Class 10 - Mathematics
Linear Equations In Two Variables - Exercise 3.2
Top Block 1
Exercise 3.2
Question : 1: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.
Answer :
(i) Let the number of boys be = x
Let the number of girls be = y
Given that total number of students is 10
Therefore, x + y = 10
⇒ x = 10 – y
Putting y = 0, 5, 10 we get
x = 10 – 0 = 10
x = 10 – 5 = 5
x = 10 – 10 = 0
|
x |
10 |
5 |
0 |
|
y |
0 |
5 |
10 |
Given: the number of girls is 4 more than the number of boys
⇒ y = x + 4
Putting x = – 4, 0, 4, and we get
y = – 4 + 4 = 0
y = 0 + 4 = 4
y = 4 + 4 = 8
|
x |
-4 |
0 |
4 |
|
y |
0 |
4 |
8 |
Graphical representation is given below:
Number of boys = 3 and number of girls = 7
(ii) Let the cost of one pencil = Rs x
Let the cost of one pen = Rs y
5 pencils and 7 pens together cost = Rs 50
⇒ 5x + 7y = 50
⇒ 5x = 50 – 7y
⇒ x = (10 – 7y) /5
Putting value of y = 5, 10 and 15 we get
x = 10 – 7 * 5/5 = 10 – 7 = 3
x = 10 – 7 * 10/5 = 10 – 14 = – 4
x = 10 – 7 * 15/5 = 10 – 21 = – 11
|
x |
3 |
-4 |
11 |
|
y |
5 |
10 |
15 |
Given: 7 pencils and 5 pens together cost Rs 46
⇒ 7x + 5y = 46
⇒ 5y = 46 – 7x
⇒ y = 46/5 – 7x/5
⇒ y = 9.2 – 1.4x
Putting x = 0, 2 and 4 we get
y = 9.2 – 1.4 * 0 = 9.2 – 0 = 9.2
y = 9.2 – 1.4 * 2 = 9.2 – 2.8 = 6.4
y = 9.2 – 1.4 * 2 = 9.2 – 5.6 = 3.6
|
x |
0 |
2 |
4 |
|
y |
9.2 |
6.4 |
3.6 |
Graphical representation is given below
From the graph, we get
Cost of one pencil = Rs 3 and cost of one pen = Rs 5.
Mddle block 1
Question : 2: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 (ii) 9x + 3y + 12 = 0 (iii) 6x – 3y + 10 = 0
7x + 6y – 9 = 0 18x + 6y + 24 = 0 2x – y + 9 = 0
Answer :
(i) 5x – 4y + 8 = 0
7x + 6y – 9 = 0
On comparing these equation with
a1 x + b1 y + c1 = 0
a2 x + b2 y + c2 = 0
We get
a1 = 5, b1 = – 4, and c1 = 8
a2 =7, b2 = 6 and c2 = – 9
a1/a2 = 5/7, b1/b2 = – 4/6 and c1/c2 = 8/(-9)
Hence, a1/a2 ≠ b1/b2
Therefore, both the lines intersect at one point.
(ii) 9x + 3y + 12 = 0
18x + 6y + 24 = 0
Comparing these equations with
a1 x + b1 y + c1 = 0
a2 x + b2 y + c2 = 0
We get
a1 = 9, b1 = 3, and c1 = 12
a2 =18, b2 = 6 and c2 = 24
a1/a2 = 9/18 = 1/2
b1/b2 = 3/6 = 1/2
and c1/c2 = 12/24 = 1/2
Hence, a1/a2 = b1/b2 = c1/c2
Therefore, both the lines are coincident
(iii) 6x – 3y + 10 = 0
2x – y + 9 = 0
Comparing these equations with
a1 x + b1 y + c1 = 0
a2 x + b2 y + c2 = 0
We get
a1 = 6, b1 = -3, and c1 = 10
a2 =2, b2 = -1 and c2 = 9
a1/a2 = 6/2 = 3/1
b1/b2 = -3/(-1) = 3/1
and c1/c2 = 12/24 = 1/2
Hence, a1/a2 = b1/b2 ≠ c1/c2
Therefore, both lines are parallel.
Question : 3: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) 3/2x + 5/3y = 7; 9x – 10y = 14
(iv) 5x – 3y = 11; -10x + 6y = –22
(v) 4/3x + 2y =8; 2x + 3y = 12
Answer :
(i) 3x + 2y = 5; 2x – 3y = 7
a1 /a2 = 3/2
b1 /b2 = -2/3 and
c1 /c2 = 5/7
Hence, a1 /a2 ≠ b1 /b2
These linear equations intersect each other at one point and therefore have only one possible
solution. Hence, the pair of linear equations is consistent.
(ii) 2x – 3y = 8; 4x – 6y = 9
a1 /a2 = 2/4 = 1/2
b1 /b2 = -3/(-6) = 1/2
c1 /c2 = 8/9
Hence, a1 /a2 = b1 /b2 ≠ c1 /c2
Therefore, these linear equations are parallel to each other and thus have no possible
solution. Hence, the pair of linear equations is inconsistent.
(iii) 3/2x + 5/3y = 7; 9x – 10y = 14
a1 /a2 = (3/2)/9 = 1/6
b1 /b2 = (5/3)/(-10) = – 1/6
c1 /c2 = 7/14 = 1/2
Hence, a1 /a2 ≠ b1 /b2
Therefore, these linear equations intersect each other at one point and thus have only one
possible solution. Hence, the pair of linear equations is consistent.
