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NCERT New Syllabus Mathematics
Class: 10
Exercise 3.1
Topic: Pair of Linear Equations in Two Variables
Exploring the Concept of Pair of Linear Equations in Two Variables: NCERT Class 10
Linear equations are everywhere in our daily lives, from calculating expenses to solving practical problems. In this chapter, we dive deeper into understanding how pairs of linear equations work in two variables. Whether finding the meeting point of two straight lines or determining the solution to complex problems, this topic opens doors to various real-world applications.
This chapter focuses on exploring how two linear equations with two variables can be represented graphically and algebraically and how to find their solutions using different methods. We will explore everything from graphical representations to algebraic solutions such as substitution, elimination, and cross-multiplication.
By this topic's end, you'll understand how to interpret, solve, and apply linear equations in two variables effectively!
EXERCISE 3.1
Q1. Form the pair of linear equations in the following problems, and find their
solutions graphically.
(i) 10 students of class X took part in a mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
(ii) 5 pencils and 7 pens together cost Rs 50, whereas 7 pencils and 5 pens together cost Rs 46. Find the cost of one pencil and that of one pen.
Explanation:
1) Let x and y represent the two variables.
2) Based on the given conditions, formulate the corresponding equations.
3) From these conditions, we will obtain two equations, which can be solved simultaneously to find the x and y values.
Solution:
(i) 10 students of class X took part in a mathematics quiz. If the number of
girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
1) Let the number of girls be x and the number of boys is y.
2) Since 10 students from class X participated in the mathematics quiz, we have:
x + y = 10
y = 10 – x ------------ equation 1
3) As the number of girls is 4 more than the number of boys, we have:
x = y + 4
x – y = 4
y = x – 4 ------------ equation 2
4) Now, let’s represent these equations graphically:
a) Take 3 points for y = 10 – x.
b) Take 3 points for y = x – 4.
5) The graphical representation is as follows.
is the number of boys. The lines intersect at the point (7,3), which means there are 7 girls and 3 boys.
(ii) 5 pencils and 7 pens together cost Rs 50, whereas 7 pencils and 5 pens
together cost Rs 46. Find the cost of one pencil and that of one pen.
1) Let the cost of the pencil be Rs x and the cost of the pen be Rs y.
2) Since the cost of 5 pencils and 7 pens is Rs 50, we can form the equation:
5 x + 7 y = 50
7 y = (50 – 5x)
y = (50 – 5x)/7 ------------ equation 1
3) Similarly, since the cost of 7 pencils and 5 pens is Rs 46, the equation becomes:
7 x + 5 y = 46
5 y = (46 – 7x)y = (46 – 7x)/5 ------------ equation 2
4) Now, to represent these equations graphically:
a) Take two points for y = (50 – 5 x)/7.
b) Take two points for y = (46 – 7x)/5.
5) The graphical representation will show the intersection of the two lines, which
gives the values of and y, representing the cost of the pencil and pen, respectively.
represent the cost (in `) of a pencil and of a pen respectively.
The lines intersect at the point (3,5). This means that the cost of a pencil is Rs 3, and the cost of a pen is Rs 5.
Q2. On comparing the ratios a1/a2, b1/b2, and c1/c2, find out whether the lines
representing the following pairs of linear equations intersect at a point, are parallel or coincident:
(i) 5x – 4y + 8 = 0, 7x + 6y – 9 = 0(ii) 9x + 3y + 12 = 0, 18x + 6y + 24 = 0(iii) 6x – 3y + 10 = 0, 2x – y + 9 = 0
Explanation:
1) For the equations, a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
a) If a1/a2 = b1/b2 = c1/c2 then the lines are coincident.b) If a1/a2 = b1/b2 ≠ c1/c2 then the lines are parallel.c) If a1/a2 ≠ b1/b2 then the lines are intersecting.
Solution:
(i) 5x – 4y + 8 = 0, 7x + 6y – 9 = 0
1) Here a1 = 5, a2 = 7, b1 = – 4, b2 = 6, c1 = 8, c2 = – 9.
a1/a2 = 5/7 ------------- 1
b1/b2 = – 4/6 ------------- 2
c1/c2 = 8/(– 9) ------------- 3
2) From 1, 2, and 3, we can say that a1/a2 ≠ b1/b2, so the lines are intersecting.
