CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables VBQs Set 03

Question. The present age of a father is three years more than three times the age of his son. Three years hence the father‘s age will be 10 years more than twice the age of the son. Determine their present ages. 
Answer: Let \(x\) (in years) be the present age of the father and \(y\) (in years) be the present age of the son. Then, according to the given question: \(x = 3y + 3\) or \(x - 3y = 3\) ...(i) After 3 years, Father’s age = \(x + 3\), son’s age = \(y + 3\) and \(x + 3 = 2 (y + 3) + 10\) or \(x - 2y = 13\) ...(ii) On solving the two equations, we get: \(y = 10\), and \(x = 33\). Thus, the father's present age is 33 years and the son's present age is 10 years.

Question. Taxi charges in a city consist of fixed charges and the remaining charges depend upon the distance travelled. For a journey of 10 km, the charge paid is ₹ 75 and for a journey of 15 km, the charge paid is ₹ 110. Find the fixed charge and charges per km. Also, find the charges of covering a distance of 35 km.
Answer: Let the fixed charge be ₹ \(x\) and charges for per km be ₹ \(y\). As per the question, \(x + 10y = 75\) and \(x + 15y = 110\) On solving the two equations, we get \(x = 5, y = 7\) Thus, the fixed charge is ₹ 5 and the charge per km is ₹ 7. Hence, charge for 35 km is ₹ [5 + 35(7)], i.e., ₹ 250.

Question. A man can row a boat downstream 20 km in 2 hours and upstream 4 km in 2 hours. Find his speed of rowing in still water. Also find the speed of the stream. 
Answer: Let the speed of the stream be \(x\) km/h and the speed of rowing in still water be \(y\) km/h. Then, the speed of rowing in the downstream is \((x + y)\) km/h. And the speed of rowing in the upstream is \((y - x)\) km/h. As per the question: \[ \frac{20}{y + x} = 2 \text{ and } \frac{4}{y - x} = 2 \] \(\Rightarrow x + y = 10\) and \(y - x = 2\) On solving the two equations, we get: \(x = 4\) and \(y = 6\) Thus, the speed of rowing in still water is 6 km/h, and the speed of the stream is 4 km/h.

Question. The angles of a triangle are \(x, y\) and 40°. The difference between the two angles \(x\) and \(y\) is 30°. Find \(x\) and \(y\). 
Answer: It is given that \(x, y\) and 40° are the angles of a triangle. We know that the sum of all the angles of a triangle is 180°. \(\Rightarrow x + y + 40 = 180\) \(\Rightarrow x + y = 140\) ...(i) Also, it is given that the difference of angles \(x\) and \(y\) is \(x - y = 30\) ...(ii) Adding eq. (i) and (ii), we get \((x + y) + (x - y) = 140 + 30\) \(\Rightarrow 2x = 170\) \(\Rightarrow x = 85\) Putting the value of \(x\) in eq. (i), we get \(\Rightarrow 85 + y = 140\) \(\Rightarrow y = 140 - 85\) \(y = 55\) Hence, the required values of \(x\) and \(y\) are 85° and 55° respectively.

Question. A part of monthly hostel charges in a college hostel are fixed and the remaining depends on the number of days one has their meals in the mess. When a student A takes food for 25 days, he has to pay ₹ 4,500, whereas a student B who takes food for 30 days has to pay ₹ 5,200. Find the fixed charges per month and the cost of food per day.
Answer: Let, the fixed charge per student = ₹ \(x\) Cost of food per day per student = ₹ \(y\) According to the given condition, \(x + 25y = 4500\) ...(i) \(x + 30y = 5200\) ...(ii) On subtracting equation (i) from equation (ii), we get \(5y = 700\) \(\Rightarrow y = \frac{700}{5} = 140\) Put the value of \(y\) in equation (i), we get: \(x + 25 \times 140 = 4500\) \(\Rightarrow x + 3500 = 4500\) \(\Rightarrow x = 1000\) Hence, the fixed charge per student is ₹ 1,000 and cost of food per day is ₹ 140.

