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.
Answer:
(i) Let number of boys = x
Let number of girls = y
According to given conditions, we have
x + y = 10
And, x = 10 – y
putting y=0,5,10,we get,
X=10-0=10
X=10-5=5,
X=10-10=0
Number of girls is 4 more than number of boys ……..
Given, so,
Y=x+4
putting x=-4,0,4 we get,
Y=-4+4=0
Y=0+4
Y=4+4=8
We plot the points for both of the equations to find the solution.
ii.Let the cost of one pencil=Rs.X
and Let the cost of one pen=Rs.Y
According to the given conditions, we have:
=5x + 7y = 50
=5x=50-7y
=x=10-7/5y
Now
7 Pencils and 5 pens together cost Rs. 46
7x+5y =46
5y =46-7x
Y= 9.2 -1.4x
Putting x =0, 2, 4 we get,
2. On comparing the ratios a1/a2,b1/b2and c1/c2, find out whether the 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
Ans:
(i) 5x − 4y + 8 = 0, 7x + 6y – 9 = 0
Comparing equation 5x − 4y + 8 = 0 with a1x + b1y + c1 = 0and 7x + 6y – 9 = 0 with a2x + b2y + c2 = 0,
We get,
= a1 = 5, b1= -4, c1 = 0
= a2 = 7, b2 = 6, c2 = -9
(ii) 9x + 3y + 12 = 0, 18x + 6y + 24 = 0
Comparing equation 9x + 3y + 12 = 0 with a1x + b1y + c1 = 0and 7x + 6y – 9 = 0 with a2x + b2y + c2 = 0,
Hence, lines are coincident.
(iii) 6x − 3y + 10 = 0, 2x – y + 9 = 0
Comparing equation 6x − 3y + 10 = 0 with a1x + b1y + c1 = 0and 7x + 6y – 9 = 0 with a2x + b2y + c2 = 0,
Hence, lines are parallel to each other
3. On comparing the ratios a1/a2,b1/b2and c1/c2, find out whether the following pair of linear equations are consistent, or inconsistent.
(i) 3x + 2y = 5, 2x − 3y = 8 (ii) 2x − 3y = 7, 4x − 6y = 9 (iii) 3x/2 + 5y/3 = 7, 9x − 10y = 14 (iv) 5x − 3y = 11, −10x + 6y = −22
Ans:
(i) 3x + 2y = 5, 2x − 3y = 7
Comparing equation 3x + 2y = 5 with a1x + b1y + c1 = 0and 7x + 6y – 9 = 0 with a2x + b2y + c2 = 0,
Therefore, these linear equations will intersect at one point only and have only one possible solution.
And, pair of linear equations is consistent
(ii) 2x − 3y = 8, 4x − 6y = 9
Comparing equation 2x − 3y = 8 with a1x + b1y + c1 = 0and 7x + 6y – 9 = 0 with a2x + b2y + c2 = 0,
Therefore, these linear equations are parallel to each other and have no possible solution.in
And,pair of linear euations is inconsistent
(iii) 3x/2 + 5y/3 = 7, 9x − 10y = 14
Comparing equation 3x/2 + 5y/3 = 7, with a1x + b1y + c1 = 0and 9x − 10y = 14 with a2x + b2y + c2 = 0,
Therefore, these linear equations will intersect at one point only and have only one possible solution.
And, pair of linear equations is consistent
(iv) 5x − 3y = 11, −10x + 6y = −22
Comparing equation 5x − 3y = 11 with a1x + b1y + c1 = 0and 7x + 6y – 9 = 0 with a2x + b2y + c2 = 0,
Therefore these pair of lines have infinite number of solutions
And, 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, 4x − 2y = 4 (iv) 2x − 2y – 2 = 0, 4x − 4y – 5 = 0
(i) x + y = 5, 2x + 2y = 10
Therefore these pair of lines have infinite number of solutions and pair of linear equation is consistent.
X + Y = 5
X = 5-Y
Putting Y= 1,2,3 we get,
And 2X + 2Y =10
X =10-2Y/2
Putting Y = 1,2,3 we get
(ii) x – y = 8, 3x − 3y = 16
Therefore, these linear equations are parallel to each other and have no possible solution.in
And,pair of linear euations is inconsistent
(iii) 2x + y = 6, 4x − 2y = 4
Therefore, these linear equations will intersect at one point only and have only one possible solution.
And, pair of linear equations is consistent
(iv) 2x − 2y – 2 = 0, 4x − 4y – 5 = 0
Therefore, these linear equations will intersect at one point only and have only one possible solution.
And, pair of linear equations is consistent
(iv) 2x − 2y – 2 = 0, 4x − 4y – 5 = 0
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 graphically.
Let the length of the garden be x m and its breadth be y m
Then according to the question,
We have:
x = y + 4
x –y =4……..(i)
And x + y =6…….(ii)
For graphical representation:
From equation (i) x – y = 4
We get
From equation (ii) x + y = 36
We get
Plotting the points of each table of solutions, we obtain the graph of two lines interesting at (20, 16).
Therefore, the solution is x= 20 and y= 16
Thus the length of the garden is 20m and its breadth is 16m.
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
(i) Given the linear equation 2x + 3y – 8 = 0.
To find another linear equation in two variables such that the geometrical representation of the pair so formed is intersecting lines, it should satisfy below condition;
(a1/a2) ≠ (b1/b2)
Thus, another equation could be 2x – 7y + 9 = 0, such that;
(a1/a2) = 2/2 = 1 and (b1/b2) = 3/-7
Clearly, you can see another equation satisfies the condition.
(ii) Given the linear equation 2x + 3y – 8 = 0.
To find another linear equation in two variables such that the geometrical representation of the pair so formed is parallel lines, it should satisfy below condition;
(a1/a2) = (b1/b2) ≠ (c1/c2)
Thus, another equation could be 6x + 9y + 9 = 0, such that;
(a1/a2) = 2/6 = 1/3
(b1/b2) = 3/9= 1/3
(c1/c2) = -8/9
Clearly, you can see another equation satisfies the condition.
(iii) Given the linear equation 2x + 3y – 8 = 0.
To find another linear equation in two variables such that the geometrical representation of the pair so formed is coincident lines, it should satisfy below condition;
(a1/a2) = (b1/b2) = (c1/c2)
Thus, another equation could be 4x + 6y – 16 = 0, such that;
(a1/a2) = 2/4 = 1/2 ,(b1/b2) = 3/6 = 1/2, (c1/c2) = -8/-16 = 1/2
Clearly, you can see another equation satisfies the condition.
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.
Ans: Given, the equations for graphs are x – y + 1 = 0 and 3x + 2y – 12 = 0.
For, x – y + 1 = 0 or x = -1+y
For, 3x + 2y – 12 = 0 or x = (12-2y)/3
