# NCERT solution for class 10 maths pair of linear equations in two variables ( Chapter 3) #### Solution for Exercise 3.1

Q1. Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically.

Ans.Let the present age of Aftab and his daughter be x and y respectively.
So, seven years ago,
Aftab's age = $$x - 7$$
Age of Aftab's daughter = $$y-7$$
According to the given statement in the question:
=>$$x-7 = 7(y-7)$$
=>$$x-7 = 7y-49$$
=>$$x-7y = -42$$............(i)
Three solutions of the equation (i) are :
$$\begin{array} {|r|r|}\hline x & -7 & 0 & 7 \\ \hline y & 5 & 6 & 7 \\ \hline \end{array}$$
Now, three years hence,
Aftab's age = $$x+3$$
Age of Aftab's daughter = $$y+3$$
According to the given statement in the question:
=>$$x + 3 = 3(y+3)$$
=>$$x + 3 = 3y + 9$$
=>$$x -3y = 6$$............(ii)
Three solutions of the equation (ii) are :
$$\begin{array} {|r|r|}\hline x & 6 & 3 & 0 \\ \hline y & 0 & -1 & -2 \\ \hline \end{array}$$
Now the graph for the above equations (i) and (ii) will be: Q2.The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later, she buys another bat and 3 more balls of the same kind for Rs 1300. Represent this situation algebraically and graphically.

Ans.Let the cost of cricket bat and cricket ball be Rs.x and Rs.y respectively.
According to the given statement in the question:
=>$$3x + 6y = 3900$$
=>$$x + 2y = 1300$$............(i)
Two solutions of the equation (i) are :
$$\begin{array} {|r|r|}\hline 0 & 1300 \\ \hline 650 & 0 \\ \hline \end{array}$$
And also according to the given statement in the question:
=>$$x + 3y = 1300$$............(ii)
Two solutions of the equation (i) are :
$$\begin{array} {|r|r|}\hline 0 & 1300 \\ \hline {{1300} \over {3}} & 0 \\ \hline \end{array}$$
Now the graph for the above equations (i) and (ii) will be: Q3. The cost of 2 kg of apples and 1 kg of grapes on a day was found to be Rs 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs 300. Represent the situation algebraically and geometrically.

Ans.Let the cost of 1 kg of apples and 1 kg of grapes be Rs.x and Rs.y respectively.
According to the given statement in the question:
=>$$2x + y = 160$$............(i)
Two solutions of the equation (i) are :
$$\begin{array} {|r|r|}\hline 50 & 45 \\ \hline 60 & 70 \\ \hline \end{array}$$
And also according to the given statement in the question:
=>$$4x + 2y = 300$$
=>$$2x + y = 150$$............(ii)
Two solutions of the equation (i) are :
$$\begin{array} {|r|r|}\hline 50 & 40 \\ \hline 50 & 70 \\ \hline \end{array}$$
Now the graph for the above equations (i) and (ii) will be: #### Solution for Exercise 3.2

Q1.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.

Ans. (i)Let the number of boys and girls who took part in quiz be x and y respectively.
According to the given statement in the question:
=>$$x + y = 10)$$............(1)
Points which lie on the equation (1) are :
$$\begin{array} {|r|r|}\hline x & 0 & 10 \\ \hline y & 10 & 0 \\ \hline \end{array}$$
According to the given statement in the question:
=>$$x - y = -4)$$............(2)
Solutions of the equation (2) are :
$$\begin{array} {|r|r|}\hline x & 0 & -4 \\ \hline y & 4 & 0 \\ \hline \end{array}$$
Now the graph for the above equations (1) and (2) will be: From the graph it is visible that equation (1) and (2) intersect at $$(3,7)$$
Therefore, number of boys who took part in quiz = 3 and, number of girls who took part in quiz = 7.
(ii)Let the cost of one pencil and cost of one pen be Rs.x and Rs.y respectively.
According to the given statement in the question:

=>$$5x + 7y = 50$$............(1)
Points which lie on the equation (1) are :
$$\begin{array} {|r|r|}\hline x & 10 & 3 \\ \hline y & 0 & 5 \\ \hline \end{array}$$
According to the given statement in the question:
=>$$7x + 5y = 46$$............(2)
Solutions of the equation (2) are :
$$\begin{array} {|r|r|}\hline x & 8 & 3 \\ \hline y & -2 & 5 \\ \hline \end{array}$$
Now the graph for the above equations (1) and (2) will be: From the graph it is visible that equation (1) and (2) intersect at $$(3,5)$$
Therefore, cost of pencil = Rs.3 and, Cost of pen = Rs.5.

Q2.On comparing the ratios $${{a_1}\over {a_2}},{{b_1}\over {b_2}},{{c_1}\over {c_2}}$$ 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) $$5x - 4y + 8 = 0, 7x + 6y – 9 = 0$$
On comparing these equations with the general form: $$ax^2 + bx + c$$
$$a_1 = 5, b_1 = -4, c_1 = 8$$,
$$a_2 = 7, b_2 = 6, c_2 = -9$$
Here $${{a_1}\over {a_2}} \ne {{b_1}\over {b_2}}$$ as,
$${{5}\over {7}} \ne {{-4}\over {6}}$$
So, these lines have a unique solution which means they intersect at one point.

(ii) $$9x + 3y + 12 = 0,18x + 6y + 24 = 0$$
On comparing these equations with the general form: $$ax^2 + bx + c$$

$$a_1 = 9, b_1 = 3, c_1 = 12$$
$$a_2 = 18, b_2 = 6, c_2 = 24$$
Here $${{a_1}\over {a_2}} = {{b_1}\over {b_2}} = {{c_1}\over {c_2}}$$ as,
$${{9}\over {18}} = {{3}\over {6}} = {{12}\over {24}}$$
So, these lines are coincident.

(iii) $$6x - 3y + 10 = 0,2x – y + 9 = 0$$
On comparing these equations with the general form: $$ax^2 + bx + c$$
$$a_1 = 6, b_1 = -3, c_1 = 10$$
$$a_2 = 2, b_2 = -1, c_2 = 9$$
Here $${{a_1}\over {a_2}} = {{b_1}\over {b_2}} \ne {{c_1}\over {c_2}}$$ as,
$${{6}\over {2}} = {{-3}\over {-1}} \ne {{10}\over {9}}$$
So, these lines are parallel to each other.

Q3. On comparing the ratios $${{a_1}\over {a_2}},{{b_1}\over {b_2}} and {{c_1}\over {c_2}}$$, 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)$${{3}\over {2}}x + {{5}\over {3}}y = 7,9x - 10y = 14$$
(iv)$$5x - 3y = 11, -10x + 6y = -22$$
(v)$$\frac{4}{3}x + 2y = 8, 2x + 3y = 12$$

Ans.(i) $$3x + 2y = 5, 2x - 3y = 7$$
On comparing these equations with the general form: $$ax^2 + bx + c$$
$$a_1 = 3, b_1 = 2, c_1 = -5$$
$$a_2 = 2, b_2 = -3, c_2 = -7$$
Here $${{a_1}\over {a_2}} \ne {{b_1}\over {b_2}}$$ as,
$${{3}\over {2}} \ne {{2}\over {3}}$$
So, these lines have a unique solution which means they intersect at one point.
Hence, they are consistent.

(ii) $$2x - 3y = 8, 4x - 6y = 9$$
On comparing these equations with the general form: $$ax^2 + bx + c$$
$$a_1 = 2, b_1 = -3, c_1 = -8$$
$$a_2 = 4, b_2 = 6, c_2 = -9$$
Here $${{a_1}\over {a_2}} = {{b_1}\over {b_2}} \ne {{c_1}\over {c_2}}$$ as,
$${{1}\over {2}} = {{1}\over {2}} \ne {{-8}\over {-9}}$$
So, these lines are parallel to each other.
Hence, they are inconsistent.

(iii) $${{3}\over {2}}x + {{5}\over {3}}y = 7,9x - 10y = 14$$
On comparing these equations with the general form: $$ax^2 + bx + c$$
$$a_1 = \frac{3}{2}, b_1 = \frac{5}{3}, c_1 = 7$$
$$a_2 = 9, b_2 = -10, c_2 = 14$$
Here $${{a_1}\over {a_2}} \ne {{b_1}\over {b_2}}$$ as,
$${{{{3}\over {2}}}\over {9}} \ne {{{{5}\over {3}}}\over {-10}}$$
=>$${{1}\over {6}} \ne {{-1}\over {6}}$$
So, these lines have a unique solution which means they intersect at one point.
Hence, they are consistent.

(iv) $$5x - 3y = 11, -10x + 6y = -22$$
On comparing these equations with the general form: $$ax^2 + bx + c$$
$$a_1 = 5, b_1 = -3, c_1 = 11$$
$$a_2 = -10, b_2 = 6, c_2 = -22$$
Here $${{a_1}\over {a_2}} = {{b_1}\over {b_2}} = {{c_1}\over {c_2}}$$ as,
$${{5}\over {-10}} = {{-3}\over {6}} = {{11}\over {-22}}$$
=>$${{-1}\over {2}} = {{-1}\over {2}} = {{-1}\over {2}}$$
So, these lines have infinite many solutions.
Hence, they are consistent.