Question. There are some students in two examination halls A and B. To make the number of students equal in each hall, 10 students are sent from A to B. But, if 20 students are sent from B to A, the number of students in A becomes double the number of students in B. Find the number of students in the two halls.
Answer: Let the number of students in hall A and B be \(x\) and \(y\) respectively. By the given condition, to make the number of students equal in each hall, 10 students are sent from A to B \(\Rightarrow x - 10 = y + 10\) \(\Rightarrow x - y = 20\) ...(i) Also, it is given that if 20 students are sent from B to A, the number of students in A becomes double the number of students in B \(\Rightarrow (x + 20) = 2(y - 20)\) \(x - 2y = -60\) ...(ii) Subtracting eq. (ii) from eq. (i), we get \((x - y) - (x - 2y) = 20 - (-60)\) \(\Rightarrow y = 80\) Putting the value of \(y\) in eq. (i), we get \(x - 80 = 20\) \(\Rightarrow x = 100\) Hence, 100 students are in hall A and 80 students are in hall B.

Question. In a competitive examination, one mark is awarded for each correct answer, while \(1/2\) mark is deducted for every wrong answer. Rahul answered 120 questions and got 90 marks. How many questions did he answer correctly? 
Answer: Let \(x\) be the number of correct answers. Then, marks awarded for correct answer = \(x \times 1 = x\) Total no. of questions attempted = 120 Number of wrong answers = \((120 - x)\) Marks deducted for wrong answers = \((120 - x) \times \frac{1}{2} = \frac{120 - x}{2}\) Total marks awarded to Rahul = 90 \(\Rightarrow x - \left( \frac{120 - x}{2} \right) = 90\) \(\Rightarrow x + \frac{x}{2} - 60 = 90\) \(\Rightarrow \frac{3x}{2} = 150\) \(\Rightarrow x = \frac{150 \times 2}{3}\) \(\Rightarrow x = 100\) Hence, Rahul answered 100 questions correctly.

Question. A father’s age is three times the sum of the ages of his children. After 5 years, his age will be two times the sum of their ages. Find the present age of the father. 
Answer: Let the sum of the ages of two children be \(x\) years and father's age be \(y\) years. According to the given condition: \(y = 3x\) or \(y - 3x = 0\) ...(i) After 5 years: Father’s age = \((y + 5)\) years Sum of the ages of children = \((x

Question. A father’s age is three times the sum of the ages of his children. After 5 years, his age will be two times the sum of their ages. Find the present age of the father. 
Answer: Let the sum of the ages of two children be \( x \) years and father’s age be \( y \) years.
According to the given condition:
\( y = 3x \) or \( y - 3x = 0 \) ...(i)
After 5 years:
Father’s age = \( (y + 5) \) years
Sum of the ages of children = \( (x + 5 + 5) \) years.
Then, \( y + 5 = 2(x + 10) \)
or \( y - 2x - 15 = 0 \) ...(ii)
On subtracting equation (i) from equation (ii), we get:
\( y - 2x - 15 = 0 \)
\( y - 3x = 0 \)
\( - + \)
________________
\( x - 15 = 0 \)
\( x = 15 \)
If we put the value of \( x \) in equation (i), we get
\( y - 3 \times 15 = 0 \)
\( \Rightarrow y = 45 \)
Hence, the present age of the father is 45 years.

Question. Solve the following system of equations: \( \frac{21}{x} + \frac{47}{y} = 110 \); \( \frac{47}{x} + \frac{21}{y} = 162, x, y \neq 0 \) 
Answer: Let \( \frac{1}{x} = a \) and \( \frac{1}{y} = b \)
\( \Rightarrow 21a + 47b = 110 \) and \( 47a + 21b = 162 \)
Adding and subtracting the two equations, we get
\( a + b = 4 \) and \( a - b = 2 \)
Solving the above two equations, we get
\( a = 3 \) and \( b = 1 \)
\( \therefore x = \frac{1}{3} \) and \( y = 1 \)

Question. The sum of reciprocals of a child’s age (in years) 3 years ago and 5 years from now is \( \frac{1}{3} \). Find his present age.
Answer: Let the present age of the child (in years) be \( x \). Then,
\( \frac{1}{x - 3} + \frac{1}{x + 5} = \frac{1}{3} \)
\( \Rightarrow \frac{(x + 5) + (x - 3)}{(x - 3)(x + 5)} = \frac{1}{3} \)
\( \Rightarrow 3(2x + 2) = (x - 3)(x + 5) \)
\( \Rightarrow 6x + 6 = x^2 + 2x - 15 \)
\( \Rightarrow x^2 - 4x - 21 = 0 \)
\( \Rightarrow (x - 7)(x + 3) = 0 \)
\( \Rightarrow x - 7 = 0 \) or \( x + 3 = 0 \)
\( \Rightarrow x = 7 \) or \( x = -3 \)
(\( x = -3 \) is not possible)
Thus, her present age is 7 years.