(ii) 9x + 3y + 12 = 0, 18x + 6y + 24 = 0
1) Here a1 = 9, a2 = 18, b1 = 3, b2 = 6, c1 = 12, c2 = 24.
a1/a2 = 9/18 = 1/2 ------------- 1
b1/b2 = 3/6 = 1/2 ------------- 2
c1/c2 = 12/24 = 1/2 ------------- 3
2) From 1, 2, and 3, we can say that a1/a2 = b1/b2 = c1/c2,
so the lines are coincident.
(iii) 6x – 3y + 10 = 0, 2x – y + 9 = 0
1) Here a1 = 6, a2 = 2, b1 = – 3, b2 = – 1, c1 = 10, c2 = 9.
a1/a2 = 6/2 = 3 ------------- 1
b1/b2 = – 3/(– 1) = 3 ------------- 2
c1/c2 = 10/9 ------------- 3
2) From 1, 2, and 3, we can say that a1/a2 = b1/b2 ≠ c1/c2,
so the lines are parallel.
Q3. On comparing the ratios a1/a2, b1/b2, and c1/c2, find out whether the
following pair of linear equations are consistent, or inconsistent.
(i) 3x + 2y = 5 ; 2x – 3y = 7(ii) 2x – 3y = 8 ; 4x – 6y = 9(iii) (3/2)x + (5/3)y =7; 9x – 10y = 14(iv) 5x – 3y = 11 ; – 10x + 6y = –22(v) (4/3)x + 2y = 8; 2x + 3y = 12
Explanation:
1) For the equations, a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
a) If a1/a2 = b1/b2 = c1/c2, the lines are coincident with infinitely many solutions
so they are consistent.
b) If a1/a2 = b1/b2 ≠ c1/c2, the lines are parallel with no solutions, so they
are inconsistent.
c) If a1/a2 ≠ b1/b2, the lines are intersecting with unique solution, so they
are consistent.
Solution:
(i) 3x + 2y = 5 ; 2x – 3y = 7
1) Write our equations as, 3x + 2y – 5 = 0; 2x – 3y – 7 = 0.
2) Here a1 = 3, a2 = 2, b1 = 2, b2 = – 3, c1 = – 5, c2 = – 7.
a1/a2 = 3/2 ------------- 1
b1/b2 = 2/(– 3) ------------- 2
c1/c2 = (– 5)/(– 7) ------------- 3
3) From 1, 2, 3, we can say that (a1/a2 ≠ b1/b2),
so the lines are intersecting with unique solution, so they are consistent.
(ii) 2x – 3y = 8 ; 4x – 6y = 9
1) Write our equations as, 2x – 3y – 8 = 0; 4x – 6y – 9 = 0.
2) Here a1 = 2, a2 = 4, b1 = – 3, b2 = – 6, c1 = – 8, c2 = – 9.
a1/a2 = 2/4 = 1/2 ------------- 1
b1/b2 = (– 3)/(– 6) = 1/2 ------------- 2
c1/c2 = (– 8)/(– 9) = 8/9 ------------- 3
3) From 1, 2, 3, we can say that (a1/a2 = b1/b2 ≠ c1/c2),
the lines are parallel with no solutions, so they are inconsistent.
(iii) (3/2)x + (5/3)y =7; 9x – 10y = 14
1) Write our equations as, (3/2)x + (5/3)y – 7 = 0; 9x – 10y – 14 = 0.
2) Here a1 = (3/2), a2 = 9, b1 = (5/3), b2 = – 10, c1 = – 7, c2 = – 14.
a1/a2 = (3/2)/(9/1) = 1/6 ------------- 1
b1/b2 = (5/3)/(– 10/1) = (– 1/6) ------------- 2
c1/c2 = (– 7)/(– 14) = 1/2 ------------- 3
3) From 1, 2, 3, we can say that (a1/a2 ≠ b1/b2),
so the lines are intersecting with unique solution, so they are consistent.
(iv) 5x – 3y = 11 ; – 10x + 6y = –22
1) Write our equations as, 5x – 3y – 11 = 0; – 10x + 6y + 22 = 0.
2) Here a1 = 5, a2 = – 10, b1 = – 3, b2 = 6, c1 = – 11, c2 = 22.
a1/a2 = 5/(– 10) = (– 1/2) ------------- 1
b1/b2 = (– 3)/(6) = (– 1/2) ------------- 2
c1/c2 = (– 11)/(22) = (– 1/2) ------------- 3
3) From 1, 2, 3, we can say that (a1/a2 = b1/b2 = c1/c2),
the lines are coincident with infinitly many solutions, so they are consistent.
(v) (4/3)x + 2y = 8; 2x + 3y = 12
1) Write our equations as, (4/3)x + 2y – 8 = 0; 2x + 3y – 12 = 0.
