Pair of Linear Equations in Two Variables

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**.**

Table of Content

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.

**Ans:** 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**.**

(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

\({\Large\frac{a_1}{a_2}}={\Large\frac{b_1}{b_2}}={\Large\frac{c_1}{c_2}}\\[4.5bp]{\Large\frac{2}{a+b}}={\Large\frac{3}{a+b-3}}\\[4.5bp] ={\Large\frac{7}{4a+b}}\\[4.5bp] {\Large\frac{2}{a+b}} = {\Large\frac{7}{4a+b}}\)

8a + 2b = 7a +7b

a - 5b = 0

Hence lines are coincident.

(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.

**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

**Ans. **Consistent

**Explanation:**We have,

\({\Large\frac{3}{2}}x+{\Large\frac{5}{3}}y - 7 = 0\)

and 9x - 10y - 14 = 0

\(\text{Here, }\space \space \space a_1 = {\Large\frac{3}{2}}, \space b_1 = {\Large\frac{5}{3}}, c_1 = -7\)

a

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

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

(i) coincident lines

(ii) intersecting lines

**Explanation:(i)** Given, x - 2y - 6 = 0

For 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

Here, a

a

\({\Large\frac{a_1}{a_2}} = {\Large\frac{1}{2}};{\Large\frac{b_1}{b_2}}= {\Large\frac{-2}{-4}}={\Large\frac{1}{2}}; {\Large\frac{c_1}{c_2}}={\Large\frac{-6}{-12}}={\Large\frac{1}{2}}\\[4.5bp] ∴ {\Large\frac{a_1}{a_2}}={\Large\frac{b_1}{b_2}}={\Large\frac{c_1}{c_2}}\)

So, it shows coincident lines.

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

a

\({\Large\frac{a_1}{a_2}}= {\Large\frac{1}{2}}, {\Large\frac{b_1}{b_2}} = {\Large\frac{-2}{-7}}={\Large\frac{2}{7}}\\[4.5bp] ∴ {\Large\frac{a_1}{a_2}} ≠ {\Large\frac{b_1}{b_2}} \)

So, they represent intersecting lines.

If you are seeking additional practice and a deeper grasp of the concepts covered in the chapter, oswal.io offers a wide-ranging collection of questions that will help you gain a more comprehensive understanding of the topic. By exploring their extensive question bank, you can reinforce your knowledge, improve your problem-solving abilities.

**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.