Question. A man wished to give ₹ 12 to each person and found that he fell short of ₹ 6 when he wanted to give to all the persons present. He, therefore, distributed ₹ 9 to each person and found that ₹ 9 were left over. How much money did he have and how many persons were there? 
Answer: Let, number of persons = \( x \)
Money share per person = \( y \)
Therefore, total money = ₹ \( xy \)
According to the question,
\( 12 \times x = xy + 6 \)
\( 12x - 6 = xy \) ...(i)
and \( 9x = xy - 9 \)
\( 9x + 9 = xy \) ...(ii)
Equating (i) and (ii), we get
\( 12x - 6 = 9x + 9 \)
\( 3x = 15 \)
\( x = 5 \)
Put the value of \( x \) in equation (i). Then
\( 12 \times 5 - 6 = x \times y \)
\( xy = 54 \)
So, he have ₹ 54 and there were 5 persons.

Question. Find the solution of the pair of equations: \( \frac{3}{x} + \frac{8}{y} = -1 \); \( \frac{1}{x} - \frac{2}{y} = 2, x, y \neq 0 \) 
Answer: Given, pair of equation is—
\( \frac{3}{x} + \frac{8}{y} = -1 \) ...(i)
\( \frac{1}{x} - \frac{2}{y} = 2 \) ...(ii)
If we multiply equation (ii) by 3 and subtract it from equation (i), we get
\( \frac{3}{x} + \frac{8}{y} = -1 \)
\( \frac{3}{x} - \frac{6}{y} = 6 \)
\( - + - \)
________________
\( \frac{14}{y} = -7 \)
\( y = -2 \)
Put the value of \( y = -2 \) in equation (i), we get
\( \frac{3}{x} + \left( \frac{8}{-2} \right) = -1 \)
\( \frac{3}{x} = -1 + 4 \)
\( x = 1 \)
Hence, the value of \( x \) and \( y \) are 1 and -2 respectively.

Question. Ratio between the girls and boys in a class of 40 students is 2 : 3. Five new students joined the class. How many of them must be boys so that the ratio between girls and boys becomes 4 : 5? 
Answer: Let number of girls = \( 2x \)
and number of boys = \( 3x \)
Total students, \( 2x + 3x = 40 \)
\( x = 8 \)
So, number of girls = \( 2 \times 8 = 16 \)
and number of boys = \( 3 \times 8 = 24 \)
Let out of 5 students, \( y \) denotes number of boys.
Then, number of girls = \( 5 - y \)
According to question,
\( \frac{16 + 5 - y}{24 + y} = \frac{4}{5} \)
\( 5(21 - y) = 4(24 + y) \)
\( 105 - 5y = 96 + 4y \)
\( 9y = 9 \)
\( y = 1 \)
Hence, 1 student must be a boy.

LONG ANSWER Type Questions

Question. Determine, algebraically, the vertices of the triangle formed by the lines \( 3x - y = 3 \), \( 2x - 3y = 2 \) and \( x + 2y = 8 \).
Answer: The given equation of lines are:
\( 3x - y = 3 \) ...(i)
\( 2x - 3y = 2 \) ...(ii)
\( x + 2y = 8 \) ...(iii)
Let lines (i), (ii) and (iii) represent the side of a \( \Delta ABC \) i.e., AB, BC and CA respectively.
On solving lines (i) and (ii), we will get the intersection point B.
Multiplying eq. (i) by 3 and then subtracting eq. (ii), we get
\( \Rightarrow (9x - 3y) - (2x - 3y) = 9 - 2 \)
\( \Rightarrow 7x = 7 \)
\( \Rightarrow x = 1 \)
Putting the value of \( x \) in eq. (i), we get
\( \Rightarrow 3 \times 1 - y = 3 \)
\( \Rightarrow y = 0 \)
Hence, the coordinate of point or vertex B is \( (1, 0) \).
On solving lines (ii) and (iii), we will get the intersection point C.
Multiplying eq. (iii) by 2 and then subtracting eq. (ii), we get
\( (2x + 4y) - (2x - 3y) = 16 - 2 \)
\( \Rightarrow 7y = 14 \)
\( \Rightarrow y = 2 \)
Putting the value of \( y \) in eq. (iii), we get
\( \Rightarrow x + 2(2) = 8 \)
\( \Rightarrow x = 4 \)
Hence, the coordinate of point or vertex C is \( (4, 2) \).
On solving lines (iii) and (i), we will get the intersecting point A.
Multiplying eq. (i) by 2 and then adding eq. (iii), we get
\( (6x - 2y) + (x + 2y) = 6 + 8 \)
\( \Rightarrow 7x = 14 \)
\( \Rightarrow x = 2 \)
Putting the value of \( x \) in eq. (i), we get
\( \Rightarrow 3 \times 2 - y = 3 \)
\( \Rightarrow y = 3 \)
Hence, the coordinate of point or vertex A is \( (2, 3) \).
Hence, the vertices of the \( \Delta ABC \) formed by the given lines are \( A(2, 3), B(1, 0) \) and \( C(4, 2) \).