(v)$$\frac{4}{3}x + 2y = 8, 2x + 3y = 12$$
On comparing these equations with the general form: $$ax^2 + bx + c$$
$$a_1 = \frac{4}{3}, b_1 = 2, c_1 = 8$$
$$a_2 = 2, b_2 = 3, c_2 = 12$$
Here $${{a_1}\over {a_2}} = {{b_1}\over {b_2}} = {{c_1}\over {c_2}}$$ as,
$${{{{4}\over {3}}}\over {2}} = {{2}\over {3}} = {{8}\over {12}}$$
=>$${{2}\over {3}} = {{2}\over {3}} = {{2}\over {3}}$$
So, these lines have infinite many solutions.
Hence, they are consistent.

Q4.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) $$x + y = 5, 2x + 2y = 10$$
For equation $$x + y -5 = 0$$, points which lie on the line:
$$\begin{array} {|r|r|}\hline x & 0 & 5 \\ \hline y & 5 & 0 \\ \hline \end{array}$$
For equation $$2x + 2y = 10$$, points which lie on the line:
$$\begin{array} {|r|r|}\hline x & 1 & 2 \\ \hline y & 4 & 3 \\ \hline \end{array}$$
Graph for the above equations is: On plotting the graph we can see that both of the lines coincide.
So, there are infinitely many solutions.
Hence, they are consistent.

(ii) $$x – y = 8, 3x - 3y = 16$$
For equation $$x - y - 8 = 0$$, points which lie on the line:
$$\begin{array} {|r|r|}\hline x & 0 & 8 \\ \hline y & -8 & 0 \\ \hline \end{array}$$
For equation $$3x - 3y = 16$$, points which lie on the line:
$$\begin{array} {|r|r|}\hline x & 0 & 16 \\ \hline y & -16 & 0 \\ \hline \end{array}$$
Graph for the above equations is: On plotting the graph we can see that both of the lines are parallel to each other.
So, there are no solutions.
Hence, they are inconsistent.

(iii) $$2x + y = 6, 4x - 2y = 4$$
For equation $$2x + y - 6 = 0$$, points which lie on the line:
$$\begin{array} {|r|r|}\hline x & 0 & 3 \\ \hline y & 6 & 0 \\ \hline \end{array}$$
For equation $$4x – 2y – 4 = 0$$, points which lie on the line:
$$\begin{array} {|r|r|}\hline x & 0 & 1 \\ \hline y & -2 & 0 \\ \hline \end{array}$$
Graph for the above equations is: On plotting the graph we can see that both of the lines are intersecting each other at exactly one point.
So, there is a unique solution.
Hence, they are consistent.

(iv) $$2x - 2y – 2 = 0, 4x - 4y – 5 = 0$$
For equation $$2x - 2y – 2 = 0$$, points which lie on the line:
$$\begin{array} {|r|r|}\hline x & 2 & 0 \\ \hline y & 0 & -2 \\ \hline \end{array}$$
For equation $$4x – 2y – 4 = 0$$, points which lie on the line:
$$\begin{array} {|r|r|}\hline x & 5 & 0 \\ \hline y & 0 & -5 \\ \hline \end{array}$$
Graph for the above equations is: On plotting the graph we can see that both of the lines are parallel to each other.
So, there is no solution.
Hence, they are inconsistent.

Q5. Half the perimeter of a rectangle garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.

Ans.Let the length of rectangular garden and width of rectangular garden be x metres and y metres respectively.
Given, the perimeter = 36m
=>$$x + y = 36$$.......(i)
Also $$x = y + 4$$
=>$$x - y = 4$$........(ii)
On adding equation (i) and (ii),
=>$$2x = 40$$
=>$$x = 20$$m
On subtracting equation (ii) from (i),
=>$$2y = 32$$
=>$$y = 16$$m
Hence , length = 20 m and width = 16 m.

Q6. 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

Ans.(i) Intersecting lines
The condition for intersecting lines is :
$${{a_1}\over {a_2}} \ne {{b_1}\over {b_2}}$$
So the second equation can be $$x + 2y = 3$$
Since, $${{a_1}\over {a_2}} \ne {{b_1}\over {b_2}}$$
=>$${{2}\over {1}} \ne {{3}\over {2}}$$

(ii) Parallel lines
The condition for parallel lines is :
$${{a_1}\over {a_2}} = {{b_1}\over {b_2}} \ne {{c_1}\over {c_2}}$$
So the second equation can be $$2x + 3y – 2 = 0$$
Since, $${{a_1}\over {a_2}} = {{b_1}\over {b_2}} \ne {{c_1}\over {c_2}}$$
=>$${{2}\over {2}} = {{3}\over {3}} \ne {{-8}\over {-2}}$$

(ii) Coincident lines
The condition for coincident lines is :
$${{a_1}\over {a_2}} = {{b_1}\over {b_2}} \ne {{c_1}\over {c_2}}$$
So the second equation can be $$4x + 6y – 16 = 0$$
Since, $${{a_1}\over {a_2}} = {{b_1}\over {b_2}} = {{c_1}\over {c_2}}$$
=>$${{2}\over {4}} = {{3}\over {6}} = {{-8}\over {-16}}$$
=>$${{1}\over {2}} = {{1}\over {2}} = {{1}\over {2}}$$

Q7. 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.For equation $$x – y + 1 = 0$$, points which lie on the line:
$$\begin{array} {|r|r|}\hline x & 0 & 1 \\ \hline y & 1 & 0 \\ \hline \end{array}$$
For equation $$3x + 2y – 12 = 0$$, points which lie on the line:
$$\begin{array} {|r|r|}\hline x & 4 & 0 \\ \hline y & 0 & 6 \\ \hline \end{array}$$
Graph for the above equations is: We can see from the graph that the point of intersection of the lines with each other is (2, 3) and the points of intersection with x - axis are :
(–1, 0) and (4, 0).

Thus, the coordinates of the vertices of the triangle formed are (2, 3), (-1, 0) and (4, 0).

#### Solution for Exercise 3.3

Q1.Solve the following pair of linear equations by the substitution method.
(i) $$x + y = 14,x – y = 4$$
(ii) $$s – t = 3,{{s}\over{3}} + {{t}\over{2}} = 6$$
(iii) $$3x – y = 3,9x - 3y = 9$$
(iv) $$0.2x + 0.3y = 1.3,0.4x + 0.5y = 2.3$$
(v) $$\sqrt{2}x + \sqrt{3}y = 0,\sqrt{3}x - \sqrt{8}y = 0$$
(vi) $${{3x}\over{2}} - {{5y}\over{3}} = -2,{{x}\over{3}} + {{y}\over{2}} = {{13}\over{6}}$$

Ans.(i)$$x + y = 14$$...........(i)
=> $$x – y = 4$$..............(ii)
Using equation (ii)
=> $$x = 4 + y$$.............(iii)
On substituting (iii) in (i):
=>$$4 + 2y = 14$$
=>$$2y = 10$$
=>$$y = 5$$...........(iv)
On substituting (iv) in (i):
=>$$x + 5 = 14$$
=>$$x = 14 - 5$$
=>$$x = 9$$
So, the value of $$y = 5$$ and $$x = 9$$.

(ii)$$s – t = 3$$...........(i)
=> $${{s}\over{3}} + {{t}\over{2}} = 6$$..............(ii)
Using equation (i)
=> $$s = 3 + t$$.............(iii)
On substituting (iii) in (ii):
=>$${{3 + t}\over{3}} + {{t}\over{2}} = 6$$
=>$${{6 + 2t + 3t}\over{6}} = 6$$
=>$$5t + 6 = 36$$
=>$$5t = 30$$
=>$$t = 6$$..............(iv)
On substituting (iv) in (i):
=>$$s – 6 = 3$$
=>$$s = 6 + 3$$
=>$$s = 9$$
So, the value of $$t = 6$$ and $$s = 9$$.

(iii) $$3x – y = 3,9x - 3y = 9$$
=> $$3x - y = 3$$..........(i)
=> $$9x - 3y = 9$$........(ii)
Using equation (i),
=> $$y = 3x - 3$$........(iii)
On substituting (iii) in (ii):
=> $$9x - 9x + 9 = 9$$
=> 9 = 9
This is always true.
Hence the given pair of equations have infinitely many solutions. The relation between them is given by :
$$y = 3x - 3$$
One such solution can be x = 1 and y = 0.

(iv)$$0.2x + 0.3y = 1.3$$...........(i)
=> $$0.4x + 0.5y = 2.3$$..............(ii)
Using equation (i)
=> $$0.2x = 1.3 - 0.3y$$
=>$$x = {{1.3 - 0.3y} \over {0.2}}$$.............(iii)
On substituting (iii) in (ii):
=>$$0.4({{1.3 - 0.3y }\over {0.2}}) + 0.5y = 2.3$$
=>$$2.6 - 0.6y + 0.5y = 2.3$$
=>$$-0.1y = -0.3$$
=>$$y = 3$$.............(iv)
On substituting (iv) in (i):
=>$$0.2x + 0.3 (3) = 1.3$$
=>$$0.2x + 0.9 = 1.3$$
=>$$0.2x = 0.4 => x = 2$$
So, the value of $$y = 3$$ and $$x = 2$$.

