The exercise explains how to solve a pair of linear equations through graphical methods. The lesson establishes how two linear equations display as straight lines on a Cartesian coordinate system, where their point of intersection reveals the solution. The visualisation enables learners to determine the solution possibilities based on the line behaviour of lines intersecting or being parallel, or coinciding. The process reveals how algebra functions together with geometric principles to solve problems that exist in the real world.
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Understandings of two-variable linear equation behaviour in graphical representations are taught in this essential section of the NCERT Solutions for Class 10 Maths. Students who perform exercise gain deeper comprehension about linear equation graphical solutions and system consistency, and how it determines unique versus infinite solutions. The clear presentation of the solutions given in the NCERT Books enables students' understanding of the connection between graphical interpretations and algebraic processes before moving on to complex multisystem topics in higher education.
Answer:
Let the number of boys be x and the number of girls be y.
Now, according to the question,
Total number of students in the class = 10, i.e.
$\Rightarrow x+y=10.....(1)$
And, given that the number of girls is 4 more than the number of boys it means; $x=y+4$
$\Rightarrow x-y=4..........(2)$
Different points (x, y) satisfying equation (1)
X | 5 | 6 | 4 |
Y | 5 | 4 | 6 |
Different points (x,y) satisfying equation (2)
X | 5 | 6 | 7 |
y | 1 | 2 | 3 |
Graph,
From the graph, both lines intersect at the point (7,3). That is x = 7 and y = 3, which means the number of boys in the class is 7 and the number of girls in the class is 3.
Answer:
Let the price of 1 pencil be x, and y be the price of 1 pen.
Now, according to the question
$5x+7y=50......(1)$
And
$7x+5y=46......(2)$
Now, the points (x,y) that satisfy the equation (1) are
X | 3 | -4 | 10 |
Y | 5 | 10 | 0 |
And, the points (x,y) that satisfy the equation (2) are
X | 3 | 8 | -2 |
Y | 5 | -2 | 12 |
The Graph,
From the graph, both lines intersect at point (3,5), that is, x = 3 and y = 5, which means the cost of 1 pencil is 3 and the cost of 1 pen is 5.
Answer:
Given Equations,
$\\5x - 4y + 8 = 0 $and$ 7x + 6y - 9 = 0$
Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get
$\frac{a_1}{a_2}=\frac{5}{7},\:\frac{b_1}{b_2}=\frac{-4}{6}\:and\:\frac{c_1}{c_2}=\frac{8}{-9}$
It is observed that;
$\frac{a_1}{a_2}\neq\frac{b_1}{b_2}$
It means that both lines intersect at exactly one point and have a unique solution.
Answer:
Given Equations,
$\\9x + 3y + 12 = 0$ and $18x + 6y + 24 = 0$
Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get
$\\\frac{a_1}{a_2}=\frac{9}{18}=\frac{1}{2},\\\:\frac{b_1}{b_2}=\frac{3}{6}=\frac{1}{2}\: and \\\:\frac{c_1}{c_2}=\frac{12}{24}=\frac{1}{2}$
It is observed that;
$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
It means that both lines are coincident and have infinitely many solutions.
Q2 (iii) On comparing the ratios $\frac{a_1}{a_2}$, $\frac{b_1}{b_2}$and $\frac{c_1}{c_2}$ , find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident: (iii) $\\6x - 3y + 10 = 0; \\ 2x - y+ 9 = 0$
Answer:
Given Equations,
$\\6x - 3y + 10 = 0$ and$ 2x - y+ 9 = 0$
Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get
$\frac{a_1}{a_2}=\frac{6}{2}=3,\:\frac{b_1}{b_2}=\frac{-3}{-1}=3\:and\:\frac{c_1}{c_2}=\frac{10}{9}$
It is observed that;
$\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$
It means that both lines are parallel and thus have no solution.
Answer:
Given Equations,
$\\3x + 2y = 5$ and $2x - 3y = 7$
Or, $3x + 2y - 5 = 0$ and $ 2x - 3y - 7 = 0$
Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get
$\frac{a_1}{a_2}=\frac{3}{2},\:\frac{b_1}{b_2}=\frac{2}{-3}\:and\:\frac{c_1}{c_2}=\frac{5}{7}$
It is observed that;
$\frac{a_1}{a_2}\neq\frac{b_1}{b_2}$
It means that the given equations have a unique solution and thus the pair of linear equations is consistent.
Answer:
Given Equations,
$\\2x - 3y = 8$ and $4x - 6y = 9$
Or, $\\2x - 3y - 8 = 0$ and $4x - 6y - 9 = 0$
Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get
$\\\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2},\\\:\frac{b_1}{b_2}=\frac{-3}{-6}=\frac{1}{2}\:and\\\:\frac{c_1}{c_2}=\frac{8}{9}$
It is observed that;
$\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$
It means the given equations have no solution, and thus the pair of linear equations is inconsistent.
