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.
Solution:-
Let the number of girls be x and the number of boys be y.
According to the question, the algebraic representation is
x + y = 10
x – y = 4
For x + y = 10,
For x – y = 4,
From the figure, it can be observed that these lines intersect each other at point (7, 3).
Therefore, the number of girls and boys in the class are 7 and 3 respectively.
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(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.
Let the cost of 1 pencil be Rs x and the cost of 1 pen be Rs y.
"5 pencils and 7 pens together cost Rs 50"
5 pencils(5x) + 7 pens(7y) =50
"7 pencils and 5 pens together cost Rs 46."
7 pencils(7x) + 5 pens(5y) =46
According to the question, the algebraic representation is
5x + 7y = 50
7x + 5y = 46
For 5x + 7y = 50,
7x + 5y = 46,
From the figure, it can be observed that these lines intersect each other at point (3, 5).
Therefore, the cost of a pencil and a pen are Rs 3 and Rs 5 respectively.
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Solution:-
(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 or -2/3(after cutting)
c1 /c2 = 8/-9
5/7 ≠ -2/3
Hence, a1/a2 ≠ b1/b2
Therefore, both are intersecting lines at one point(intersecting lines).
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(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 (after cutting)
b1 /b2 = 3/6 = 1/2 (after cutting)
c1 /c2 = 12/24 = 1/2 (after cutting)
Hence, a1/a2 = b1/b2 =c1/c2
Therefore, both lines are coincident.
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(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 (after cutting)
b1 /b2 = -3/-1 = 3/1 (after cutting)
c1 /c2 =12/24 = 1/2 (after cutting)
Hence, a /a = b /b ≠c /c
Therefore, both lines are parallel.
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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.
Solution:-
i) 3x + 2y = 5 ;
2x – 3y = 7
These linear equations have unique solution.
Hence, the pair of linear equations is consistent.
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(ii) 2x – 3y = 8
4x – 6y = 9
here we have no possible solution.
Hence, the pair of linear equations is inconsistent.
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(iii) 3/2x + 5/3y = 7
9x – 10y = 14
here we have a unique solution.
Hence, the pair of linear equations is consistent.
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(iv) 5x – 3y = 11
–10x + 6y = –22
here we have infinite number of solutions.
Hence, the pair of linear equations is consistent.
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(v) 4/3x + 2y =8 ;
2x + 3y = 12
here we have infinite number of solutions.
Hence, the pair of linear equations is consistent.
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Question-4: which of the following pairs of linear equations are consistent/inconsistent? If consistent ,obtain the solution graphically.
Solution:-
(i) x + y = 5
2x + 2y =10
a1 = 1 , b1 = 1 , c1 = 5
a2 = 2 , b2 = 2 , c2 = 10
So, equations are consistent.
Now, we have to solve it graphycally.
For x + y = 5
For 2x + 2y = 10
Now create a table using these values.
(ii) x - y = 8
3x - 3y =16
a1 = 1 , b1 = -1 , c1 = 8
a2 = 3 , b2 = -3 , c2 = 16
given equations are parallel lines
so there is no solution.
It means equations are inconsistent.
(iii) 2x + y - 6 = 0
4x - 2y - 4 = 0
or
2x + y = 6
4x - 2y = 4
a1 = 2 , b1 = 1 , c1 = 6
a2 = 4 , b2 = -2 , c2 = 4
here we have unique solutions.
So, equations are consistent.
Now, we have to solve it graphycally
For, 2x + y = 6
(iv) 2x - 2y - 2 = 0
4x - 4y - 5 = 0
or
2x - 2y = 2
4x - 4y = 5
a1 = 2 , b1 = -2 , c1 = 2
a2 = 4 , b2 = -4 , c2 = 5
here the given equations are parallel lines.
So, the equations are inconsistent.
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Question-5: Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36m. Find the dimensions of the garden.
Solution:
Let the length of the garden be x.
and the width of the garden be y.
Half the perimeter of a rectangular garden is 36m.
length is 4 m more than its width
length = width + 4
x = y + 4 or
x - y = 4
So, we have two equations:
x + y = 36 ..............(i)
x - y = 4 ...............(ii)
For first equation
x + y = 36
For
x - y = 4
now make tables
Now put all the values on graph
here both lines intersect each other at point (20,16). hence x = 20 and y = 16 is the solution of given linear equations.
Therefore, Length of the garden = 20 and Breadth of the garden = 16.
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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
Solution-:
(i) intersecting lines
first equation is : 2x + 3y - 8 = 0
so if we assume second equation as 3x + 2y + 4 = 0 then it will satisfy the condition
(ii)Parallel lines
so we assume 2x + 3y - 12 = 0 as second equation
(iii)Coincident lines
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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.
Solution-:
For equation x - y + 1 = 0 or x - y = -1
For equation 3x +2y -12 = 0 or 3x + 2y = 12
now we prepare table with these values
put these values on graph
Here lines of equations formed a triangle with the help of x-axis.
So, the coordinates of the triangle ABC are:
A (2,3)
B (-1,0)
C (4,0)
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