(v)$$\sqrt{2}x + \sqrt{3}y = 0$$...........(i)
=> $$\sqrt{3}x - \sqrt{8}y = 0$$..............(ii)
Using equation (i)
=> $$x = y\sqrt{{-3}\over {2}}$$.............(iii)
On substituting (iii) in (ii):
=>$$\sqrt{3}(y\sqrt{{-3}\over {2}}) - \sqrt{8}y =0$$
=>$${{-3y}\over {\sqrt 2}} - \sqrt{8}y =0$$
=>$$y({{-3}\over {\sqrt 2}} - \sqrt{8}) =0$$
=>$$y = 0$$.............(iv)
On substituting (iv) in (iii):
=>$$x = 0$$
So, the value of $$y = 0$$ and $$x = 0$$.

(vi)$${{3x}\over{2}} - {{5y}\over{3}} = -2$$...........(i)
=> $${{x}\over{3}} + {{y}\over{2}} = {{13}\over{6}}$$..............(ii)
Using equation (ii)
=> $$x = ({{13}\over{6}} - {{y}\over{2}}) × 3$$
=> $$x = {{13}\over{2}} - {{3y}\over{2}}$$.............(iii)
On substituting (iii) in (i):
=>$${3({{13}\over{2}} - {{3y}\over{2}})\over{2}} - {{5y}\over{3}} = -2$$
=>$${{39}\over{4}} - {{9y}\over{4}} - {{5y}\over{3}} = -2$$
=>$${{-27y -20y}\over{12}} = -2 - {{39}\over{4}}$$
=>$${{-47y}\over{12}} = {{-8-39}\over{4}}$$
=>$${{-47y}\over{12}} = {{-47}\over{4}}$$
=>$$y = 3$$
On substituting (iv) in (ii):
=>$${{x}\over{3}} + {{3}\over{2}} = {{13}\over{6}}$$
=>$${{x}\over{3}} = {{13}\over{6}} - {{3}\over{2}}$$
=>$${{x}\over{3}} = {{2}\over{3}}$$
=>$$x = 2$$
So, the value of $$y = 3$$ and $$x = 2$$.

Q2. Solve $$2x + 3y = 11$$ and $$2x - 4y = -24$$ and hence find the value of ‘m’ for which $$y = mx + 3$$.

Ans. $$2x + 3y = 11$$...........(i)
=> $$2x - 4y = -24$$..............(ii)
Using equation (ii)
=> $$2x = -24 + 4y$$
=> $$x = -12 + 2y$$.............(iii)
On substituting (iii) in (i):
=>$$2(-12 + 2y) + 3y = 11$$
=>$$-24 + 4y + 3y = 11$$
=>$$7y = 35$$
=>$$y = 5$$...........(iv)
On substituting (iv) in (i):
=>$$2x + 3 (5) = 11$$
=>$$2x = 11 – 15 = -4$$
=>$$x = -2$$
So, the value of $$y = 5$$ and $$x = -2$$.
On substituting the values of x and y in $$y = mx + 3$$,
=> $$5 = m (-2) + 3$$
=> $$5 = -2m + 3$$
=> $$-2m = 2$$
$$=> m = -1$$

Q3.Form a pair of linear equations for the following problems and find their solution by substitution method.
(i) The difference between two numbers is 26 and one number is three times the other. Find them.
(ii)The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
(iii)The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.
(iv)The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
(v) A fraction becomes $${{9}\over{11}}$$ , if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and denominator it becomes $${{5}\over{6}}$$. Find the fraction.
(vi)Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

Ans (i)Let first number and second number be x and y respectively.
According to given question
$$x – y = 26$$ (assuming x > y) … (1)
$$x = 3y$$(Since,x > y)… (2)
On putting equation (2) in (1),
$$3y – y = 26$$
=>$$2y = 26$$
=>$$y = 13$$
On putting value of y in equation (2),
=>$$x = 3y = 3(13) = 39$$
Therefore, two numbers are 13 and 39.

(ii) Let smaller angle and larger angle be x and y respectively.
According to given question
$$y = x + 18$$ … (1)
$$x + y = 180$$(Sum of supplementary angle)… (2)
On putting equation (1) in (2),
$$x + x + 18 = 180$$
=>$$2x = 180 - 18 = 162$$
=>$$x = 81^\circ$$
On putting value of x in equation (1),
=>$$y = x + 18 = 81 + 18 = 99^\circ$$
Therefore, two numbers are $$81^\circ$$ and $$99^\circ$$

(iii) Let cost of one bat and cost of one ball be x and y respectively.
According to given question
$$7x + 6y = 3800$$ .....(1)
$$3x + 5y = 1750$$ .....(2)
Using equation (1):
$$7x = 3800 - 6y$$
=>$$x = {{3800 - 6y} \over{7}}$$ .....(3)
On putting equation (3) in (1),
$$3({{3800 - 6y} \over{7}}) + 5y = 1750$$
=>$${{11400 - 18y} \over{7}} + 5y = 1750$$
=>$${{5y} \over{1}} - {{18y} \over{7}} = {{1750} \over{1}} -{{11400 } \over{7}}$$
=>$${{35y - 18y} \over{7}} = {{12250 - 11400} \over{7}}$$
=>$$17y = 850$$
=>$$y = 50$$
On putting value of y in equation (2),
=>$$3x + 250 = 1750$$
=>$$3x = 1500$$
=>$$x = 500$$
Therefore, cost of one bat and cost of one ball is Rs.500 and Rs.50 respectively.

(iv) Let fixed charge and charge for every km be Rs.x and Rs.y respectively.
According to given question
$$x + 10y = 105$$ … (1)
$$x + 15y = 155$$ … (2)
Using equation (1)
$$x = 105 - 10y$$ … (3)
On putting equation (3) in (2),
$$105 - 10y + 15y = 155$$
=>$$5y = 50$$
=>$$y = 10$$
On putting value of y in equation (1),
=>$$x + 10 (10) = 105$$
=>$$x = 105 – 100 = 5$$
Therefore, fixed charge and charge for every km is Rs.5 and Rs.10 respectively.
To travel distance of 25 Km, person will have to pay

=> Rs $$x + 25y$$
=> Rs $$5 + 25 × 10$$
=> Rs $$5 + 250$$ = Rs 255

(v) Let numerator and denominator be x and y respectively.
According to given question
$${{x + 2} \over{y + 2}} = {{9} \over{11}}$$ .........(1)
$${{x + 3} \over{y + 3}} = {{5} \over{6}}$$ .........(2)

Using equation (2):
$$6(x + 3) = 5(y + 3)$$.......(4)
Using equation (1):
$$11(x + 2) = 9(y + 2)$$
=>$$11x + 22 = 9y + 18$$
=>$$11x = 9y - 4$$
=>$$x = {{9y - 4} \over{11}}$$........(3)
On putting equation (3) in (4),
$$6({{9y - 4} \over{11}} + 3) = 5(y + 3)$$
=>$${{54y} \over{11}} - {{24} \over{11}} + 18 = 5y + 15$$
=>$$-{{24} \over{11}} + {{33} \over{11}} = {{55y} \over{11}} - {{54y} \over{11}}$$
=>$$-{{24 + 33} \over{11}} = {{55y - 54y} \over{11}}$$
=>$$y = 9$$
On putting value of y in equation (1),
=>$${{x + 2} \over{9 + 2}} = {{9} \over{11}}$$
=>$$x + 2 = 9$$
=>$$x = 7$$
Therefore, numerator and denominator is 7 and 9 respectively.
Fraction = $${{7} \over{9}}$$

(vi) Let present age of Jacob and his son be x and y respectively.
According to given question
$$x + 5 = 3(y + 5)$$ .....(1)
$$x - 5 = 7(y - 5)$$ .....(2)
Using equation (1):
$$x + 5 = 3y + 15$$
=>$$x = 10 + 3y$$ .....(3)
On putting equation (3) in (2),
$$10 + 3y – 5 = 7y - 35$$
=>$$-4y = -40$$
=>$$y = 10$$
On putting value of y in equation (3),
=>$$x = 10 + 3(10)$$
=>$$x = 40$$
Therefore, present age of Jacob and his son be 40 years and 10 years respectively.

