Exercise - 3.2
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.
(i) Let the no. of girls be x and no. of boys be y.
The algebraic representation is given by the following equations:
x + y = 10 ......(i)
x – y = 4 ......(ii)
Lets represent these equations in graphically:
x + y = 10 .......(i)
x – y = 4 ......(ii)
Plot the points A(0, 10), B(10, 0) and P(0, -4), Q(4, 0), corresponding to the solutions in Table.
Now draw the lines AB and PQ, representing the equations x + y = 10 and x – y = 4, as shown in following Figure:
Now, Observe that the two lines representing the two equations are intersecting at the point (7, 3).
Therefore, the no. of girls and no. of boys in the class are 7 and 3 respectively.
So, it is consistent.
(ii) Let the cost of pencil be x and cost of pen be y.
The algebraic representation is given by the following equations:
5x + 7y = 50 ......(i)
7x + 4y = 46 ......(ii)
Lets represent these equations in graphically:
5x + 7y = 50 ......(i)
7x + 4y = 46 ......(ii)
Plot the points A(3, 5), B(10, 0) and P(3, 5), Q(8, -2), corresponding to the solutions in Table.
Now draw the lines AB and PQ, representing the equations 5x + 7y = 50 and 7x + 4y = 46, as shown in following Figure:
.png)
Now, Observe that the two lines representing the two equations are intersecting at the point (3, 5).
Therefore, the cost of one pencil and cost of one pen are 3 and 5 respectively.
So, it is consistent.
On comparing the ratios a₁/a₂, b₁/b₂ and c₁/c₂ find out whether 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
(i) 5x – 4y + 8 = 0; 7x + 6y – 9 = 0
a₁ = 5, b₁ = -4, c₁ = 8
a₂ = 7, b₂ = 6, c₂ = -9
Here, a₁/a₂ = 5/7,
Also, b₁/b₂ = -4/6 and
c₁/c₂ = 8/(-9)
Therefore, a₁/a₂ ≠ b₁/b₂ ≠ c₁/c₂
Now, these linear equations are intersecting each other at one point and its have only one solution.
(ii) 9x + 3y + 12 = 0; 18x + 6y + 24 = 0
a₁ = 9, b₁ = 3, c₁ = 12
a₂ = 18, b₂ = 6, c₂ = 24
Here, a₁/a₂ = 9/18 = 1/2,
Also, b₁/b₂ = 3/6 = 1/2 and
c₁/c₂ = 12/24 = 1/2
Therefore, a₁/a₂ = b₁/b₂ = c₁/c₂
Now, these linear equations are coincident and its have many solution.
(iii) 6x – 3y + 10 = 0; 2x – y + 9 = 0
a₁ = 6, b₁ = -3, c₁ = 10
a₂ = 2, b₂ = -1, c₂ = 9
Here, a₁/a₂ = 6/2 = 3,
Also, b₁/b₂ = -3/(-1) = 3 and
c₁/c₂ = 10/9
Therefore, a₁/a₂ = b₁/b₂ ≠ c₁/c₂
Now, these linear equations are parallel each other and its have no solution.
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) 3x/2 + 5y/3 = 7; 9x - 10y = 14
(iv) 5x – 3y = 11 ; – 10x + 6y = –22
(v) 4x/3 + 2y = 8; 2x + 3y = 12
(i) 3x + 2y = 5 ; 2x – 3y = 7
3x + 2y - 5 = 0 ; 2x – 3y - 7 = 0
a₁ = 3, b₁ = 2, c₁ = -5
a₂ = 2, b₂ = -3, c₂ = -7
Here, a₁/a₂ = 3/2,
Also, b₁/b₂ = 2/(-3) and
c₁/c₂ = -5/(-7)= 5/7
Therefore, a₁/a₂ ≠ b₁/b₂ ≠ c₁/c₂
Now, these linear equations are intersecting each other at one point and its have only one solution.
So, It is consitent.
