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
Since ,
Therefore, these linear equations are parallel to each other and thus have no possible solution. Hence, the pair of linear equations is inconsistent.
(iii)
Since ,
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 − 3 y = 11
− 10x + 6y = − 22
Since ,
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)
Since
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.
Question 4:
Which of the following pairs of linear equations are consistent/ inconsistent? If consistent, obtain the solution graphically:
Answer:
(i)x + y = 5
2x + 2y = 10
Since ,
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 |
4 |
3 |
2 |
y |
1 |
2 |
3 |
And, 2x + 2y = 10
x |
4 |
3 |
2 |
y |
1 |
2 |
3 |
Hence, the graphic representation is as follows.
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
Since ,
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
Since ,
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
x |
1 |
2 |
3 |
y |
0 |
2 |
4 |
Hence, the graphic representation is as follows.
From the figure, it can be observed that these lines are intersecting each other at the only point i.e., (2, 2) and it is the solution for the given pair of equations.
(iv)2x − 2y − 2 = 0
4x − 4y − 5 = 0
Since ,
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 the width of the garden be x and length be y.
According to the Question,
y − x = 4 (1)
y + x = 36 (2)
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 |
Hence, the graphic representation is as follows.
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.
Question 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,
The second line such that it is intersecting the given line is .
(ii) Parallel lines:
For this condition,
Hence, the second line can be
4x + 6y − 8 = 0
(iii)Coincident lines:
For coincident lines,
Hence, the second line can be
6x + 9y − 24 = 0
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:
x − y + 1 = 0
x = y − 1
x |
0 |
1 |
2 |
y |
1 |
2 |
3 |
3x + 2y − 12 = 0
x |
4 |
2 |
0 |
y |
0 |
3 |
6 |
Hence, the graphic representation is as follows.
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).