#### Solution for Exercise 3.4

Q1.Solve the following pair of linear equations by the elimination method and the substitution method:
(i) $$x + y = 5,2x – 3y = 4$$
(ii) $$3x + 4y = 10, 2x – 2y = 2$$
(iii) $$3x - 5y – 4 = 0, 9x = 2y + 7$$
(iv) $${{x}\over{2}} + {{2y}\over{3}} = -1, x - {{y}\over{3}} = 3$$

Ans.(i) $$x + y = 5$$ … (1)
$$2x – 3y = 4$$ … (2)
Elimination method:
On multiplying equation (1) by 2, we get equation (3)
$$2x + 2y = 10$$ … (3)
$$2x - 3y = 4$$ … (2)
On subtracting equation (2) from (3), we get
$$5y = 6$$
=> $$y = {{6}\over{5}}$$
On putting value of y in (1), we get
$$x + {{6}\over{5}} = 5$$
$$x = 5 - {{6}\over{5}} = {{19}\over{5}}$$
Therefore, x = $${{6}\over{5}}$$ and y = $${{19}\over{5}}$$
Substitution method:
$$x + y = 5$$ … (1)
$$2x - 3y = 4$$ … (2)
From equation (1), we get,
$$x = 5 - y$$
On putting this in equation (2), we get
$$2 (5 - y) - 3y = 4$$
=>$$10 - 2y - 3y = 4$$
=>$$5y = 6$$
=>$$y = {{6}\over{5}}$$
On putting value of y in (1), we get
$$x = 5 - {{6}\over{5}} = {{19}\over{5}}$$
Therefore, y = $${{6}\over{5}}$$ and x = $${{19}\over{5}}$$

(ii) $$3x + 4y = 10$$ … (1)
$$2x – 2y = 2$$ … (2)
Elimination method:
On multiplying equation (2) by 2, we get equation (3)
$$4x - 4y = 4$$ … (3)
$$3x + 4y = 10$$ … (1)
On adding equation (1) and (3), we get
$$7x = 14 => x = 2$$
On putting value of x in (1), we get
$$3 (2) + 4y = 10$$
$$4y = 10 – 6 = 4$$
$$=> y = 1$$
Therefore, x = $$2$$ and y = $$1$$
Substitution method:
$$3x + 4y = 10$$ … (1)
$$2x - 2y = 2$$ … (2)
From equation (2), we get,
$$2x = 2 + 2y$$
=> $$x = 1 + y$$..........(3)
On putting this in equation (1), we get
$$3 (1 + y) + 4y = 10$$
=>$$3 + 3y + 4y = 10$$
=>$$7y = 7 => y = 1$$
On putting value of y in (3), we get
$$x = 1 + 1 = 2$$
Therefore, x = $$2$$ and y = $$1$$

(iii) $$3x - 5y – 4 = 0$$ … (1)
$$9x = 2y + 7$$… (2)
Elimination method:
Multiplying (1) by 3, we get (3)
$$9x - 15y – 12 = 0$$… (3)
$$9x - 2y – 7 = 0$$… (2)
On subtracting (2) from (3), we get
$$-13y – 5 = 0$$
=>$$-13y = 5$$
=> $$y = {{-5} \over {13}}$$
On putting value of y in (1), we get
$$3x – 5({{-5} \over {13}}) - 4 = 0$$
=>$$3x = 4 - {{25} \over {13}}= {{52 - 25} \over {13}} = {{27} \over {13}}$$
=> $$x = {{27} \over {13(3)}} = {{9} \over {13}}$$
Therefore, x = $${{9} \over {13}}$$ and y = $${{-5} \over {13}}$$
Substitution Method:
$$3x - 5y – 4 = 0$$ … (1)
$$9x = 2y + 7$$… (2)
From equation (1), we can say that
$$3x = 4 + 5y => x = {{4 + 5y} \over {3}}$$
On putting this in equation (2), we get
$$9({{4 + 5y} \over {3}}) - 2y = 7$$
=>$$12 + 15y - 2y = 7$$
=>$$13y = -5$$
$$y = {{-5} \over {13}}$$
On putting value of y in (1), we get
$$3x – 5 ({{-5} \over {13}})= 4$$
=>$$3x = 4 - {{25} \over {13}} = {{52 - 25} \over {13}} = {{27} \over {13}}$$
=>$$x = {{27} \over {13(3)}} = {{9} \over {13}}$$
Therefore, x = $${{9} \over {13}}$$ and y = $${{-5} \over {13}}$$

(iv)$${{x}\over{2}} + {{2y}\over{3}} = -1$$ … (1)
$$x - {{y}\over{3}} = 3$$… (2)
Elimination method:
On multiplying equation (2) by 2, we get (3)
$$2x - {{2y}\over{3}} = 6$$… (3)
$${{x}\over{2}} + {{2y}\over{3}} = -1$$… (1)
Adding (3) and (1), we get
$${{5x}\over{2}} = 5 => x = 2$$
On putting value of x in (2), we get
$$2 - {{y}\over{3}} = 3$$
=>$$y = -3$$
Therefore, x = 2 and y = -3
Substitution method:
$${{x}\over{2}} + {{2y}\over{3}} = -1$$ … (1)
$$x - {{y}\over{3}} = 3$$… (2)
From equation (2), we can say that
=> $$x = 3 + {{y}\over{3}} = {{9 + y}\over{3}}$$
Putting this in equation (1), we get
=>$${{9 + y}\over{6}} + {{2y}\over{3}} = -1$$
=>$${{9 + y + 4y}\over{6}} = -1$$
=>$$5y + 9 = -6$$
=>$$5y = -15$$
=>$$y = -3$$
Putting value of y in (1), we get
=>$${{x}\over{2}} + {{2(-3)}\over{3}} = -1$$
=>$$x = 2$$
So, the value of $$y = -3$$ and $$x = 2$$.

Q2. Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:
(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes ½ if we only add 1 to the denominator. What is the fraction?
(ii)Five years ago, Nuri was thrice as old as sonu. Ten years later, Nuri will be twice as old as sonu. How old are Nuri and Sonu?
(iii)The sum of the digits of a two–digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
(iv) Meena went to a bank to withdraw Rs 2000. She asked the cashier togive her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how many notes of Rs 50 and Rs 100 she received.
(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs 27 for a book kept for seven days, while Susy paid Rs 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Ans.(i)Let numerator =x and let denominator =y
According to given condition,
=>$${{x + 1}\over{y - 1}} = 1$$ and $${{x}\over{y + 1}} = {{1}\over{2}}$$
=>$$x + 1 = y – 1$$ and $$2x = y + 1$$
=>$$x – y = -2$$….......(1) and,
$$2x – y = 1$$…...... (2)
So, we have equations (1) and (2), multiplying equation (1) by 2 we get (3)
$$2x - 2y = -4$$… (3)
$$2x – y = 1$$… (2)
Subtracting equation (2) from (3), we get
$$-y = -5 => y = 5$$
Putting value of y in (1), we get
$$x – 5 = -2 => x = -2 + 5 = 3$$
Therefore, fraction = $${{x}\over{y}} = {{3}\over{5}}$$

(ii) Let present age of Nuri = x years and let present age of Sonu = y years
5 years ago, age of Nuri = $$(x – 5)$$ years
5 years ago, age of Sonu = $$(y – 5)$$ years
According to given condition,
$$(x - 5) = 3 (y - 5)$$
=> $$x – 5 = 3y – 15$$
=> $$x - 3y = -10$$… (1)
10 years later from present, age of Nuri = $$(x + 10)$$ years
10 years later from present, age of Sonu = $$(y + 10)$$ years
According to given condition,
$$(x + 10) = 2 (y + 10)$$
=> $$x + 10 = 2y + 20$$
=> $$x - 2y = 10 … (2)$$
Subtracting equation (1) from (2),
$$y = 10 - (-10) = 20$$years
Putting value of y in (1),
$$x – 3 (20) = -10$$
=> $$x – 60 = -10$$
=> $$x = 50$$years
Therefore, present age of Nuri = 50 years and present age of Sonu = 20 years

(iii) Let digit at ten’s place = x and Let digit at one’s place = y
According to given condition,
$$x + y = 9$$… (1)
And 9 (10x + y) = 2 (10y + x)
=>$$90x + 9y = 20y + 2x$$
=>$$88x = 11y$$
=>$$8x = y$$
=>$$8x – y = 0$$… (2)
$$9x = 9 => x = 1$$
Putting value of x in (1),
$$1 + y = 9$$
=>$$y = 9 – 1 = 8$$
Therefore, number = $$10x + y = 10 (1) + 8 = 10 + 8 = 18$$

(iv)Let number of Rs 100 notes = x and let number of Rs 50 notes = y
According to given conditions,
$$x + y = 25$$ … (1)
and $$100x + 50y = 2000$$
=>$$2x + y = 40$$ … (2)
Subtracting (2) from (1),
$$-x = -15 => x = 15$$
Putting value of x in (1),
$$15 + y = 25$$
=>$$y = 25 – 15 = 10$$
Therefore, number of Rs 100 notes = 15 and number of Rs 50 notes = 10

(v) Let fixed charge for 3 days = Rs x
Let additional charge for each day thereafter = Rs y
According to given condition,
$$x + 4y = 27$$… (1)
$$x + 2y = 21$$… (2)
Subtracting (2) from (1),
$$2y = 6 => y = 3$$
Putting value of y in (1),
$$x + 4 (3) = 27$$
=>$$x = 27 – 12 = 15$$
Therefore, fixed charge for 3 days = Rs 15 and additional charge for each day after 3 days = Rs 3

#### Solution for Exercise 3.5

Q1. Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions? In case there is a unique solution, find it by using cross multiplication method.
(i) $$x – 3y – 3 = 0,3x – 9y – 2 = 0$$
(ii) $$2x + y = 5,3x + 2y = 8$$
(ii) $$3x - 5y = 20,6x - 10y = 40$$
(ii) $$x - 3y – 7 = 0,3x - 3y – 15 = 0$$

Ans.(i) $$x – 3y – 3 = 0,3x – 9y – 2 = 0$$
On comparing these equations with the general form : $$ax^2 + bx + c$$
$$a_1 = 1, b_1 = -3, c_1 = -3,a_2 = 3, b_2 = -9, c_2 = -2$$
Here $${{a_1}\over {a_2}} = {{b_1}\over {b_2}} \ne {{c_1}\over {c_2}}$$
=>$${{1}\over {3}} = {{-3}\over {-9}} \ne {{-3}\over {-2}}$$
So, these lines are parallel to each other.
Thus, the following pairs of linear equations will have no solution.