(ii) 2x – 3y = 8 ; 4x – 6y = 9
2x – 3y - 8 = 0 ; 4x – 6y - 9 = 0
a₁ = 2, b₁ = -3, c₁ = -8
a₂ = 4, b₂ = -6, c₂ = -9
Here, a₁/a₂ = 2/4 = 1/2,
Also, b₁/b₂ = -3/(-6) = 1/2 and
c₁/c₂ = -8/(-9) = 8/9
Therefore, a₁/a₂ = b₁/b₂ ≠ c₁/c₂
Now, these linear equations are parallel to each other and its have only no possible solution.
So, It is inconsitent.
(iii) 3x/2 + 5y/3 = 7; 9x - 10y = 14
3x/2 + 5y/3 - 7 = 0; 9x - 10y - 14 = 0
a₁ = 3/2, b₁ = 5/3, c₁ = -7
a₂ = 9, b₂ = -10, c₂ = -14
Here, a₁/a₂ = (3/2)/9 = 1/6,
Also, b₁/b₂ = (5/3)/(-10) = 1/(-6) and
c₁/c₂ = -7/(-14) = 1/2
Therefore, a₁/a₂ ≠ b₁/b₂ ≠ c₁/c₂
Now, these linear equations are intersecting each other at one point and its have only one solution.
So, It is consitent.
(iv) 5x – 3y = 11 ; – 10x + 6y = –22
5x – 3y - 11 = 0 ; – 10x + 6y + 22 = 0
a₁ = 5, b₁ = -3, c₁ = -11
a₂ = -10, b₂ = 6, c₂ = 22
Here, a₁/a₂ = 5/(-10) = 1/(-2),
Also, b₁/b₂ = (-3)/6 = 1/(-2) and
c₁/c₂ = -11/22 = 1/(-2)
Therefore, a₁/a₂ = b₁/b₂ = c₁/c₂
Now, these linear equations are coincident and its have many solution.
So, It is consitent.
(v) 4x/3 + 2y = 8; 2x + 3y = 12
4x/3 + 2y - 8 = 0; 2x + 3y - 12 = 0
a₁ = 4/3, b₁ = 2, c₁ = -8
a₂ = 2, b₂ = 3, c₂ = -12
Here, a₁/a₂ = (4/3)/2 = 2/3,
Also, b₁/b₂ = 2/3 and
c₁/c₂ = -8/(-12) = 2/3
Therefore, a₁/a₂ = b₁/b₂ = c₁/c₂
Now, these linear equations are coincident and its have many solution.
So, It is consitent.
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
(i) x + y = 5, 2x + 2y = 10
x + y - 5 = 0 ; 2x + 2y - 10 = 0
a₁ = 1, b₁ = 1, c₁ = -5
a₂ = 2, b₂ = 2, c₂ = -10
Here, a₁/a₂ = 1/2,
Also, b₁/b₂ = 1/2 and
c₁/c₂ = -5/(-10)= 1/2
Therefore, a₁/a₂ = b₁/b₂ = c₁/c₂
Now, these linear equations are coincident and its have many solution.
So, It is consitent.
Lets represent these equations in graphically:
x + y = 5 ...........(i)
2x + 2y = 10 ...........(ii)
Plot the points A(0, 5), B(5, 0) and P(0, 5), Q(5, 0), corresponding to the solutions in Table.
Now draw the lines AB and PQ, representing the equations x + y = 5 and 2x + 2y = 10, as shown in following Figure:
.png)
(ii) x – y = 8, 3x – 3y = 16
x – y - 8 = 0 ; 3x – 3y - 16 = 0
a₁ = 1, b₁ = -1, c₁ = -8
a₂ = 3, b₂ = -3, c₂ = -16
Here, a₁/a₂ = 1/3,
Also, b₁/b₂ = -1/(-3) = 1/3 and
c₁/c₂ = -8/(-16) = 1/2
Therefore, a₁/a₂ = b₁/b₂ ≠ c₁/c₂
Now, these linear equations are parallel to each other and its have only no possible solution.
So, It is inconsitent.
