unique solution if they intersect (a1/a2 ≠ b1/b2), and infinitely many solutions if they coincide (a1/a2 = b1/b2 = c1/c2). Examples include x + 2y = 4 and 2x + 4y = 12 for no solution, x + y = 5 and 2x - y = 4 for unique solution, and x + y = 2 and 2x + 2y = 4 for infinitely many solutions.
Champa purchased some pants and skirts. The number of skirts is two less than twice the number of pants. Also, the number of skirts is four less than four times the number of pants. Represent this situation algebraically and find the number of pants and skirts she bought.
Set up two equations based on the given conditions and solve them simultaneously.
Solution
Let the number of pants be x and skirts be y. The equations are y = 2x - 2 and y = 4x - 4. Solving these gives x = 1 and y = 0, meaning she bought 1 pant and no skirts.
The sum of a two-digit number and the number obtained by reversing its digits is 66. If the digits differ by 2, find the number. How many such numbers exist?
Express the two-digit number and its reverse in terms of x and y, then form equations based on the given conditions.
Solution
Let the digits be x and y. The equations are 10x + y + 10y + x = 66 and |x - y| = 2. Solving gives two numbers: 42 and 24.
A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ` 27 for seven days, and Susy paid ` 21 for five days. Find the fixed charge and the charge for each extra day.
Set up equations based on the total charges for Saritha and Susy, considering the fixed and additional charges.
Solution
Let the fixed charge be ` x and the additional charge per day be ` y. The equations are x + 4y = 27 and x + 2y = 21. Solving gives x = 15 and y = 3.
The ratio of incomes of two persons is 9:7, and the ratio of their expenditures is 4:3. If each saves ` 2000 per month, find their monthly incomes.
Express savings as income minus expenditure and set up proportional relationships.
Solution
Let the incomes be 9x and 7x, and expenditures be 4y and 3y. The equations are 9x - 4y = 2000 and 7x - 3y = 2000. Solving gives x = 2000, so incomes are ` 18000 and ` 14000.
A fraction becomes 9/11 if 2 is added to both numerator and denominator. If 3 is added to both, it becomes 5/6. Find the fraction.
Cross-multiply to eliminate denominators and solve the resulting linear equations.
Solution
Let the fraction be x/y. The equations are (x + 2)/(y + 2) = 9/11 and (x + 3)/(y + 3) = 5/6. Solving gives x = 7 and y = 9, so the fraction is 7/9.
Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. Find their present ages.
Set up equations based on age relationships in the past and future.
Solution
Let Nuri's age be x and Sonu's age be y. The equations are x - 5 = 3(y - 5) and x + 10 = 2(y + 10). Solving gives x = 50 and y = 20.
The taxi charges in a city consist of a fixed charge plus a charge per km. For 10 km, the charge is ` 105, and for 15 km, it's ` 155. Find the fixed charge and the charge per km. What will be the charge for 25 km?
Set up linear equations based on the total charges for different distances.
Solution
Let the fixed charge be ` x and the charge per km be ` y. The equations are x + 10y = 105 and x + 15y = 155. Solving gives x = 5 and y = 10. The charge for 25 km is ` 255.
A cricket team's coach buys 7 bats and 6 balls for ` 3800. Later, she buys 3 bats and 5 balls for ` 1750. Find the cost of each bat and ball.
Set up two equations based on the total costs for different purchases and solve them simultaneously.
Solution
Let the cost of a bat be ` x and a ball be ` y. The equations are 7x + 6y = 3800 and 3x + 5y = 1750. Solving gives x = 500 and y = 50.
Real Numbers encompass all rational and irrational numbers, forming a complete and continuous number line essential for various mathematical concepts.
Explore the world of Polynomials, understanding their types, degrees, and operations to solve algebraic expressions and equations effectively.
Explore the world of quadratic equations, learning to solve them using various methods like factoring, completing the square, and the quadratic formula.
A chapter that explores sequences where each term after the first is obtained by adding a constant difference, focusing on their properties, nth term, and sum formulas.