Question. It can take 12 hours to fill a swimming pool using two pipes. If the pipe of larger diameter is used for four hours and the pipe of smaller diameter for 9 hours, only half of the pool can be filled. How long would it take for each pipe to fill the pool separately? 
Answer: Let two pipes A and B of diameter \( d_1 \) and \( d_2 \) (\( d_1 > d_2 \)) take \( x \) and \( y \) hours to fill the pool, respectively. Then,
\( \frac{1}{x} + \frac{1}{y} = \frac{1}{12} \) ...(i)
and \( \frac{4}{x} + \frac{9}{y} = \frac{1}{2} \) ...(ii)
Let \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \)
\( \therefore u + v = \frac{1}{12} \) ...(iii)
and \( 4u + 9v = \frac{1}{2} \) ...(iv)
If we multiply equation (iii) by 4 and subtract it from (iv)
\( 4u + 4v = \frac{1}{3} \)
\( 4u + 9v = \frac{1}{2} \)
\( - 5v = -\frac{1}{6} \)
\( \therefore v = \frac{1}{30} \)
and \( u = \frac{1}{20} \)
\( \therefore x = 20, y = 30 \)
Thus, the pipe with diameter \( d_1 \) takes 20 hours and the pipe with diameter \( d_2 \) takes 30 hours to fill the pool alone.

Question. Ankita travels 14 km to her home partly by rickshaw and partly by bus. She takes half an hour if she travels 2 km by rickshaw and the remaining distance by bus. On the other hand, if she travels 4 km by rickshaw and the remaining distance by bus, she takes 9 minutes longer. Find the speed of the rickshaw and of the bus. 
Answer: Let the speed of the rickshaw and the bus be \( x \) km/hr and \( y \) km/hr, respectively.
We know that \( \text{speed} = \frac{\text{distance}}{\text{time}} \) and \( \text{time} = \frac{\text{distance}}{\text{speed}} \)
Case I:
Time taken by Ankita to travel 2 km by rickshaw,
\( t_1 = \frac{2}{x} \) hr
Remaining distance \( = 14 - 2 = 12 \) km
Time taken by Ankita to travel remaining distance, i.e., 12 km by bus,
\( t_2 = \frac{12}{y} \) hr
It is given that:
Total time taken by rickshaw and bus \( = \frac{1}{2} \) hr
\( \Rightarrow t_1 + t_2 = \frac{1}{2} \)
\( \Rightarrow \frac{2}{x} + \frac{12}{y} = \frac{1}{2} \) ...(i)
Case II:
Time taken by Ankita to travel 4 km by rickshaw,
\( t_3 = \frac{4}{x} \) hr
Remaining distance \( = 14 - 4 = 10 \) km
Time taken by Ankita to travel remaining distance i.e., 10 km by bus,
\( t_4 = \frac{10}{y} \) hr
It is given that:
Total time taken by rickshaw and bus \( = \frac{1}{2} + \frac{9}{60} = \frac{1}{2} + \frac{3}{20} = \frac{13}{20} \) hr
\( t_3 + t_4 = \frac{13}{20} \)
\( \Rightarrow \frac{4}{x} + \frac{10}{y} = \frac{13}{20} \) ...(ii)
Let \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \)
Then eq. (i) and (ii) become
\( 2u + 12v = \frac{1}{2} \) ...(iii)
\( 4u + 10v = \frac{13}{20} \) ...(iv)
Multiplying eq. (iii) by 2 and then subtracting eq. (iv), we get
\( (4u + 24v) - (4u + 10v) = 1 - \frac{13}{20} \)
\( \Rightarrow 14v = \frac{7}{20} \)
\( \Rightarrow v = \frac{1}{40} \)
Putting the value of \( v \) in eq. (iii),
\( \Rightarrow 2u + 12\left(\frac{1}{40}\right) = \frac{1}{2} \)
\( \Rightarrow 2u = \frac{1}{2} - \frac{3}{10} = \frac{2}{10} \)
\( \Rightarrow u = \frac{1}{10} \)
\( \Rightarrow x = \frac{1}{u} = 10 \text{ km/hr} \)
\( \Rightarrow y = \frac{1}{v} = 40 \text{ km/hr} \)
Hence, the speed of the rickshaw is 10 km/hr and the speed of bus is 40 km/hr.