(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0
a₁ = 2, b₁ = 1, c₁ = -6
a₂ = 4, b₂ = -2, c₂ = -4
Here, a₁/a₂ = 2/4 = 1/2,
Also, b₁/b₂ = 1/(-2) and
c₁/c₂ = -6/(-4) = 3/2
Therefore, a₁/a₂ ≠ b₁/b₂ ≠ c₁/c₂
Now, these linear equations are intersecting each other at one point and its have only one solution.
So, It is consitent.
Lets represent these equations in graphically:
2x + y = 6 ...........(i)
4x – 2y = 4 ...........(ii)
Plot the points A(0, 6), B(3, 0) and P(0, -2), Q(1, 0), corresponding to the solutions in Table.
Now draw the lines AB and PQ, representing the equations 2x + y = 6 and 4x – 2y = 4, as shown in following Figure:
.png)
(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0
a₁ = 2, b₁ = -2, c₁ = -2
a₂ = 4, b₂ = -4, c₂ = -5
Here, a₁/a₂ = 2/4 = 1/2,
Also, b₁/b₂ = (-2)/(-4) = 1/2 and
c₁/c₂ = -2/(-5) = 2/5
Therefore, a₁/a₂ = b₁/b₂ ≠ c₁/c₂
Now, these linear equations are parallel to each other and its have only no possible solution.
So, It is inconsitent.
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.
Let the width of garden and length of garden be araxand y respectively.
According to question linear equations are:
y = x + 4
x - y = -4 ........(i)
y + x = 36
x + y = 36 ........(ii)
Lets represent these equations in graphically:
x - y = -4 ...........(i)
x + y = 36 ...........(ii)
Plot the points A(0, 4), B(-4, 0) and P(0, 36), Q(36, 0), corresponding to the solutions in Table.
Now draw the lines AB and PQ, representing the equations x - y = -4 and x + y = 36, as shown in following Figure:
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
(i) For intersecting lines,
a₁/a₂ ≠ b₁/b₂ ≠ c₁/c₂
Given : 2x + 3y - 8 = 0 ....(i)
According to condition the second line is :
3x + 5y - 15 = 0 ....(ii)
a₁ = 2, b₁ = 3, c₁ = -8
a₂ = 3, b₂ = 5, c₂ = -15
Here, a₁/a₂ = 2/3,
Also, b₁/b₂ = 3/5 and
c₁/c₂ = -8/(-15) = 8/15
Hence, a₁/a₂ ≠ b₁/b₂ ≠ c₁/c₂.
(ii) For parallel lines,
a₁/a₂ = b₁/b₂ ≠ c₁/c₂
Given : 2x + 3y - 8 = 0 ....(i)
According to condition the second line is :
4x + 6y - 18 = 0 ....(ii)
a₁ = 2, b₁ = 3, c₁ = -8
a₂ = 4, b₂ = 6, c₂ = -18
Here, a₁/a₂ = 2/4,
Also, b₁/b₂ = 3/6 and
c₁/c₂ = -8/(-18) = 4/9
Hence, a₁/a₂ = b₁/b₂ ≠ c₁/c₂.
(iii) For coincident lines,
a₁/a₂ = b₁/b₂ = c₁/c₂
Given : 2x + 3y - 8 = 0 ....(i)
According to condition the second line is :
6x + 9y - 24 = 0 ....(ii)
a₁ = 2, b₁ = 3, c₁ = -8
a₂ = 6, b₂ = 9, c₂ = -24
Here, a₁/a₂ = 2/6 = 1/3,
Also, b₁/b₂ = 3/9 = 1/3 and
c₁/c₂ = -8/(-24) = 1/3
Hence, a₁/a₂ = b₁/b₂ = c₁/c₂.
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.
x – y + 1 = 0
x - y = -1 ........(i)
3x + 2y – 12 = 0
3x + 2y = 12 ........(ii)
Lets represent these equations in graphically:
x - y = -1 ........(i)
3x + 2y = 12 ........(ii)
Plot the points A(0, 1), B(-1, 0) and P(0, 6), Q(4, 0), corresponding to the solutions in Table.
Now draw the lines AB and PQ, representing the equations x - y = -1 and 3x + 2y = 12, as shown in following Figure:
