GRAPHICAL METHOD OF SOLUTION OF A PAIR OF LINEAR EQUATIONS:
As we are showing two equations, there will be two lines on the graph.
CASE 1: If the two lines intersect each other at one particular point then that point will be the only solution of that pair of Linear Equations. It is said to be a consistent pair of equations.
CASE 2: If the two lines coincide with each other, then there will be infinite solutions as all the points on the line will be the solution for the pair of Linear Equations. It is said to be dependent or consistent pair of equations.
CASE 3: If the two lines are parallel then there will be no solution as the lines are not intersecting at any point. It is said to be an inconsistent pair of equations.
Representing linear equations for a word problem:
To represent a word problem as a linear equation
Identify unknown quantities and denote them by variables.
Represent the relationships between quantities in a mathematical form,
replacing the unknowns with variables.
EXERCISE 3.2
1. Form the pair of linear equations of 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.
Solution: (i) Let number of boys who took part in the quiz = x
Let number of girls who took part in the quiz = y
According to given conditions, we have x + y =10… (1)
We plot the points for both of the equations to find the solution.
We can clearly see that the intersection point of two lines is (3, 7).
Therefore, number of boys who took park in the quiz = 3
and, number of girls who took part in the quiz = 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.
Solution : Let cost of one pencil = Rs x and Let cost of one pen = Rs y
According to given conditions, we have
5 x + 7 y = 50 … (1)
7 x + 5 y = 46 … (2)
For equation 5 x + 7 y = 50, we have
The point of intersection of two lines is (3, 5).
Therefore, cost of pencil = Rs 3 and, cost of pen = Rs 5.
( i ) 3 x + 2 y = 5 , 2 x − 3 y = 8
( ii ) 2 x − 3 y = 7 , 4x −6 y = 9
( iii ) ( 3 / 2 ) x + ( 5 / 3 ) y = 7 , 9 x − 10 y = 14
( iv ) 5 x − 3 y = 11 , − 10 x + 6 y = −22
( i ) Given : 3 x + 2 y = 5 or 3 x + 2 y – 5 = 0
and 2 x – 3 y = 7 or 2 x –3 y – 7 = 0
Solution : Comparing these equations with a1 x + b1 y + c1 = 0
And a 2 x + b 2 y + c 2 = 0. We get,
a1 = 3, b1 = 2, c1 =– 5
a2 = 2, b2 = – 3, c2 = –7
( a1 / a2 ) = 3 / 2
( b1 / b2 ) = 2 /–3
( c1 / c2 ) = – 5 /–7 = 5 / 7
Since, ( a1 / a2 ) ≠ ( b1 / b2 )
So, the given equations intersect each other at one point and they have only one possible solution. The equations are consistent.
( ii ) Given 2 x – 3 y = 8 and 4 x – 6 y = 9
2 x –3 y –8 = 0 and 4 x – 6 y – 9 = 0
Solution : a 1 = 2, b 1 = – 3, c 1 = – 8
a 2 = 4, b 2 = – 6, c 2 = – 9
( a1 / a2 ) = 2 / 4 = 1 / 2
( b1 / b2 ) = – 3 /– 6 = 1 / 2
( c1 / c2 ) = – 8 /– 9 = 8 / 9
Since , ( a1 / a2 ) = ( b1 / b2 ) ≠ ( c 1 / c2 )
So, the equations are parallel to each other and they have no possible solution. Hence, the equations are inconsistent.
( iii ) Given ( 3 / 2 ) x + ( 5 / 3 ) y = 7 and 9 x – 10 y = 14
3/2 x + 5/3y – 7 =0 and 9 x – 10 y – 14 =0
Solution : a1= 3 / 2, b1= 5 / 3, c1 = –7 , a2 = 9, b2 = –10, c2 = –14
( a1 / a2 ) = 3 / ( 2 × 9 ) = 1 /6
( b1 / b2 ) = 5 / ( 3 ×– 10 ) = – 1 /6
( c1 / c2 ) = – 7 /– 14 = 1 /2
Since, ( a1 / a2 ) ≠ ( b1 / b2 )
So, the equations are intersecting each other at one point and they have only one possible solution. Hence, the equations are consistent.
( iv ) Given, 5 x – 3 y = 11 and –10 x + 6 y = – 22
5 x – 3 y – 11 = 0 and – 10 x + 6 y + 22 =0
Solution : a1 = 5 b1 = –3, c1 = –11 , a2 = –10, b2 = 6, c2 = 22
( a 1 / a 2 ) = 5 / (–10 ) = –5 / 10 = –1 / 2
( b 1 / b 2 ) = –3 / 6 = –1 / 2
( c 1 / c 2 ) = –11 / 22 = –1 / 2
Since ( a 1 / a 2 ) = ( b 1 / b 2 ) = ( c 1 / c 2 )
These linear equations are coincident lines and have infinite number of possible solutions. Hence, the equations are consistent.