Question. A motorboat can travel 30 km upstream and 28 km downstream in 7 hrs. It can travel 21 km upstream and return in 5 hrs. Find the speed of the boat in still water and the speed of the stream.
Answer: Let, the speed of the boat in still water = \( x \) km/hr
the speed of the stream = \( y \) km/hr
\( \therefore \) the speed of the motorboat upstream = \( (x - y) \) km/hr
And the speed of the motorboat downstream = \( (x + y) \) km/hr
Case I:
We know that \( \text{time} = \frac{\text{distance}}{\text{speed}} \)
Time taken by motorboat to travel 30 km upstream,
\( t_1 = \frac{30}{x - y} \) hrs
Time taken by motorboat to travel 28 km downstream,
\( t_2 = \frac{28}{x + y} \) hrs
According to the given condition,
\( t_1 + t_2 = 7 \) hrs
\( \Rightarrow \frac{30}{x - y} + \frac{28}{x + y} = 7 \) ...(i)
Case II:
Time taken by motorboat to travel 21 km upstream,
\( t_3 = \frac{21}{x - y} \) hrs
Time taken by motorboat to travel 21 km downstream,
\( t_4 = \frac{21}{x + y} \) hrs
According to the given condition,
\( t_3 + t_4 = 5 \) hrs
\( \Rightarrow \frac{21}{x - y} + \frac{21}{x + y} = 5 \) ...(ii)
Let \( p = \frac{1}{x - y} \) and \( q = \frac{1}{x + y} \)
Putting these values in eq. (i) and eq. (ii) we get
\( 30p + 28q = 7 \) ...(iii)
and \( 21p + 21q = 5 \)
\( \Rightarrow p + q = \frac{5}{21} \) ...(iv)
Multiplying eq. (iv) by 28 and subtracting from eq. (iii), we get
\( (30p + 28q) - (28p + 28q) = 7 - \frac{140}{21} \)
\( \Rightarrow 2p = 7 - \frac{20}{3} \)
\( \Rightarrow 2p = \frac{1}{3} \)
\( \Rightarrow p = \frac{1}{6} \)
Putting the value of \( p \) in eq. (iv), we get
\( \Rightarrow \frac{1}{6} + q = \frac{5}{21} \)
\( \Rightarrow q = \frac{5}{21} - \frac{1}{6} = \frac{10 - 7}{42} = \frac{3}{42} \)
\( \Rightarrow q = \frac{1}{14} \)
We know that
\( p = \frac{1}{x - y} \) and \( q = \frac{1}{x + y} \)
\( \frac{1}{x - y} = \frac{1}{6} \Rightarrow x - y = 6 \) ...(v)
\( \frac{1}{x + y} = \frac{1}{14} \Rightarrow x + y = 14 \) ...(vi)
Adding eq. (v) and (vi), we get
\( 2x = 20 \Rightarrow x = 10 \)
Putting the value of \( x \) in eq. (v), we get
\( 10 - y = 6 \Rightarrow y = 4 \)
Hence, the speed of the motorboat in still water is 10 km/hr and the speed of the stream is 4 km/hr.

Question. A shopkeeper sells a saree at a profit of 8% and a sweater at a discount of 10%, thereby getting a sum ₹ 1008. If she had sold the saree at a profit of 10% and the sweater at a discount of 8%, she would have got ₹ 1028. Find the cost of the saree and the list price (price before discount) of the sweater. 
Answer: Let the cost price of a saree = ₹ \( x \)
and the list price of sweater = ₹ \( y \)
Case I:
(S. P. of saree at 8% profit) + (S.P. of a sweater at 10% discount) = ₹ 1008
\( \Rightarrow (100 + 8)\% \text{ of } x + (100 - 10)\% \text{ of } y = 1008 \)
\( \Rightarrow \frac{108x + 90y}{100} = 1008 \)
\( \Rightarrow 108x + 90y = 100800 \)
\( \Rightarrow 6x + 5y = 5600 \)
Multiplying above eq. by 46, we get
\( \Rightarrow 276x + 230y = 257600 \) ...(i)
Case II:
(S.P. of saree at 10% profit) + (S.P. of a sweater at 8% discount) = ₹ 1028
\( \Rightarrow (100 + 10)\% \text{ of } x + (100 - 8)\% \text{ of } y = 1028 \)
\( \Rightarrow 110x + 92y = 102800 \)
\( \Rightarrow 55x + 46y = 51400 \)
Multiplying above eq. by 5, we get
\( \Rightarrow 275x + 230y = 257000 \) ...(ii)
Subtracting eq. (ii) from eq. (i), we get
\( \Rightarrow (276x + 230y) - (275x + 230y) = 257600 - 257000 \)
\( x = 600 \)
Putting the value of \( x \) in the above equation, we get
\( 6x + 5y = 5600 \)
\( \Rightarrow 6(600) + 5y = 5600 \)
\( \Rightarrow 5y = 5600 - 3600 \)
\( \Rightarrow y = \frac{2000}{5} = 400 \)
Hence, the cost price of the saree and the list price (price before discount) of the sweater are ₹ 600 and ₹ 400, respectively.

Question. Ruhi invested a certain amount of money in two schemes A and B, which offer interest at the rate of 8% per annum and 9% per annum, respectively. She received ₹ 1860 as annual interest. However, had she interchanged the amount of investments in the two schemes, she would have received ₹ 20 more as annual interest. How much money did she invest in each scheme?
Answer: Let the money invested in scheme A = ₹ \( x \)
and the money invested in scheme B = ₹ \( y \)
Case I:
Ruhi invested ₹ \( x \) at 8% p.a. + Ruhi invested ₹ \( y \) at 9% p.a. and received ₹ 1860 as annual interest.
We know that simple interest, \( SI = \frac{\text{Principle} \times \text{Rate} \times \text{Time}}{100} \)
Interest earned when \( x \) invested at 8% per annum on scheme A,
\( SI_1 = \frac{x \times 8 \times 1}{100} = \frac{8x}{100} \)
Interest earned when \( y \) invested at 9% per annum on scheme B,
\( SI_2 = \frac{y \times 9 \times 1}{100} = \frac{9y}{100} \)
Interest at 8% per annum on scheme A + Interest at 9% per annum on scheme B = 1860
\( \Rightarrow \frac{8x}{100} + \frac{9y}{100} = 1860 \)
\( \Rightarrow 8x + 9y = 186000 \) ...(i)
Case II:
Ruhi invested ₹ \( y \) at 8% p.a. + Ruhi invested ₹ \( x \) at 9% p.a. and received ₹ (1860 + 20) as annual interest.
Interest earned when \( y \) invested at 8% per annum on scheme A,
\( SI_1 = \frac{y \times 8 \times 1}{100} = \frac{8y}{100} \)
Interest earned when \( x \) invested at 9% per annum on scheme B,
\( SI_2 = \frac{x \times 9 \times 1}{100} = \frac{9x}{100} \)
Interest at 8% per annum on scheme A + Interest at 9% per annum on scheme B = 1880
\( \Rightarrow \frac{8y}{100} + \frac{9x}{100} = 1880 \)
\( \Rightarrow 9x + 8y = 188000 \) ...(ii)
Multiplying eq. (i) by 9 and eq. (ii) by 8 and then subtracting them, we get
\( (72x + 81y) - (72x + 64y) = 1674000 - 1504000 \)
\( \Rightarrow 81y - 64y = 170000 \)
\( \Rightarrow 17y = 170000 \)
\( \Rightarrow y = 10000 \)
Putting the value of \( y \) in eq. (i), we get
\( \Rightarrow 8x + 9(10000) = 186000 \)
\( \Rightarrow 8x = 186000 - 90000 \)
\( \Rightarrow 8x = 96000 \)
\( \Rightarrow x = 12000 \)
Hence, Ruhi invested ₹ 12000 and ₹ 10000 in schemes A and B respectively.