2) Here a1 = (4/3), a2 = 2, b1 = 2, b2 = 3, c1 = – 8, c2 = – 12.
a1/a2 = (4/3)/(2/1) = 2/3 ------------- 1
b1/b2 = 2/3 ------------- 2
c1/c2 = (– 8)/(– 12) = 2/3 ------------- 3
3) From 1, 2, 3, we can say that (a1/a2 = b1/b2 = c1/c2),
the lines are coincident with infinitly many solutions, so they are consistent.
Q4. Which of the following pairs of linear equations are
consistent/inconsistent? If consistent, obtain the solution graphically:
(i) x + y = 5, 2x + 2y = 10(ii) x – y = 8, 3x – 3y = 16(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0
Explanation:
1) For the equations, a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
a) If a1/a2 = b1/b2 = c1/c2, the lines are coincident with infinitly many solutions
so they are consistent.
b) If a1/a2 = b1/b2 ≠ c1/c2, the lines are parallel with no solutions, so they
are inconsistent.
c) If a1/a2 ≠ b1/b2, the lines are intersecting with unique solution, so they
are consistent.
Solution:
(i) x + y = 5, 2x + 2y = 10
1) Write our equations as, x + y – 5 = 0; 2x + 2y – 10 = 0.
2) Here a1 = 1, a2 = 2, b1 = 1, b2 = 2, c1 = – 5, c2 = – 10.
a1/a2 = 1/2 ------------- 1
b1/b2 = 1/2 ------------- 2
c1/c2 = (– 5)/(– 10) = 1/2 ------------- 3
3) From 1, 2, 3, we can say that (a1/a2 = b1/b2 = c1/c2),
the lines are coincident with infinitly many solutions, so they are consistent.
4) We will simplify our equations x + y = 5 and 2x + 2y = 10,
x + y = 5
so, y = (5 – x) ------------- 4
2x + 2y = 10
x + y = 5
so, y = (5 – x) ------------- 5
5) Now, we will represent these equations graphically.
a) We will take 3 points for y = (5 – x).
b) We will take 3 points for y = (5 – x).
(ii) x – y = 8, 3x – 3y = 16
1) Write our equations as, x – y – 8 = 0; 3x – 3y – 16 = 0.
2) Here a1 = 1, a2 = 3, b1 = – 1, b2 = – 3, c1 = – 8, c2 = – 16.
a1/a2 = 1/3 ------------- 1
b1/b2 = (– 1)/(– 3) = 1/3------------- 2
c1/c2 = (– 8)/(– 16) = 1/2 ------------- 3
3) From 1, 2, 3, we can say that (a1/a2 = b1/b2 ≠ c1/c2),
the lines are parallel with no solutions, so they are inconsistent.
4) We will simplify our equations x – y = 8 and 3x – 3y = 16,
x – y = 8
so, y = (x – 8) ------------- 4
3x – 3y = 16
3(x – y) = 16
x – y = 16/3
so, y = x – (16/3) ------------- 5
5) Now, we will represent these equations graphically.
a) We will take 3 points for y = (x – 8).
b) We will take 3 points for y = x – (16/3).
6) The graphical representation will be as follows.
(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0
1) Write our equations as, 2x + y – 6 = 0; 4x – 2y – 4 = 0.
2) Here a1 = 2, a2 = 4, b1 = 1, b2 = – 2, c1 = – 6, c2 = – 4.
a1/a2 = 2/4 = 1/2 ------------- 1
b1/b2 = 1/(– 2) = – (1/2)------------- 2
c1/c2 = (– 6)/(– 4) = 3/2 ------------- 3
3) From 1, 2, 3, we can say that (a1/a2 ≠ b1/b2),
the lines are intersecting with unique solution, so they are consistent.
4) We will simplify our equations 2x + y – 6 = 0 and 4x – 2y – 4 = 0,
2x + y = 6
so, y = (6 – 2x) ------------- 4
4x – 2y = 4
2(2x – y) = 4
(2x – y) = 2
y = 2x – 2 ------------- 5
5) Now, we will represent these equations graphically.
a) We will take 3 points for y = (6 – 2x).
b) We will take 3 points for y = (2x – 2).
6) The graphical representation will be as follows.
7) The lines are intersecting with a unique solution, so they are consistent.
The solution of these equations is (2, 2).
(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0
1) Write our equations as, 2x – 2y – 2 = 0; 4x – 4y – 5 = 0.