(ii) $$2x + y = 5,3x + 2y = 8$$
On comparing these equations with the general form : $$ax^2 + bx + c$$
$$a_1 = 2, b_1 = 1, c_1 = -5,a_2 = 3, b_2 = 2, c_2 = 8$$
Here $${{a_1}\over {a_2}} \ne {{b_1}\over {b_2}}$$
=>$${{2}\over {3}} \ne {{1}\over {2}}$$
So, these lines have a unique solution.
We can find the unique solution using cross multiplication:
=>$${{x}\over {(-8)(-1)-(2)(-5)}} = {{y}\over {(-5)(3)-(-8)(2)}} = {{1}\over {(2)(2)-(3)(1)}}$$
=>$${{x}\over {-8 + 10}} = {{y}\over {-15 + 16}} = {{1}\over {4 - 3}}$$
=>$${{x}\over {2}} = {{y}\over {1}} = {{1}\over {1}}$$
=> x = 2 and y = 1.

(iii) $$3x - 5y = 20,6x - 10y = 40$$
On comparing these equations with the general form : $$ax^2 + bx + c$$
$$a_1 = 3, b_1 = -5, c_1 = -20,a_2 = 6, b_2 = -10, c_2 = -40$$
Here $${{a_1}\over {a_2}} = {{b_1}\over {b_2}} = {{c_1}\over {c_2}}$$
=>$${{3}\over {6}} = {{-5}\over {-10}} = {{-20}\over {-40}}$$
So, these lines coincide with each other.
Hence, there are infinite many solutions.

(iv) $$x - 3y – 7 = 0,3x - 3y – 15 = 0$$
On comparing these equations with the general form : $$ax^2 + bx + c$$
$$a_1 = 1, b_1 = -3, c_1 = -7,a_2 = 3, b_2 = -3, c_2 = -15$$
Here $${{a_1}\over {a_2}} \ne {{b_1}\over {b_2}}$$
=>$${{1}\over {3}} \ne {{-3}\over {-3}}$$
So, these lines have a unique solution.
We can find the unique solution using cross multiplication:
=>$${{x}\over {(-3)(-15)-(-3)(-7)}} = {{y}\over {(-7)(3)-(-15)(1)}} = {{1}\over {(-3)(1)-(-3)(3)}}$$
=>$${{x}\over {45 - 21}} = {{y}\over {-21 + 15}} = {{1}\over {-3 + 9}}$$
=>$${{x}\over {24}} = {{y}\over {-6}} = {{1}\over {6}}$$
=> x = 4 and y = -1.

Q2(i) For which values of a and b does the following pair of linear equations have an infinite number of solutions?
2x + 3y = 7
(a - b) x + (a + b) y = 3a + b – 2

(ii) For which value of k will the following pair of linear equations have no solution?
3x + y = 1
(2k - 1) x + (k - 1) y = 2k + 1

Ans.(i) $$2x + 3y = 7$$
$$(a - b) x + (a + b) y = 3a + b – 2$$
On comparing these equations with the general form : $$ax^2 + bx + c$$
$$a_1 = 2, b_1 = 3, c_1 = -7,a_2 = (a-b), b_2 = (a + b), c_2 = 2 - b -3a$$
Here $${{a_1}\over {a_2}} = {{b_1}\over {b_2}} = {{c_1}\over {c_2}}$$ since it has infinite solutions
=>$${{2}\over {a-b}} = {{3}\over {a+b}} = {{-7}\over {2 - b -3a}}$$
=>$${{2}\over {a-b}} = {{3}\over {a+b}}$$ and $${{3}\over {a+b}} = {{-7}\over {2 - b -3a}}$$
=>$$2a + 2b = 3a - 3b$$ and $$6 - 3b - 9a = -7a - 7b$$
=>$$a = 5b$$.....(i) and $$-2a = -4b – 6$$......(ii)
Putting (i) in (ii), we get
$$-2 (5b) = -4b – 6$$
=>$$-10b + 4b = -6$$
=> $$-6b = –6 - b = 1$$
Putting value of b in (i), we get
$$a = 5b = 5 (1) = 5$$
Therefore, a = 5 and b = 1
Hence, for the values a = 5 and b = 1, the given pair of equations have infinite solutions.

(ii) $$3x + y = 1$$
$$(2k - 1) x + (k - 1) y = 2k + 1$$
On comparing these equations with the general form : $$ax^2 + bx + c$$
$$a_1 = 3, b_1 = 1, c_1 = -1,a_2 = (2k - 1), b_2 = (k - 1), c_2 = -(2k + 1)$$
Here $${{a_1}\over {a_2}} = {{b_1}\over {b_2}} \ne {{c_1}\over {c_2}}$$ since it has no solutions
=>$${{3}\over {2k - 1}} = {{1}\over {k - 1}} \ne {{-1}\over {-2k-1}}$$
=>$${{3}\over {2k - 1}} = {{1}\over {k - 1}}$$
=>$$3(k - 1) = 2k - 1$$
=>$$3k - 3 = 2k - 1$$
=>$$k = 2$$

Q3. Solve the following pair of linear equations by the substitution and cross-multiplication methods:
$$8x + 5y = 9$$
$$3x + 2y = 4$$

Ans.(x + y = 14\)...........(i)
=> $$x – y = 4$$..............(ii)
Using equation (ii)
=> $$x = 4 + y$$......................(iii)
On substituting (iii) in (i):
=>$$4 + 2y = 14$$
=>$$2y = 10$$
=>$$y = 5$$...........(iv)
On substituting (iv) in (i):
=>$$x + 5 = 14$$
=>$$x = 14 - 5$$
=>$$x = 9$$
So, the value of $$y = 5$$ and $$x = 9$$.
$$x - 3y – 7 = 0,3x - 3y – 15 = 0$$
On comparing quations with general form of equation $$ax^2 + bx + c$$
$$a_1 = 1, b_1 = -3, c_1 = -7,a_2 = 3, b_2 = -3, c_2 = -15$$
Here $${{a_1}\over {a_2}} \ne {{b_1}\over {b_2}}$$
=>$${{1}\over {3}} \ne {{-3}\over {-3}}$$
So, these lines have a unique solution.
We can find the unique solution using cross multiplication:
=>$${{x}\over {(-3)(-15)-(-3)(-7)}} = {{y}\over {(-7)(3)-(-15)(1)}} = {{1}\over {(-3)(1)-(-3)(3)}}$$
=>$${{x}\over {45 - 21}} = {{y}\over {-21 + 15}} = {{1}\over {-3 + 9}}$$
=>$${{x}\over {24}} = {{y}\over {-6}} = {{1}\over {6}}$$
=> x = 4 and y = -1.

Q4.Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method:
(i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay Rs 1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs 1180 as hostel charges. Find the fixed charges and the cost of food per day.
(ii)\A fraction becomes $${{1}\over {3}}$$ when 1 is subtracted from the numerator and it becomes $${{1}\over {4}}$$ when 8 is added to its denominator. Find the fraction.
(iii)Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?
(iv)Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?
(v)The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

Ans.(i)Let fixed monthly charge = Rs x and let charge of food for one day = Rs y
According to given conditions,
$$x + 20y = 1000$$… (1),
$$x + 26y = 1180$$… (2)
Subtracting equation (1) from equation (2),
$$6y = 180$$
=>$$y = 30$$
Putting value of y in (1),
$$x + 20 (30) = 1000$$
=>$$x = 1000 – 600 = 400$$
Therefore, fixed monthly charges = Rs 400 and, charges of food for one day = Rs 30

(ii) Let numerator = x and let denominator = y
According to given conditions,
$${{x - 1}\over {y}} = {{1}\over {3}}$$… (1) and $${{x}\over {y + 8}} = {{1}\over {4}}$$… (2)
=>$$3x – 3 = y$$… (1) $$4x = y + 8$$ … (2)
=>$$3x – y = 3$$… (1) $$4x – y = 8$$… (2)
Subtracting equation (1) from (2),
$$4x – y - (3x - y) = 8 – 3$$
$$x = 5$$
Putting value of x in (1),
$$3 (5) – y = 3$$
$$=> 15 – y = 3$$
$$=> y = 12$$
Therefore, numerator = 5 and, denominator = 12
It means fraction = $${{5}\over {12}}$$

(iii) Let number of correct answers = x and let number of wrong answers = y
According to given conditions,
$$3x – y = 40$$ … (1)
And, $$4x - 2y = 50$$ … (2)
From equation (1), $$y = 3x - 40$$
Putting this in (2),
$$4x – 2 (3x - 40) = 50$$
=>$$4x - 6x + 80 = 50$$
=>$$-2x = -30$$
=>$$x = 15$$
Putting value of x in (1),
$$3 (15) – y = 40$$
=>$$45 – y = 40$$
=>$$y = 45 – 40 = 5$$
Therefore, number of correct answers = x = 15 and number of wrong answers = y = 5
Total questions = x + y = 15 + 5 = 20

(iv)Let speed of car which starts from part A = x km/hr
Let speed of car which starts from part B = y km/hr
According to given conditions,(Assuming x > y)
=>$${{100} \over {x-y}} = 5$$
=>$$5x - 5y = 100$$
=>$$x – y = 20$$… (1)
And, $${{100} \over {x+y}} = 1$$
=>$$x + y = 100$$ … (2)
$$2x = 120$$
=>$$x = 60$$
Putting value of x in (1),
$$60 – y = 20$$
=>$$y = 60 – 20 = 40$$ km/hr
Therefore, speed of car starting from point A = 60 km/hr
And, Speed of car starting from point B = 40 km/hr

