Pair of Linear Equations in Two Variables

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Class 10 Maths Chapter 3
Pair of Linear Equation in Two Variables
Important Questions

Here are some important Class 10 Mathematics questions for Chapter 3, Pair of Linear Equations in Two Variables. These thoughtfully selected questions are designed to support students in their preparation for the CBSE Class 10 Mathematics Examination 2023-24. Practicing various question types will not only clarify doubts but also improve exam readiness. By solving these questions, students can boost their confidence in problem-solving and excel in the Pair of Linear Equations in Two Variables chapter.

Introduction

In Chapter 3, “Pair of Linear Equations in Two Variables,” we explore the graphical method of solving a system of two linear equations. The concept involves representing both equations graphically on the Cartesian plane and finding the point of intersection, which corresponds to the solution of the equations.

What are pair of linear equations in two Variables?

A pair of linear equations in two variables is a set of two equations of the form: ax + by = c px + qy = r where ‘x’ and ‘y’ are the variables, and ‘a’, ‘b’, ‘p’, ‘q’, ‘c’, and ‘r’ are constants with at least one of ‘a’ and ‘ B ‘ or ‘p’ and ‘q’ is not equal to zero.

Class 10 Pair of Linear Equations in Two Variables Important Questions and Answers

Q 1. If 2x - 3y = 7 and (a + b) x - (a + b - 3)y = 4a+b represent______ lines, then a and b satisfy the equation a - 5b = 0.
Options
(a) Coincident
(b) Parallel
(c) Unique
(d) None of these

Ans. (a)

Explanation:
Given, 2x - 3y = 7 and (a + b) x - (a + b - 3)y = 4a+b
\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \\ \frac{2}{a+b} = \frac{3}{a+b–3} = \frac{7}{4a+b} \\ \frac{2}{a+b} = \frac{7}{4a+b}

8a + 2b = 7a + 7b
a – 5b = 0
Hence lines are coincident.

Q 2. The pair equations y = 0 and y = - 7 has:
Options
(a) One solution
(b) Two solution
(c) Infinitely many solutions
(d) No solution

Ans. (d)

Explanation:
The given pair of equations y = 0 and y = - 7

Graphically both lines are parallel and have no solution.

Q 3. Cost of 2 pens and 3 pencils together is ₹ 40 and the cost of 6 pens and 9 pencils together is ₹ 130. Express above statement in the form of linear equations.

Ans. 2x + 3y=40
and 6x + 9y=130
Explanation:
Given, cost of 2 pens and 3 pencils together is ₹ 40 and the cost of 6 pens and 9 pencils together is ₹ 130.
Let the cost of 1 pen be ₹ x and cost of 1 pencil be ₹ y
Then, 2x + 3y=40
and 6x + 9y=130

Q 4. Find whether the following system of equations is consistent or inconsistent :
{\Large\frac{3}{2}}x+{\Large\frac{5}{3}}y = 7 \text{ and 9x - 10y = 14.}

Ans. Consistent
Explanation:
‍We have, {\Large\frac{3}{2}}x+{\Large\frac{5}{3}}y - 7 = 0 \space and \space 9x - 10y - 14 = 0 \\[4.5bp] \text{Here,} \\ a_1 = \frac{3}{2}, \space b_1 = \frac{5}{3}, c_1 = -7, a_2 = 9, b_2 = - 10, c_2 = - 14 \\[4.5bp]

\text{Thus, } \space {\Large\frac{a_1}{a_2}}={\Large \frac{3}{2×9}} = {\Large\frac{1}{6}}, {\Large\frac{b_1}{b_2}}={\Large\frac{5}{3(-10)}}=-{\Large\frac{1}{6}}\\[4.5bp] \text{Since, }\space {\Large\frac{a_1}{a_2}}\not= {\Large\frac{b_1}{b_2}}

So, a given system of equations is unique and it is consistent.

Q 5. Given the linear equation x – 2y – 6 = 0, write another linear equation in these two variables, such that the geometrical representation of the pair so formed is:
(i) coincident lines
(ii) intersecting lines

Explanation:
(i) Given, x - 2y - 6 = 0For lines to be coincident, the condition is:
{\Large\frac{a_1}{a_2}}={\Large\frac{b_1}{b_2}}={\Large\frac{c_1}{c_2}}

Thus, one possible option can be:
2x - 4y - 12 = 0 \space \\ \text{Here,} \space a_1 = 1, b_1 = - 2, c_1 = - 6. \space a_2 = 2, b_2 = - 4, c_2 = - 12. \\ {\Large\frac{a_1}{a_2}} = {\Large\frac{1}{2}};{\Large\frac{b_1}{b_2}}= {\Large\frac{-2}{-4}}={\Large\frac{1}{2}}; {\Large\frac{c_1}{c_2}}={\Large\frac{-6}{-12}}={\Large\frac{1}{2}}\\[4.5bp] ∴ {\Large\frac{a_1}{a_2}}={\Large\frac{b_1}{b_2}}={\Large\frac{c_1}{c_2}}

So, it shows coincident lines.‍

(ii) Given, 2x - 2y - 6 =0 For intersecting  lines {\Large\frac{a_1}{a_2}}\not={\Large\frac{b_1}{b_2}} Thus, one possible can be: 2x - 7y - 13 = 0
Here, a_1 = 1, \space b_1 = - 2, \space c_1 = - 6. \space a_2 = 2, \space b_2 = - 7, \space c_2 = - 13.\\ {\Large\frac{a_1}{a_2}}= {\Large\frac{1}{2}}, {\Large\frac{b_1}{b_2}} = {\Large\frac{-2}{-7}}={\Large\frac{2}{7}}\\[4.5bp] ∴ {\Large\frac{a_1}{a_2}} ≠ {\Large\frac{b_1}{b_2}} So, they represent intersecting lines.

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CBSE Class 10 Maths Chapter wise Important Questions

Conclusion

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Frequently Asked Questions

Ans: The two common algebraic methods are the substitution method and the elimination method.
Ans: Graphical method involves plotting both equations on a Cartesian plane as lines and finding the point of intersection, which represents the solution to the system.
Ans: Yes, if the two lines representing the equations coincide (overlap), the system has infinitely many solutions.
Ans: The graphical method involves representing each equation on the Cartesian plane as a straight line by plotting its intercepts or using the slope-intercept form. The solution to the pair of equations is the point of intersection of these lines on the graph.
Ans: Yes, if the two lines representing the equations are parallel and do not intersect, the system has no solution.