2) Here a1 = 2, a2 = 4, b1 = – 2, b2 = – 4, c1 = – 2, c2 = – 5.
a1/a2 = 2/4 = 1/2 ------------- 1
b1/b2 = (– 2)/(– 4) = 1/2------------- 2
c1/c2 = (– 2)/(– 5) = 2/5 ------------- 3
3) From 1, 2, 3, we can say that (a1/a2 = b1/b2 ≠ c1/c2),
the lines are parallel with no solutions, so they are inconsistent.
4) We will simplify our equations 2x – 2y – 2 = 0 and 4x – 4y – 5 = 0,
2x – 2y = 2
so, y = (x – 1) ------------- 4
4x – 4y = 5
4(x – y) = 5
(x – y) = 5/4
y = x – (5/4) ------------- 5
5) Now, we will represent these equations graphically.
a) We will take 3 points for y = (x – 1).
b) We will take 3 points for y = (x – (5/4)).
6) The graphical representation will be as follows.
7) The lines are parallel with no solutions, so they are inconsistent.
Q5. Half the perimeter of a rectangular garden, whose length is 4 m more than
its width, is 36 m. Find the dimensions of the garden.
Explanation:
1) Let the length and width of the rectangular garden be x and y respectively.
2) Using the given conditions, we can form two equations.
3) Solve these two equations to find the values of x and y.
Solution:
1) Let the length and width of the rectangle be x and y respectively.
2) Since the length is 4 meters more than the width, we can write:
x = y + 4
y = x – 4 ------------ equation 1
3) The perimeter of the rectangle is 2(x + y),
so half of perimeter is (x + y).
4) Given that half the perimeter is 36 meters, we have:
x + y = 36
y = 36 – x ------------ equation 2
5) Now, we will represent these equations graphically.
a) We will plot 3 points for y = (x – 4).
b) We will plot 3 points for y = (36 – x).
6) The graphical representation will be as follows.
7) The lines are intersecting with a unique solution, so they are consistent.
8) Here, the dimensions of the rectangular garden are as follows.
a) length is 20 m.
b) width is 16 m.
Q6. Given the linear equation 2x + 3y – 8 = 0, write another linear equation in
two variables such that the geometrical representation of the pair so formed is:
(i) intersecting lines (ii) parallel lines (iii) coincident lines
Explanation:
1) For the equations, a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
a) If a1/a2 = b1/b2 = c1/c2, the lines are coincident with infinitly many solutions
so they are consistent.
b) If a1/a2 = b1/b2 ≠ c1/c2, the lines are parallel with no solutions, so they
are inconsistent.
c) If a1/a2 ≠ b1/b2, the lines are intersecting with unique solution, so they
are consistent.
Solution:
(i) Intersecting lines
1) Consider the two equations of lines:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0,
a) If a1/a2 ≠ b1/b2, the lines are intersecting and have a unique solution.
2) Let's take the first equation as a1x + b1y + c1 = 0, and the second equation as
2x + 3y – 8 = 0.
3) Here a1 = a1, a2 = 2, b1 = b1, b2 = 3, c1 = c1, c2 = – 8.
a1/a2 = a1/2 ------------- 1
b1/b2 = b1/3------------- 2
c1/c2 = c1/(– 8) ------------- 3
4) Since the lines are intersecting, we have a1/a2 ≠ b1/b2,
so, we have a1/2 ≠ b1/3. Let's choose values for a1, b1 and c1 which satisfies, a1/2 ≠ b1/3
let a1 = 3, b1 = (– 2), and c1= 14.
5) Therefore, our first equation becomes: 3x – 2y + 14 = 0, and
the second equation is: 2x + 3y – 8 = 0.
6) Now, let's simplify both equations: 2x + 3y – 8 = 0 and 3x – 2y + 14 = 0,
2x + 3y = 8
so, 3y = (8 – 2x)
y = (8 – 2x)/3 ------------- 4
3x – 2y = – 14
2y = (3x + 14)
y = (3x + 14)/2 ------------- 5
7) Graphical Representation:
a) We will take 3 points for y = (8 – 2x)/3.
b) We will take 3 points for y = (3x + 14)/2.
8) The graphical representation will be as follows.
The point of intersection is (– 2, 4).
(ii) parallel lines
1) For the equations, a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
a) If a1/a2 = b1/b2 ≠ c1/c2, the lines are parallel with no solutions, so they
are inconsistent.
2) Let our equation be a1x + b1y + c1 = 0, and the given equation is 2x + 3y – 8 = 0.