(v) Let length of rectangle = x units and Let breadth of rectangle = y units
Area =xy square units. According to given conditions,
$$xy – 9 = (x - 5) (y + 3)$$
=>$$xy – 9 = xy + 3x - 5y – 15$$
=>$$3x - 5y = 6$$… (1)
And, $$xy + 67 = (x + 3) (y + 2)$$
=>$$xy + 67 = xy + 2x + 3y + 6$$
=>$$2x + 3y = 61$$… (2)
From equation (1),
=>$$3x = 6 + 5y$$
=>$$x = {{6 + 5y} \over {3}}$$
Putting this in (2),
=>$$2({{6 + 5y} \over {3}}) + 3y = 61$$
=>$$12 + 10y + 9y = 183$$
=>$$19y = 171$$
=>$$y = 9$$ units
Putting value of y in (2),
$$2x + 3 (9) = 61$$
=>$$2x = 61 – 27 = 34$$
=>$$x = 17$$ units
Therefore, length = 17 units and, breadth = 9 units

#### Solution for Exercise 3.6

Q1. Solve the following pairs of equations by reducing them to a pair of linear equations:
(i) $${{1} \over {2x}} + {{1} \over {3y}} = 2$$
$${{1} \over {3x}} + {{1} \over {2y}} = {{13} \over {6}}$$

(ii) $${{2} \over {\sqrt{x}}} + {{3} \over {\sqrt{y}}} = 2$$
$${{4} \over {\sqrt{x}}} - {{9} \over {\sqrt{y}}} = -1$$

(iii) $${{4} \over {x}} + 3y = 14$$
$${{3} \over {x}} - 4y = 23$$

(iv) $${{5} \over {x - 1}} + {{1} \over {y - 2}} = 2$$
$${{6} \over {x - 1}} - {{3} \over {y - 2}} = 1$$

(v) $$7x - 2y = 5xy$$
$$8x + 7y = 15xy$$

(vi)$$6x + 3y - 6xy = 0$$
$$2x + 4y - 5xy = 0$$

(vii) $${{10} \over {x + y}} + {{2} \over {x - y}} = 4$$
$${{15} \over {x + y}} - {{5} \over {x - y}} = -2$$

(viii) $${{1} \over {3x + y}} + {{1} \over {3x - y}} = {{3} \over {4}}$$
$${{1} \over {2(3x + y)}} - {{1} \over {2(3x - y)}} = {{-1} \over {8}}$$

Ans.(i) $${{1} \over {2x}} + {{1} \over {3y}} = 2$$… (1)
$${{1} \over {3x}} + {{1} \over {2y}} = {{13} \over {6}}$$ … (2)
Let $${{1} \over {x}}$$ = p and $${{1} \over {y}}$$ = q
Putting this in equation (1) and (2),
=>$${{p} \over {2}} + {{q} \over {3}} = 2$$ and $${{p} \over {3}} + {{q} \over {2}} = {{13} \over {6}}$$
=>$$3p + 2q = 12$$ and $$2p + 3q =13$$
=>$$3p + 2q – 12 = 0$$… (3) and $$2p + 3q – 13 = 0$$ … (4)
Using cross multiplication
=>$${{p} \over {(2)(-13) - 3(-12)}} = {{q} \over {(-12)(2) - (-13)(3)}} = {{1} \over {(3)(3) - (2)(2)}}$$
=> $${{p} \over {-26 + 36}} = {{q} \over {-24 + 39}} = {{1} \over {9 - 4}}$$
=>$${{p} \over {10}} = {{q} \over {15}} = {{1} \over {5}}$$
=> p = 2 and q = 3
But $${{1} \over {x}}$$ = p and $${{1} \over {y}}$$ = q
Putting value of p and q in this
x = $${{1} \over {2}}$$ and y = $${{1} \over {3}}$$

(ii)$${{2} \over {\sqrt{x}}} + {{3} \over {\sqrt{y}}} = 2$$… (1)
$${{4} \over {\sqrt{x}}} - {{9} \over {\sqrt{y}}} = -1$$ … (2)
Let $${{1} \over {\sqrt{x}}}$$ = p and $${{1} \over {\sqrt{y}}}$$= q
Putting this in (1) and (2),
$$2p + 3q = 2$$ … (3)
$$4p - 9q = -1$$… (4)
Multiplying (3) by 2 and subtracting it from (4),
$$4p - 9q + 1 – 2 (2p + 3q - 2) = 0$$
=>$$4p - 9q + 1 - 4p - 6q + 4 = 0$$
=>$$-15q + 5 = 0$$
=>$$q = {{5} \over {15}} = {{1} \over {3}}$$
Putting value of q in (3),
$$2p + 1 = 2 => 2p = 1 => p = {{1} \over {2}}$$
Putting values of p and q in ($${{1} \over {\sqrt{x}}}$$= p and $${{1} \over {\sqrt{y}}}$$= q), we get
$${{1} \over {\sqrt{x}}} = {{1} \over {2}}$$ and $${{1} \over {\sqrt{y}}} = {{1} \over {3}}$$
=>$${{1} \over {x}} = {{1} \over {4}}$$ and $${{1} \over {y}} = {{1} \over {9}}$$
=>$$x = 4$$ and $$y = 9$$

(iii) $${{4} \over {x}} + 3y = 14$$ … (1)
$${{3} \over {x}} - 4y = 23$$ … (2) and Let $${{1} \over {x}}= p$$ … (3)
Putting (3) in (1) and (2),
$$4p + 3y = 14$$… (4)
$$3p - 4y = 23$$… (5)
Multiplying (4) by 3 and (5) by 4,
$$3 (4p + 3y – 14 = 0)$$ and, $$4 (3p - 4y – 23 = 0)$$
=>$$12p + 9y – 42 = 0$$… (6) $$12p - 16y – 92 = 0$$ … (7)
Subtracting (7) from (6),
$$9y - (-16y) – 42 - (-92) = 0$$
$$=> 25y + 50 = 0$$
$$=> y = \frac{-50}{25} = -2$$
Putting value of y in (4),
$$4p + 3 (-2) = 14$$
$$=> 4p – 6 = 14$$
$$=> 4p = 20$$
$$p = 5$$
Putting value of p in (3),
$${{1} \over {x}} = 5 => x = {{1} \over {5}}$$
Therefore, x = $${{1} \over {5}}$$ and y = -2

(iv) $${{5} \over {x - 1}} + {{1} \over {y - 2}} = 2$$ … (1)
$${{6} \over {x - 1}} - {{3} \over {y - 2}} = 1$$… (2)
Let $${{1} \over {x - 1}} = p$$ and $${{1} \over {y - 2}} = q$$
Putting this in (1) and (2),
$$5p + q = 2$$
$$5p + q – 2 = 0$$ … (3)
And, $$6p - 3q = 1$$
$$6p - 3q – 1 = 0$$… (4)
Multiplying (3) by 3 and adding it to (4),
$$3 (5p + q - 2) + 6p - 3q – 1 = 0$$
$$15p + 3q – 6 + 6p - 3q – 1 = 0$$
$$21p – 7 = 0$$
$$p = {{1} \over {3}}$$
Putting this in (3),
$$5 ({{1} \over {3}}) + q – 2 = 0$$
$$5 + 3q = 6$$
$$3q = 6 – 5 = 1$$
$$q = {{1} \over {3}}$$
Putting values of p and q in ($${{1} \over {x - 1}} = p$$ and $${{1} \over {y - 2}} = q$$),
$${{1} \over {x - 1}} = {{1} \over {3}}$$ and $${{1} \over {y - 2}} = {{1} \over {3}}$$
$$3 = x - 1$$ and $$3 = y – 2$$
$$x = 4$$ and $$y = 5$$

(v) $$7x - 2y = 5xy$$ … (1)
$$8x + 7y = 15xy$$… (2)
Dividing both the equations by xy,
$${{7} \over {y}} - {{2} \over {x}} = 5$$… (3)
$${{8} \over {y}} + {{7} \over {x}} = 15$$… (4)
Let $${{1} \over {x}}$$ = p and $${{1} \over {y}}$$ = q
Putting these in (3) and (4),
$$7q - 2p = 5$$… (5)
$$8q + 7p = 15$$… (6)
From equation (5),
$$2p = 7q – 5$$
$$p = {{7q – 5} \over {2}}$$
Putting value of p in (6),
$$8q + 7 (\frac{7q – 5}{2}) = 15$$
$$16q + 49q – 35 = 30$$
$$65q = 30 + 35 = 65$$
$$q = 1$$
Putting value of q in (5),
$$7 (1) - 2p = 5$$
$$- 2p = -2$$
$$p = 1$$
Putting value of p and q in ($${{1} \over {x}} = p$$ and $${{1} \over {y}} = q$$), we get x = 1 and y = 1