Question. Two water taps together can fill a tank in \( 1\frac{7}{8} \) hours. The tap with a larger diameter takes 2 hours less than the tap with the smaller one to fill the tank separately. Find the time in which each tap can fill the tank. 
Answer: Let the smaller tap fills the tank in \( x \) hrs.
\( \therefore \) the larger tap fills the tank in \( (x - 2) \) hrs.
Time taken by both the taps together \( = \frac{15}{8} \) hrs.
Now, work done by the smaller tap in an hour \( = \frac{1}{x} \)
Work done by the larger tap in an hour \( = \frac{1}{x - 2} \)
Now, according to the given condition:
\( \frac{1}{x} + \frac{1}{x - 2} = \frac{8}{15} \)
\( \Rightarrow \frac{(x - 2) + x}{x(x - 2)} = \frac{8}{15} \)
\( \Rightarrow \frac{2x - 2}{x^2 - 2x} = \frac{8}{15} \)
\( \Rightarrow 8x^2 - 16x = 30x - 30 \)
\( \Rightarrow 8x^2 - 46x + 30 = 0 \)
\( \Rightarrow 4x^2 - 23x + 15 = 0 \)
\( \Rightarrow (4x - 3) (x - 5) = 0 \)
\( x \neq \frac{3}{4} \) (not possible)
\( \therefore x = 5 \)
Hence, the smaller tap will fill the tank in 5 hours.
The larger tap will fill the tank in \( 5 - 2 = 3 \) hours.

Question. Rahul had some bananas and he divided them into two lots A and B. He sold the first lot at the rate of ₹ 2 for 3 bananas and the second lot at the rate of ₹ 1 per banana and got a total of ₹ 400. If he had sold the first lot at the rate of ₹ 1 per banana and the second lot at the rate of ₹ 4 for 5 bananas, his total collection would have been ₹ 460. Find the total number of bananas he had.
Answer: Let the number of bananas in lot A = \( x \)
and the number of bananas in lot B = \( y \)
Case I:
Sold the first lot at the rate of ₹ 2 for 3 bananas + Sold the second lot at the rate of ₹ 1 per banana = Amount received
S.P. of 3 bananas of lot A = ₹ 2
\( \Rightarrow \) S.P. of 1 banana of lot A = ₹ \( \frac{2}{3} \)
\( \Rightarrow \) S.P. of \( x \) bananas of lot A = ₹ \( \frac{2x}{3} \)
S.P. of 1 banana of lot B = ₹ 1
\( \Rightarrow \) S.P. of \( y \) bananas of lot B = ₹ \( y \)
As per given condition
\( \frac{2x}{3} + y = 400 \)
\( 2x + 3y = 1200 \) ...(i)
Case II:
Sold the first lot at the rate of ₹ 1 per banana + Sold the second lot at the rate of ₹ 4 for 5 bananas = Amount received
S.P. of 1 banana of lot A = ₹ 1
\( \Rightarrow \) S.P. of \( x \) bananas of lot A = ₹ \( x \)
S.P. of 5 bananas of lot B = ₹ 4
\( \Rightarrow \) S.P. of 1 banana of lot B = ₹ \( \frac{4}{5} \)
\( \Rightarrow \) S.P. of \( y \) bananas of lot B = ₹ \( \frac{4y}{5} \)
As per the given condition
\( x + \frac{4y}{5} = 460 \)
\( 5x + 4y = 2300 \) ...(ii)
Multiplying eq. (i) by 4 and eq. (ii) by 3 and then subtracting them, we get
\( (8x + 12y) - (15x + 12y) = 4800 - 6900 \)
\( \Rightarrow -7x = -2100 \)
\( \Rightarrow x = 300 \)
Putting the value of \( x \) in eq. (i), we get
\( 2x + 3y = 1200 \)
\( \Rightarrow 2(300) + 3y = 1200 \)
\( \Rightarrow 3y = 1200 - 600 \)
\( \Rightarrow 3y = 600 \)
\( \Rightarrow y = 200 \)
Total number of bananas
= Number of bananas in lot A + Number of bananas in lot B
\( = (x + y) \)
\( = (300 + 200) \)
\( = 500 \)
Hence, the total number of bananas he had is 500.