3) So, here a1 = a1, a2 = 2, b1 = b1, b2 = 3, c1 = c1, c2 = – 8.
a1/a2 = a1/2 ------------- 1
b1/b2 = b1/3------------- 2
c1/c2 = c1/(– 8) ------------- 3
4) As our lines are parallel, we have a1/a2 = b1/b2 ≠ c1/c2,
so, we have a1/2 = b1/3. Let us take any values for a1, b1 and c1 which satisfies, a1/2 = b1/3
let a1 = 4, b1 = 6, and c1= 9.
5) So, our equation will be 4x + 6y + 9 = 0, the given equation is 2x + 3y – 8 = 0.
6) We will simplify our equations 2x + 3y – 8 = 0 and 4x + 6y + 10 = 0,
2x + 3y = 8
so, 3y = (8 – 2x)
y = (8 – 2x)/3 ------------- 4
4x + 6y = 10
6y = (10 – 4x)
y = (10 – 4x)/6 ------------- 5
7) Now, we will represent these equations graphically.
a) We will take 3 points for y = (8 – 2x)/3.
b) We will take 3 points for y = (10 – 4x)/6.
(iii) coincident lines
1) For the equations, a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
a) If a1/a2 = b1/b2 = c1/c2, the lines are coincident with infinitly many solutions.
2) Let our equation be a1x + b1y + c1 = 0, and the given equation is 2x + 3y – 8 = 0.
3) So, here a1 = a1, a2 = 2, b1 = b1, b2 = 3, c1 = c1, c2 = – 8.
a1/a2 = a1/2 ------------- 1
b1/b2 = b1/3------------- 2
c1/c2 = c1/(– 8) ------------- 3
4) As our lines are parallel, we have a1/a2 = b1/b2 = c1/c2,
so, we have a1/2 = b1/3. Let us take any values for a1, b1 and c1 which satisfies, a1/2 = b1/3
let a1 = 4, b1 = 6, and c1= –16.
5) So, our equation will be 4x + 6y – 16 = 0, the given equation is 2x + 3y – 8 = 0.
6) We will simplify our equations 2x + 3y – 8 = 0 and 4x + 6y – 16 = 0,
2x + 3y = 8
so, 3y = (8 – 2x)
y = (8 – 2x)/3 ------------- 4
4x + 6y = 16
6y = (16 – 4x)
y = (16 – 4x)/6 ------------- 5
7) Now, we will represent these equations graphically.
a) We will take 3 points for y = (8 – 2x)/3.
b) We will take 3 points for y = (16 – 4x)/6.
9) Another linear equation: 4x + 6y – 16 = 0. The lines are coincident.
Q7. Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0.
Determine the coordinates of the vertices of the triangle formed by these lines and the x–axis, and shade the triangular region.
Explanation:
1) For the equations, a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
a) If a1/a2 ≠ b1/b2, the lines are intersecting with unique solution, so they
are consistent.
Solution:
1) Write our equations as, x – y + 1 = 0; 3x + 2y – 12 = 0.
2) Here a1 = 1, a2 = 3, b1 = – 1, b2 = 2, c1 = 1, c2 = – 12.
a1/a2 = 1/3 ------------- 1
b1/b2 = (– 1)/2 = – (1/2)------------- 2
c1/c2 = 1/(– 12) = – (1/12) ------------- 3
3) From 1, 2, 3, we can say that (a1/a2 ≠ b1/b2),
the lines are intersecting with unique solution, so they are consistent.
4) We will simplify our equations x – y + 1 = 0 and 3x + 2y – 12 = 0,
x – y + 1 = 0
so, y = x + 1 ------------- 4
3x + 2y = 12
2y = 12 – 3x
y = (12 – 3x)/2 ------------- 5
5) Now, we will represent these equations graphically.
a) We will take 3 points for y = x + 1.
b) We will take 3 points for y = (12 – 3x)/2.
7) The lines are intersecting with unique solutions, so they are consistent.
8) Here, the coordinates of the vertices of the triangle formed by these lines and the
x-axis, and shade the triangular region are A(– 1, 0), B (2, 3), and C(4, 0).
Conclusion: Mastering Pair of Linear Equations in Two Variables
As we reach the end of this fascinating topic on Pair of Linear Equations in Two Variables, it’s clear how these equations form the foundation for solving real-world problems involving relationships between two variables. By learning how to graph, calculate, and interpret their solutions, you’re honing your mathematical skills and building your problem-solving mindset. Remember, mastering this concept will open doors to more advanced mathematical techniques in the future. So keep practicing, explore different substitution, elimination, and graphical representation methods, and enjoy the learning journey!
Stay curious and keep solving!
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