(vi)$$6x + 3y - 6xy = 0$$… (1)
$$2x + 4y - 5xy = 0$$ … (2)
Dividing both the equations by xy,
$${{6} \over {y}} + {{3} \over {x}} - 6 = 0$$… (3)
$${{2} \over {y}} + {{4} \over {x}} - 5 = 0$$… (4)
Let $${{1} \over {x}} = p$$ and $${{1} \over {y}} = q$$
Putting these in (3) and (4),
$$6q + 3p – 6 = 0$$ … (5)
$$2q + 4p – 5 = 0$$ … (6)
From (5),
$$3p = 6 - 6q$$
$$p = 2 - 2q$$
Putting this in (6),
$$2q + 4 (2 - 2q) – 5 = 0$$
$$2q + 8 - 8q – 5 = 0$$
$$-6q = -3$$
$$q = {{1} \over {2}}$$
Putting value of q in (p = 2 – 2q), we get
$$p = 2 – 2 ({{1} \over {2}}) = 2 – 1 = 1$$
Putting values of p and q in ($${{1} \over {x}} = p$$ and $${{1} \over {y}} = q$$), we get x = 1 and y = 2

(vii) $${{10} \over {x + y}} + {{2} \over {x - y}} = 4$$… (1)
$$\frac{15}{(x + y)} - \frac{5}{(x - y)} = -2$$…(2)
Let $${{1} \over {(x + y)}} = p$$ and $${{1} \over {(x - y)}} = q$$
Putting this in (1) and (2),
$$10p + 2q = 4$$ … (3)
$$15p - 5q = -2$$ … (4)
From equation (3),
$$2q = 4 - 10p$$
$$q = 2 - 5p$$ … (5)
Putting this in (4),
$$15p – 5 (2 - 5p) = -2$$
$$15p – 10 + 25p = -2$$
$$40p = 8$$ => $$p = {{1} \over {5}}$$
Putting value of p in (5),
$$q = 2 – 5 ({{1} \over {5}}) = 2 – 1 = 1$$
Putting values of p and q in ($${{1} \over {x + y}} = p$$ and $${{1} \over {x - y}} = q$$), we get
$$x + y = 5$$ … (6) and $$x – y = 1$$ … (7)
$$2x = 6$$ => $$x = 3$$
Putting x = 3 in (7), we get
$$3 – y = 1$$
$$y = 3 – 1 = 2$$
Therefore, x = 3 and y = 2

(viii)$${{1} \over {3x + y}} + {{1} \over {3x - y}} = {{3} \over {4}}$$ … (1)
$${{1} \over {2(3x + y)}} - {{1} \over {2(3x - y)}} = {{-1} \over {8}}$$ … (2)
Let $${{1} \over {(3x + y)}} = p$$ and $${{1} \over {(3x - y)}} = q$$ … (2)
Putting this in (1) and (2),
$$p + q = {{3} \over {4}}$$ and $${{p} \over {2}} - {{q} \over {2}} = {{1} \over {8}}$$
$$4p + 4q = 3$$ … (3) and $$4p - 4q = -1$$ … (4)
$$8p = 2 - p = {{1} \over {4}}$$
Putting value of p in (3),
$$4 ({{1} \over {4}}) + 4q = 3$$
$$1 + 4q = 3$$
$$4q = 3 – 1 = 2$$
$$q = {{1} \over {2}}$$
Putting value of p and q in $${{1} \over {(3x + y)}} = p$$ and $${{1} \over {(3x - y)}} = q$$, we get
$${{1} \over {(3x + y)}} = {{1} \over {4}}$$ and $${{1} \over {(3x - y)}} = {{1} \over {2}}$$
$$3x + y = 4$$… (5) and $$3x – y = 2$$… (6)
$$6x = 6 - x = 1$$
Putting x = 1 in (5) ,
$$3 (1) + y = 4$$
$$y = 4 – 3 = 1$$
Therefore, x = 1 and y = 1

Q2. Formulate the following problems as a part of equations, and hence find their solutions.
(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.
(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days.Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.
(iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately

Ans.(i) Let speed of rowing in still water = x km/h
Let speed of current = y km/h
So, speed of rowing downstream = (x + y) km/h
And, speed of rowing upstream = (x - y) km/h
According to given conditions,
$${{20} \over {x + y}} = 2$$ and $${{4} \over {x - y}} = 2$$
$$2x + 2y = 20$$ and $$2x - 2y = 4$$
$$x + y = 10$$ … (1) and $$x – y = 2$$ … (2)
$$2x = 12$$ => $$x = 6$$
Putting x = 6 in (1),
$$6 + y = 10$$
$$y = 10 – 6 = 4$$
Therefore, speed of rowing in still water = 6 km/h
Speed of current = 4 km/h

(ii) Let time taken by 1 woman alone to finish the work = x days
Let time taken by 1 man alone to finish the work = y days
So, 1 woman’s 1-day work = ($${{1} \over {x}}$$)th part of the work
And, 1 man’s 1-day work = ($${{1} \over {y}}$$)th part of the work
So, 2 women’s 1-day work = ($${{2} \over {x}}$$)th part of the work
And, 5 men’s 1-day work = ($${{5} \over {y}}$$)th part of the work
Therefore, 2 women and 5 men’s 1-day work = ($${{2} \over {x}} + {{5} \over {y}}$$)th part of the work....… (1)
It is given that 2 women and 5 men complete work in = 4 days
It means that in 1 day, they will be completing ($${{1} \over {4 }}$$)th part of the work ….... (2)
Since, (1) = (2)
$${{2} \over {x}} + {{5} \over {y}} = {{1} \over {4 }}$$….... (3)
Similarly, $${{3} \over {x}} + {{6} \over {y}} = {{1} \over {3}}$$ .....… (4)
Let $${{1} \over {x}} = p$$ and $${{1} \over {y}} = q$$
Putting this in (3) and (4),
$$2p + 5q = {{1} \over {4 }}$$ and $$3p + 6q = {{1} \over {3}}$$
$$8p + 20q = 1$$ … (5) and $$9p + 18q = 1$$ … (6)
Multiplying (5) by 9 and (6) by 8,
$$72p + 180q = 9$$ … (7)
$$72p + 144q = 8$$ … (8)
Subtracting (8) from (7), we get
$$36q = 1$$ => $$q = {{1} \over {36}}$$
Putting this in (6), we get
$$9p + 18 ({{1} \over {36}}) = 1$$
$$9p = {{1} \over {2}}$$ => $$p = {{1} \over {18}}$$
Putting values of p and q in , we get x = 18 and y = 36
Therefore, 1 woman completes work in = 18 days
And, 1 man completes work in = 36 days

(iii) Let speed of train = x km/h and let speed of bus = y km/h
According to given conditions,
$${{60} \over {x}} + {{240} \over {y}} = 4$$ and $${{100} \over {x}} + {{200} \over {y}} = 4 + {{10} \over {60}}$$
Let $${{1} \over {x}} = p$$ and $${{1} \over {y}} = q$$
Putting this in the above equations, we get
$$60p + 240q = 4$$ … (1)
And $$100p + 200q = {{25} \over {6}}$$ … (2)
Multiplying (1) by 5 and (2) by 3,
$$300p + 1200q = 20$$… (3)
$$300p + 600q = {{25} \over {2}}$$ … (4)
Subtracting (4) from (3),
$$600q = 20 - {{25} \over {2}} = 7.5$$
$$q = {{7.5} \over {600}}$$
Putting value of q in (2),
$$100p + 200 ({{7.5} \over {600}}) = {{25} \over {6}}$$
$$100p + 2.5 = {{25} \over {6}}$$
$$100p = {{25} \over {6}} – 2.5$$
$$p = {{10} \over {600}}$$
But $${{1} \over {x}} = p$$ and $${{1} \over {y}} = q$$
Therefore, $$x = {{600} \over {10}} = 60$$ km/h and $$y = {{600} \over {7.5}} = 80$$ km/h
Therefore, speed of train = 60 km/h
And, speed of bus = 80 km/h

#### Solution for Exercise 3.7

Q1. The age of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.

Ans. Let the age of Ani and Biju be x years and y years respectively.
Age of Dharam = 2x years and Age of Cathy = $${{y}\over{2}}$$ years
Case 1 : Ani's age is greater than Biju's age
According to question, x – y = 3… (1)
And $$2x - {{y}\over{2}} = 30$$
$$4x – y = 60$$… (2)
Subtracting (1) from (2),:
$$3x = 60 – 3 = 57$$
x = Age of Ani = 19 years
Age of Biju = 19 – 3 = 16 years
Case 2 : Ani's age is less than Biju's age
Thus, $$y – x = 3$$… (3)
And $$2x - {{y}\over{2}} = 30$$
$$4x – y = 60$$… (4)
Adding (3) and (4), we obtain:
3x = 63
x = 21
Age of Ani = 21 years
Age of Biju = 21 + 3 = 24 years

Q2. One says, “Give me a hundred, friend! I shall then become twice as rich as you.” The other replies, “If you give me ten, I shall be six times as rich as you.” Tell me what is the amount of their (respective) capital? [From the Bijaganita of Bhaskara II]

Ans.Let the money with the first person and second person be Rs x and Rs y respectively.
According to the question,
x + 100 = 2(y – 100)
x + 100 = 2y – 200
x – 2y = – 300… (1)
Also, 6(x – 10) = (y + 10)
6x – 60 = y + 10
6x – y = 70… (2)
Multiplying equation (2) by 2, we obtain:
12x – 2y = 140… (3)
Subtracting equation (3) from equation (1), we obtain:
11x = 140 + 300
=> 11x = 440
=> x = 40
Putting the value of x in equation (1), we obtain:
40 – 2y = –300
=> 40 + 300 = 2y
=> 2y = 340
=> y = 170
Thus, the two friends had Rs 40 and Rs 170 with them.

