Pair of Linear Equations in two Variables Exercise 3.2
Page No: 49
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.
Answer
Let number of boys = x
Number of girls = y
Given that total number of student is 10 so that
x + y = 10
Subtract y both side we get
x = 10 – y
Putting y = 0 , 5, 10 we get
x = 10 – 0 = 10
x = 10 – 5 = 5
x = 10 – 10 = 0
x 105y 05
Given that If the number of girls is 4 more than the number of boys
So that
y = x + 4
Putting x = -4, 0, 4, and we get
y = - 4 + 4 = 0
y = 0 + 4 = 4
y = 4 + 4 = 8
x -404y 048
Graphical representation
Therefore, number of boys = 3 and number of girls = 7.
(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
Let cost of pencil = Rs x
Cost of pens = Rs y
5 pencils and 7 pens together cost Rs 50,
So we get
5x + 7y = 50
Subtracting 7y both sides we get
5x = 50 – 7y
Dividing by 5 we get
x = 10 - 7 y /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 3-4-11y 51015
Given that 7 pencils and 5 pens together cost Rs 46
7x + 5y = 46
Subtracting 7x both side we get
5y = 46 – 7x
Dividing by 5 we get
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 (4) = 9.2 – 5.6 = 3.6
x 024y 9.26.43.6
Graphical representation
Therefore, cost of one pencil = Rs 3 and cost of one pen = Rs 5.
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.
Answer
(i) 5x – 4y + 8 = 0
7x + 6y – 9 = 0
Comparing these equation with
a1x + b1y + c1 = 0
a2x + b2y + 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 are intersecting lines at one point.
(ii) 9x + 3y + 12 = 0
18x + 6y + 24 = 0
Comparing these equations with
a1x + b1y + c1 = 0
a2x + b2y + 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 lines are coincident
(iii) 6x – 3y + 10 = 0
2x – y + 9 = 0
Comparing these equations with
a1x + b1y + c1 = 0
a2x + b2y + 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
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 are intersecting each other at one point and thus 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 and
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 and
c1/c2 = 7/14 = 1/2
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.
(iv) 5x – 3y = 11 ; – 10x + 6y = –22
a1/a2 = 5/-10 = -1/2
b1/b2 = -3/6 = -1/2 and
c1/c2 = 11/-22 = -1/2
Hence, a1/a2 = b1/b2 = c1/c2
Therefore, these linear equations are coincident pair of lines and thus have infinite number of possible solutions. Hence, the pair of linear equations is consistent.
(v) 4/3x + 2y =8 ; 2x + 3y = 12
a1/a2 = 4/3/2 = 2/3
b1/b2 = /3 and
c1/c2 = 8/12 = 2/3
Hence, a1/a2 = b1/b2 = c1/c2
Therefore, these linear equations are coincident pair of lines and thus have infinite number of possible solutions. Hence, the pair of linear equations is consistent.
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 pair of lines and thus have infinite number of possible solutions. Hence, the pair of linear equations is consistent.
x + y = 5
x = 5 - y
x 432y 123
And, 2x + 2y = 10
x = 10-2y/2
x 432y 123
Graphical representation
From the figure, it can be observed that these lines are overlapping each other. Therefore, infinite 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 012y 642
And, 4x - 2y -4 = 0
y = 4x - 4/2
x 123y 024
Graphical representation
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.
Page No: 50
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 rectangle = x m
Width of the rectangle = y m
According to the question,
y - x = 4 ... (i)
y + x = 36 ... (ii)
y - x = 4
y = x + 4
x 0812y 41216
y + x = 36
x 03616y 36020
Graphical representation
From the figure, it can be observed that these lines are intersecting each other at only point i.e., (16, 20). Therefore, the length and width of the given garden is 20 m and 16 m respectively.
6. Given the linear equation 2x + 3y - 8 = 0, write another linear equations 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:
For this 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
For this 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
For coincident lines,
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
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
x - y + 1 = 0
x = y - 1
x 012y 123
3x + 2y - 12 = 0
x = 12 - 2y/3
x 420y 036
Graphical representation
From the figure, it can be observed that these lines are intersecting each other at point (2, 3) and x-axis at ( - 1, 0) and (4, 0). Therefore, the vertices of the triangle are (2, 3), ( - 1, 0), and (4, 0).