Answer:
Given Equations,
$\\\frac{3}{2}x + \frac{5}{3}y = 7$ and $9x -10y = 14$
Or, $\\\frac{3}{2}x + \frac{5}{3}y - 7 = 0$ and $9x -10y - 14 = 0$
Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get
$\\\frac{a_1}{a_2}=\frac{3/2}{9}=\frac{3}{18}=\frac{1}{6},\\\:\frac{b_1}{b_2}=\frac{5/3}{-10}=\frac{5}{-30}=-\frac{1}{6}\:and\\\:\frac{c_1}{c_2}=\frac{7}{14}=\frac{1}{2}$
It is observed that;
$\frac{a_1}{a_2}\neq\frac{b_1}{b_2}$
It means the given equations have exactly one solution, and thus the pair of linear equations is consistent.
Answer:
Given Equations,
$5x - 3y = 11$ and $-10x + 6y =-22$
Or, $5x - 3y - 11 = 0$ and $-10x + 6y + 22 = 0$
Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get
$\\\frac{a_1}{a_2}=\frac{5}{-10}=-\frac{1}{2},\\\:\frac{b_1}{b_2}=\frac{-3}{6}=-\frac{1}{2}\:and\\\:\frac{c_1}{c_2}=\frac{11}{-22}=-\frac{1}{2}$
It is observed that;
$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
It means the given equations have an infinite number of solutions, and thus a pair of linear equations is consistent.
Answer:
Given Equations,
$\\\frac{4}{3}x + 2y = 8$ and $2x + 3y = 12$
Or, $\\\frac{4}{3}x + 2y - 8 = 0$ and $2x + 3y - 12 = 0$
Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get
$\\\frac{a_1}{a_2}=\frac{4/3}{2}=\frac{4}{6}=\frac{2}{3},\\\:\frac{b_1}{b_2}=\frac{2}{3}\:\:and\\\:\frac{c_1}{c_2}=\frac{8}{12}=\frac{2}{3}$
It is observed that;
$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
It means the given equations have an infinite number of solutions, and thus a pair of linear equations is consistent.
Answer:
Given Equations,
$\\x + y = 5$ and $2x + 2 y = 10$
Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get
$\\\frac{a_1}{a_2}=\frac{1}{2},\\\:\frac{b_1}{b_2}=\frac{1}{2}\:and\\\:\frac{c_1}{c_2}=\frac{5}{10}=\frac{1}{2}$
It is observed that;
$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
It means the given equations have an infinite number of solutions, and thus a pair of linear equations is consistent.
The points (x,y) which satisfy both equations are
X | 1 | 3 | 5 |
Y | 4 | 2 | 0 |
Answer:
Given Equations,
$\\x - y = 8$ and $3x - 3y = 16$
Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get
$\\\frac{a_1}{a_2}=\frac{1}{3},\\\:\frac{b_1}{b_2}=\frac{-1}{-3}=\frac{1}{3}\:and\\\:\frac{c_1}{c_2}=\frac{8}{16}=\frac{1}{2}$
It is observed that:
$\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$
It means the given equations have no solution, and thus the pair of linear equations is inconsistent.
Answer:
Given Equations,
$\\2x + y - 6 =0$ and $4x - 2 y - 4 = 0$
Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get
$\\\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2},\\\:\frac{b_1}{b_2}=\frac{1}{-2}=-\frac{1}{2}\:and\\\:\frac{c_1}{c_2}=\frac{-6}{-4}=\frac{3}{2}$
It is observed that;
$\frac{a_1}{a_2}\neq\frac{b_1}{b_2}$
It means the given equations have exactly one solution, and thus the pair of linear equations is consistent.
The points(x, y) satisfying the equation $\\2x + y - 6 =0$ are,
X | 0 | 2 | 3 |
Y | 6 | 2 | 0 |
And The points(x,y) satisfying the equation $\\4x - 2y - 4 = 0$ are,
X | 0 | 1 | 2 |
Y | -2 | 0 | 2 |
GRAPH:
As we can see, both lines intersect at point (2,2) and hence the solution of both equations is x = 2 and y = 2.
Answer:
Given Equations,
$\\2x - 2y - 2 =0, \qquad\\ 4x - 4y -5 = 0$
Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get
$\\\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2},\\\:\frac{b_1}{b_2}=\frac{-2}{-4}=\frac{1}{2}\:and\\\:\frac{c_1}{c_2}=\frac{-2}{-5}=\frac{2}{5}$
It is observed that;
$\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$
It means the given equations have no solution, and thus the pair of linear equations is inconsistent.
Answer:
Let $ l$ be the length of the rectangular garden and $b$ be the width.
Now, according to the question, the length is 4 m more than its width, so we can write it as $l=b+4$
Or, $l-b=4....(1)$
Also given Half Parameter of the rectangle = 36 it means $l+b=36....(2)$
Now, as we have two equations, add both equations, and we get,
$l+b+l-b=4+36$
$\Rightarrow 2l=40$
$\Rightarrow l=20$
We get the value of $l$, which is 20m
Now, putting this in equation (1), we get;
$\Rightarrow 20-b=4$
$\Rightarrow b=20-4$
$\Rightarrow b=16$
Hence, the Length and width of the rectangle are 20m and 16m, respectively.