( v ) Given, ( 4 / 3 ) x + 2 y = 8 and 2 x + 3y = 12
( 4 / 3 ) x + 2 y – 8 = 0 and 2 x + 3y – 12 = 0
Solution : a1 = 4 / 3 , b1 = 2 , c1 = –8 , a2 = 2, b2 = 3 , c2 = – 12
( a1 / a2 ) = 4 / ( 3 × 2 ) = 4 / 6 = 2 / 3
( b1 / b2 ) = 2 / 3
( c1 / c2 ) = – 8 / –12 = 2 / 3
Since ( a1 / a2 ) = ( b1 / b2 ) = ( c1 / c2 )
These linear equations are coincident lines and have infinite number of possible solutions. Hence, the equations are consistent.
4. Which of the following pairs of linear equations are consistent / inconsistent. If consistent, obtain the solution graphically:
( i ) x + y = 5, 2 x + 2 y = 10
( ii ) x – y = 8, 3 x − 3 y = 16
( iii ) 2 x + y = 6, 4 x − 2 y = 4
( iv ) 2 x − 2 y – 2 = 0, 4 x − 4 y – 5 = 0
Solution : ( i ) x + y = 5,
2 x + 2 y = 10
Their graph lines are coincident, as shown
x + y = 5
Since, the graphs are coinciding
∴ they have infinitely many solutions.
We plot both equations on the graph, the two lines intersect at ( 2, 2 )
Therefore x = 2 ; y = 2
( iv ) 2 x – 2 y – 2 = 0,
4 x – 4 y – 5 = 0
on comparing with ax + by + c + 0
Therefore, these linear equations are parallel to each other and thus, have no possible solution.
Hence, the pair of linear equations is inconsistent.
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.
Solution: Perimeter = 2 ( l + b )
Half perimeter is 36 m = ( l + b )
let breadth = x and length = 4x
Perimeter = 2 × 36 = 72 m
Perimeter = 2 ( l + b )
72 = 2 ( 4 + x + x )
⇒ 36 = 4 + 2 x
36 – 4 = 2 x
⇒ 32 = 2 x
∴ x = 16
∴ breadth = 16
length = x + 4 = 16 + 4 = 20 m.
6. Given the linear equation 2 x + 3 y – 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
Solution: 2 x + 3 y – 8 = 0 ………………. ( 1 )
∴ Any line intersecting with equation (i) may be taken as 3 x + 2 y – 9 = 0 or 3 x + 2 y – 7 = 0
∴ Any line parallel with equation (1) may be taken as 6x + 9 y + 7 = 0 or 2 x + 3 y – 12 = 0
∴Any line coincident with equation ( 1 ) may be taken as
4 x + 6 y – 16 = 0 or 6 x + 9 y – 24 = 0
7. Draw the graphs of the equations x – y + 1 = 0 and 3 x + 2 y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x axis, and shade the triangular region.
Solution :
x – y + 1 = 0 ——– ( 1 )
If we put x = 0, then y = 1
If we put y = 0, then x = – 1
So, we can draw a line with points ( 0, 1 ) and ( – 1, 0 ).
3 x + 2 y – 12 = 0 ——– ( 2 )
If we put x = 0, then y = 6
If we put y = 0, then x = 4
So, we can draw a line with points ( 0,6 ) and ( 4,0 ).
Now, from the graph, we can see the coordinates of the vertices of triangles are,
(– 1, 0 ), ( 2, 3 ), ( 4, 0 ).
Additional Questions:
1.Form the pair of linear equations in the following problems, and find their solutions graphically.
( i )12 students of Class XII took part in a Science 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 ) 15 Notebooks and 7 Textbooks together cost Rs 550, whereas 7 Notebooks and 5 Textbooks together cost ` 400. Find the cost of one Notebook and that of one Textbook.
2. On comparing the ratios 
and find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:
( i ) 5 x + 4 y – 60 = 0 : 6 x + 3 y – 18 = 0
( ii ) 8 x – 3 y – 28 = 0 : 4 x – 2 y – 36 = 0
3. On comparing the ratios and , find out whether the following pair of linear equations are consistent, or inconsistent.
( i ) 4 x + 3 y = 15, 5 x − 3 y = 1 8
( ii ) 6 x − 5 y = 17, 7 x − 5 y = 19
4. Which of the following pairs of linear equations are consistent / inconsistent? If consistent, obtain the solution graphically:
( i ) 2 x + 3 y = 15, 3 x + 4 y = 16
( ii ) 4 x – 2 y = 1 8, 9 x − 6 y = 26
5. Half the perimeter of a rectangular garden, whose length is 5 m more than its width, is 40 m. Find the dimensions of the garden.
6.Draw the graphs of the equations 2 x – 3 y + 5 = 0 and 3 x + 2 y – 16 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x – axis, and shade the triangular region.