Q3. A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h, it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.

Ans.Let the speed of the train be x km/h and the time taken by train to travel the given distance be t hours and the distance to travel be d km.
Since Speed = Distance travelled/Time taken to travel the distance
=> $$x = {{d} \over {t}}$$
=> d = xt … (1)
According to the question
=> $$x + 10 = {{d} \over {t - 2}}$$
=>$$(x + 10) (t - 2) = d$$
=> $$xt + 10t - 2x - 20 = d$$
=> $$-2x + 10t = 20$$……(2)[Using eq. (1)]
Again, => $$x - 10 = {{d} \over {t + 3}}$$
=>$$(x - 10) (t + 3) = d$$
=> $$xt - 10t + 3x - 30 = d$$
=> $$3x - 10t = 30$$……(3)[Using eq. (1)]
x = 50
Substituting the value of x in equation (2):
$$(-2) (50) + 10t =20$$
$$=> –100 + 10t = 20$$
$$=> 10t = 120t$$
=> $$t = 12$$
From equation (1):
d = xt = (50)(12) = 600
Thus, the distance covered by the train is 600 km.

Q4. The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.

Ans.Let the number of rows be x and number of students in a row be y.
Total number of students in the class = Number of rows x Number of students in a row = xy
Case 1 :
Total number of students = (x – 1) (y + 3)
$$xy = (x – 1) (y + 3)$$
$$=> xy = xy – y + 3x – 3$$
$$=> 3x – y – 3 = 0$$
$$=> 3x – y = 3$$… (1)
Case 2 :
Total number of students = (x + 2) (y – 3)
$$xy = xy + 2y – 3x – 6$$
$$=> 3x – 2y = –6$$… (2)
Subtracting equation (2) from (1):
y = 9
Substituting the value of y in equation (1):
$$3x – 9 = 3$$
$$3x = 9 + 3 = 12$$
x = 4
Number of rows = x = 4
Number of students in a row = y = 9
Hence, Total number of students in a class = xy = 36

Q5. In a ABC, $$\angle C = \angle 3B = 2(\angle A + \angle B)$$. Find three angles.

Ans.$$\angle C = \angle 3B = 2(\angle A + \angle B)$$
Taking $$3\angle B = 2 ( \angle A + \angle B )$$
$$\angle B = 2\angle A$$
$$2\angle A – \angle B = 0$$ …….(1)
We know that the sum of the measures of all angles of a triangle is 180°.
$$\angle A + \angle B + \angle C = 180°$$
$$\angle A + \angle B + 3\angle B = 180°$$
$$\angle A + 4\angle B = 180°$$ …….(2)
Multiplying equation (1) by 4:
$$8\angle A – 4\angle B = 0$$ …….(3)
Adding equations (2) and (3), we get
$$9\angle A = 180°$$
$$\angle A = 20°$$
From eq. (2), we get,
$$20° + 4\angle B = 180°$$
$$\angle B = 40°$$
And $$\angle C = 3 (40°) = 120°$$
Hence the measures of $$\angle A, \angle B$$ and $$\angle C$$ are respectively.

Q6. Draw the graphs of the equations and Determine the co-ordinate of the vertices of the triangle formed by these lines and the axis.

Ans.According to the given statement in the question:
=>$$y = 5x - 5$$............(1)
Three solutions of this equation can be written in a table as follows:
$$\begin{array} {|r|r|}\hline x & 0 & 1 & 2 \\ \hline y & -5 & 0 & 5 \\ \hline \end{array}$$
According to the given statement in the question:
=>$$y = 3x - 3$$............(2)
Three solutions of the equation (2) are :
$$\begin{array} {|r|r|}\hline x & 0 & 1 & 2 \\ \hline y & -3 & 0 & 3 \\ \hline \end{array}$$
Now the graph for the above equations (1) and (2) will be: It can be observed that the required triangle is ABC.
The coordinates of its vertices are A (1, 0), B (0, –3), C (0, –5).

Q7.Solve the following pair of linear equations:
(i) $$px + py = p - q$$
$$qx –py = p + q$$

(ii) $$ax + by = c$$
$$bx + ay = 1 + c$$

(iii) $${{x}\over{a}} - {{y}\over{b}} = 0$$
$$ax + by = a ^2 + b^2$$

(iv) $$(a - b)x + (a + b)y = a^2 -2ab - b^2$$
$$(a + b)(x + y) = a^2 + b^2$$

(v) $$152x - 378y = -74$$
$$-378x + 152y = -604$$

(i)$$px + py = p - q$$ … (1)
$$qx –py = p + q$$… (2)
Multiplying equation (1) by p and equation (2) by q:
$$p^2 x + pqy = p^2 - pq$$… (3) $$q^2 x - pqy = pq + q^2$$… (4)
Adding equations (3) and (4), :
=>$$p^2 x + q^2 x = p^2 + q^2$$
=>$$(p^2 + q^2)x = p^2 + q^2$$

=> x = $${{p^x + q^x} \over {p^2 + q^2}}$$= 1
Substituting the value of in equation (1):
$$p(1) + qy = p - q$$
=>$$qy = -q => y = -1$$
Hence the required solution is x = 1 and y = –1.

(ii)$$ax + by = c$$… (1)
$$bx + ay = 1 + c$$ … (2)
Multiplying equation (1) by a and equation (2) by b:
$$a^2x + aby = ac$$… (3)
$$b^2x + aby = b + bc$$ … (4)
Subtracting equation (4) from equation (3),
$$(a^2 - b^2)x = ac - bc -b$$
=>$$x = {{c(a - b) - b} \over {a^2 - b^2}}$$
Substituting the value of x in equation (1):
$$a ({{c(a - b) - b} \over {a^2 - b^2}}) + by = c$$
=>$${{ac(a - b) - ab} \over {a^2 - b^2}} + by = c$$
=>$$by = c -{{ac(a - b) - ab} \over {a^2 - b^2}}$$
=>$$by = {{a^2c - b^2c - a^2c + abc + ab} \over {a^2 - b^2}}$$
=> $$by = {{ abc - b^2 c+ ab} \over {a^2 - b^2}}$$
=>$$y = {{ c(a - b)+ a} \over {a^2 - b^2}}$$

(iii)$${{x}\over{a}} - {{y}\over{b}} = 0$$
=>$$bx - ay = 0$$……..(1)
=>$$ax + by = a^2 - b^2$$……..(2)
Multiplying equation (1) and (2) by b and a respectively:
=> $$xb^2 - aby = 0$$……..(3)
=> $$xa^2 + aby = a^3 - ab^2$$……..(4)
Adding equations (3) and (4), we obtain:
$$xb^2 + xa^2 = a^3 + ab^2$$
=>$$x(b^2 + a^2) = a (b^2 + a^2)$$
=> $$x = a$$
Substituting the value of in equation (1):
$$b(a) - ay = 0$$
=>$$ab - ay = 0$$
=> $$y = b$$

(iv) $$(a - b)x + (a + b)y = a^2 -2ab - b^2$$ … (1)
$$(a + b)(x + y) = a^2 + b^2$$……..(2)
Subtracting equation (2) from (1):
$$(a - b)x - (a + b)x = (a^2 - 2ab - b^2) - ( a^2 + b^2)$$
=>$$(a - b - a - b)x = -2ab - 2b^2$$
=>$$-2bx = -2b(a + b)$$
=> $$x = a + b$$
Substituting the value of in equation (1):
$$(a - b)(a + b) + (a + b)y = a^2 -2ab - b^2$$
=>$$a^2 - b^2 + (a + b)y = a^2 -2ab - b^2$$
=>$$(a + b)y = -2ab$$
=>$$y = {{-2ab} \over {a + b}}$$

(v)$$152x – 378y = –74$$… (1)
$$–378x + 152y = –604$$ … (2)
Adding the equations (1) and (2):
$$–226x – 226y = –678$$
$$=> x + y = 3$$ ………(3)
Subtracting the equation (2) from equation (1):
$$530x – 530y = 530$$
$$=> x – y = 1$$ ……..(4)
$$2x = 4$$
$$=> x = 2$$
Substituting the value of x in equation (3):
$$y = 1$$

Q8. ABCD is a cyclic quadrilateral (see figure). Find the angles of the cyclic quadrilateral. We know that the sum of the measures of opposite angles in a cyclic quadrilateral is :
$$\angle$$A + $$\angle$$C = 180°
=>$$4y + 20 - 4x = 180°$$
=>$$-4x + 4y = 160°$$
=>$$x - y = -40°$$ ………(1)
Also $$\angle$$B + $$\angle$$D = 180°
=>$$3y - 5 - 7x + 5 = 180°$$
=>$$-7x + 3y = 180°$$………(2)
Multiplying equation (1) by 3:
$$3x - 3y = - 120°$$ ……….(3)
$$-4x = 60° => x = -15°$$
$$-15 - y = -40°$$
=>$$y = -15 + 40 =25$$
$$\angle$$A = $$4y + 20 = 4(25) + 20 = 120°$$
$$\angle$$B = $$3y - 5 = 3(25) - 5 = 70°$$
$$\angle$$C = $$-4x = -4(-15) = 60°$$
$$\angle$$D = $$-7x + 5 = -7(-15) + 5 = 110°$$