Answer:
Given the equation,
$2x + 3y -8 =0$
We know that the condition for the intersection of lines for the equations in the form $ a_1x+b_1y+c_1=0$ and $ a_2x+b_2y+c_2=0$ is,
$\frac{a_1}{a_2}\neq\frac{b_1}{b_2}$
So any line with this condition can be $4x+3y-16=0$
Proof,
$\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}$
$\frac{b_1}{b_2}=\frac{3}{3}=1$
Hence, $\frac{1}{2}\neq1$ it means $\frac{a_1}{a_2}\neq\frac{b_1}{b_2}$
Therefore, the pair of lines has a unique solution, thus forming intersecting lines.
Answer:
Given the equation,
$2x + 3y -8 =0$
As we know that the condition for the parallel lines for the equations in the form $ a_1x+b_1y+c_1=0$ and $ a_2x+b_2y+c_2=0$ is,
$\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$
So any line with this condition can be $4x+6y-8=0$
Proof,
$\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}$
$\frac{b_1}{b_2}=\frac{3}{6}=\frac{1}{2}$
$\frac{c_1}{c_2}=\frac{-8}{-8}=1$
Hence, $\frac{1}{2}=\frac{1}{2}\neq1$ it means $\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$
Therefore, the pair of lines has no solutions; thus lines are parallel.
Answer:
Given the equation,
$2x + 3y -8 =0$
As we know that the condition for the coincidence of the lines for the equations in the form $ a_1x+b_1y+c_1=0$ and $ a_2x+b_2y+c_2=0$ is,
$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
So any line with this condition can be $4x+6y-16=0$
Proof,
$\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}$
$\frac{b_1}{b_2}=\frac{3}{6}=\frac{1}{2}$
$\frac{c_1}{c_2}=\frac{-8}{-16}=\frac{1}{2}$
Hence, $\frac{1}{2}=\frac{1}{2}=\frac{1}{2}$ it means $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
Therefore, the pair of lines has infinitely many solutions; thus lines are coincident.
Answer:
Given two equations,
$x - y + 1=0.........(1)$
And
$3x +2 y - 12=0.........(2)$
The points (x,y) satisfying (1) are
X | 0 | 3 | 6 |
Y | 1 | 4 | 7 |
And The points(x,y) satisfying (2) are,
X | 0 | 2 | 4 |
Y | 6 | 3 | 0 |
GRAPH:
From the graph, we can see that both lines intersect at the point (2,3), and therefore the vertices of the Triangle are ( -1,0), (2,3) and (4,0). The area of the triangle is shaded with a green colour.
Also Read-
1. Graphical Representation of Linear Equations: The process includes plotting linear equations on Cartesian graphs by using tables of values to draw their resulting straight lines.
2. Finding Solutions through Intersection Points: The intersection point between two lines indicates all solutions of the system while offering accurate insights into the equations.
3. Understanding Consistency of Systems: Knowing about System Consistency involves examining the graph to identify the number of solutions, along with determining whether the system follows a consistent or inconsistent or dependent pattern.
4. Types of Solutions: Unique solutions when lines intersect, No solutions when lines remain parallel and Infinite solutions occur with coinciding lines.
5. Verification of Solutions: The final verification involves testing whether graphically obtained solutions match both equations through substitution and transforming these results into calculations for accuracy verification.
Also see-
Students must check the NCERT solutions for class 10 of Mathematics and Science Subjects.
Students must check the NCERT Exemplar solutions for class 10 of Mathematics and Science Subjects.
Frequently Asked Questions (FAQs)
The concepts related to linear equation is discussed in ex 3.2 class 10. Practice the problems discussed in this exercise to command the concepts. For this question, because of the term xy is of degree 2, xy - 9 = 3 is not a linear equation in two variables.
To understand the concepts of quadrants go through the problems discussed in class 10 maths ex 3.2. In a graph, there are four quadrants. a point can be represent in in a plain using the (x, y) coordinates.
To get in depth understanding of related concepts practice problems enumerated in the class 10 ex 3.2. as per these concepts, when two lines are in a plane, there are three alternative solutions. They really are.
In Class 10th Maths chapter 3 exercise 3.2, there are seven questions based on the notion of graphical representation of a system of equations.
You can go through the 10th class maths exercise 3.2 answers to get deeper understanding of the concepts related to equations are consistent and dependent. The requirement for linear equations in two variables is:
a1/a2 = b1/b2 = c1/c2
If the equations are consistent and dependent, there are no solutions to linear equations in two variables.
The basic strategy to represent the linear equations on the graph and determine the point of intersection is the graphical method of solution of a pair of linear equations.
On Question asked by student community
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