(iv) 5x – 3y = 11; – 10x + 6y = –22
a1 /a2 = 5/–10 = – 1/2
b1 /b2 = – 3/6 = – 1/2
c1 /c2 = 11/–22 = – 1/2
Hence, a1 /a2 = b1 /b2 = c1 /c2
Therefore, these linear equations are coincident and have infinite number of possible
solutions. Therefore, the pair of linear equations is consistent.
(v) 4/3x + 2y =8; 2x + 3y = 12
a1 /a2 = 4/3/2 = 2/3
b1 /b2 = 2 /3 and
c1 /c2 = 8/12 = 2/3
Hence, a1 /a2 = b1 /b2 = c1 /c2
Therefore, these linear equations are coincident and have infinite number of possible
solutions. Therefore, the pair of linear equations is consistent.
Question : 4: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
Answer :
(i) x + y = 5; 2x + 2y = 10
a1 /a2 = 1/2
b1 /b2 = 1/2 and
c1 /c2 = 5/10 = 1/2
Hence, a1 /a2 = b1 /b2 = c1 /c2
Therefore, these linear equations are coincident and have infinite number of possible
solutions. Therefore, the pair of linear equations is consistent.
x + y = 5
|
x |
4 |
3 |
2 |
|
y |
1 |
2 |
3 |
x = 5 – y
and, 2x + 2y = 10
x = (10 – 2y)/2
|
x |
4 |
3 |
2 |
|
y |
1 |
2 |
3 |
Graphical representation
infinite number of solutions are possible for the given pair of equations.
(ii) x – y = 8, 3x – 3y = 16
a1 /a2 = 1/3
b1 /b2 = – 1/–3 = 1/3 and
c1 /c2 = 8/16 = 1/2
Hence, a1 /a2 = b1 /b2 ≠ c1 /c2
Therefore, these linear equations are parallel to each other and thus have no possible
solution. Hence, the pair of linear equations is inconsistent.
(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0
a1 /a2 = 2/4 = 1/2
b1 /b2 = – 1/2 and
c1 /c2 = – 6/–4 = 3/2
Hence, a1 /a2 ≠ b1 /b2
Therefore, these linear equations are intersecting each other at one point and thus have only
one possible solution. Hence, the pair of linear equations is consistent.
2x + y – 6 = 0
y = 6 – 2x
|
x |
0 |
1 |
2 |
|
y |
6 |
4 |
2 |
and, 4x – 2y – 4 = 0
y = (4x – 4)/2
|
x |
1 |
2 |
3 |
|
y |
0 |
2 |
4 |
Graphical representation is given below:
From the figure, it can be observed that these lines are intersecting each other at the only one
point i.e., (2,2) which is the solution for the given pair of equations.
(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0
a1 /a2 = 2/4 = 1/2
b1 /b2 = – 2/–4 = 1/2 and
c1 /c2 = 2/5
Hence, a1 /a2 = b1 /b2 ≠ c1 /c2
Therefore, these linear equations are parallel to each other and thus, have no possible
solution. Hence, the pair of linear equations is inconsistent.
Question : 5: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.
Answer :
Let length of the rectangle be = x m
Let Width of the rectangle be = y m
According to the question,
y – x = 4 … (i)
y + x = 36 … (ii)
y – x = 4
y = x + 4
|
x |
0 |
8 |
12 |
|
y |
4 |
12 |
16 |
y + x = 36
|
x |
0 |
36 |
16 |
|
y |
36 |
0 |
20 |
Graphical representation is given below:
(16, 20). Therefore, the length and width of the given garden is 20 m and 16 m respectively.
Question : 6: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
Answer :
(i) Intersecting lines:
Condition,
a1 /a2 ≠ b1 /b2
The second line such that it is intersecting the given line is
2x + 4y – 6 = 0 as
a1 /a2 = 2/2 = 1
b1 /b2 = 3/4 and
a1 /a2 ≠ b1 /b2
(ii) Parallel lines
Condition,
a1 /a2 = b1 /b2 ≠ c1 /c2
Hence, the second line can be
4x + 6y – 8 = 0 as
a1 /a2 = 2/4 = 1/2
b1 /b2 = 3/6 = 1/2 and
c1 /c2 = – 8/–8 = 1
and a1 /a2 = b1 /b2 ≠ c1 /c2
(iii) Coincident lines
Condition,
a1 /a2 = b1 /b2 = c1 /c2
Hence, the second line can be
6x + 9y – 24 = 0 as
a1 /a2 = 2/6 = 1/3
b1 /b2 = 3/9 = 1/3 and
c1 /c2 = – 8/–24 = 1/3
and a1 /a2 = b1 /b2 = c1 /c2
Question : 7: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.
Answer :
From equation 1, we get
x = y – 1
|
x |
0 |
1 |
2 |
|
y |
1 |
2 |
3 |
From equation 2, we get
x = (12 – 2y)/3
|
x |
4 |
2 |
0 |
|
y |
0 |
3 |
6 |
The coordinate of the vertices of the triangle formed by these lines and the x-axis are:
(-1, 0), (4, 0) and (2, 3).