Question. The angles of a cyclic quadrilateral ABCD are \( \angle A = (6x + 10)^\circ \), \( \angle B = (5x)^\circ \), \( \angle C = (x + y)^\circ \) and \( \angle D = (3y - 10)^\circ \). Find \( x \) and \( y \) and hence the values of the four angles. 
Answer: It is given that,
\( \angle A = (6x + 10)^\circ \)
\( \angle B = (5x)^\circ \)
\( \angle C = (x + y)^\circ \)
\( \angle D = (3y - 10)^\circ \)
We know that by property of cyclic quadrilateral:
Sum of opposite angles = 180°
\( \angle A + \angle C = 180^\circ \)
\( \Rightarrow (6x + 10) + (x + y) = 180 \)
\( \Rightarrow 6x + 10 + x + y = 180 \)
\( \Rightarrow 7x + y = 170 \) ...(i)
Also, \( \angle B + \angle D = 180^\circ \)
\( \Rightarrow 5x + (3y - 10) = 180 \)
\( \Rightarrow 5x + 3y = 180 + 10 \)
\( \Rightarrow 5x + 3y = 190 \) ...(ii)
Multiplying eq. (i) by 3 and then subtracting eq. (ii) from it, we get
\( 3(7x + y) - (5x + 3y) = 3(170) - 190 \)
\( \Rightarrow 16x = 320 \)
\( \Rightarrow x = 20 \)
Putting the value of \( x = 20 \) in eq. (i), we get
\( 7(20) + y = 170 \)
\( \Rightarrow y = 30 \)
\( \angle A = (6x + 10)^\circ = (6 \times 20 + 10)^\circ = 130^\circ \)
\( \angle B = (5x)^\circ = (5 \times 20)^\circ = 100^\circ \)
\( \angle C = (x + y)^\circ = (20 + 30)^\circ = 50^\circ \)
\( \angle D = (3y - 10)^\circ = (3 \times 30 - 10)^\circ = 80^\circ \)
Hence, the required values of \( x \) and \( y \) are 20 and 30 respectively, and the values of the four angles i.e., \( \angle A, \angle B, \angle C \) and \( \angle D \) are 130°, 100°, 50°, and 80° respectively.

SOLVED EXAMPLES

Question. Solve the following system of linear equations by using the method of elimination (substitution)
\( 2x - 3y = 7 \)
\( x + y = 1 \)

Answer: The given system of equations is
\( 2x - 3y = 7 \) ...(1)
\( x + y = 1 \) ...(2)
From (2), we get \( y = 1 - x \) ...(3)
Substituting \( y = 1 - x \) in (1), we get
\( 2x - 3(1 - x) = 7 \)

\( \implies \) \( 2x - 3 + 3x = 7 \)

\( \implies \) \( 5x = 10 \)

\( \implies \) \( x = \frac{10}{5} = 2 \)
Putting \( x = 2 \) in (3), we get
\( y = 1 - 2 = -1 \)
Hence, the solution of the given system is \( x = 2, y = -1 \)
Verification. On putting \( x = 2, y = -1 \), both the equations are satisfied. Hence, the answer is correct.

Question. Solve the following pair of linear equations by the substitution method:
(i) \( x + y = 14, x - y = 4 \)
(ii) \( s - t = 3, \frac{s}{3} + \frac{t}{2} = 6 \)
(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) \( \frac{3x}{2} - \frac{5y}{3} = -2, \frac{x}{3} + \frac{y}{2} = \frac{13}{6} \)

Answer: (i) The given system of equations is
\( x + y = 14 \) ...(1)
and \( x - y = 4 \) ...(2)
From (1), \( y = 14 - x \)
Substituting \( y = 14 - x \) in (2), we get
\( x - (14 - x) = 4 \)

\( \implies \) \( x - 14 + x = 4 \)

\( \implies \) \( 2x = 4 + 14 \)

\( \implies \) \( 2x = 18 \)

\( \implies \) \( x = 9 \)
Putting \( x = 9 \) in (1), we get
\( 9 + y = 14 \)

\( \implies \) \( y = 14 - 9 \)

\( \implies \) \( y = 5 \)
Hence, the solution of the given system of equations is \( x = 9, y = 5 \).

(ii) The given system of equations is
\( s - t = 3 \) ...(1)
\( \frac{s}{3} + \frac{t}{2} = 6 \) ...(2)
From (1), \( s = 3 + t \)
Substituting \( s = 3 + t \) in (2), we get
\( \frac{3 + t}{3} + \frac{t}{2} = 6 \)

\( \implies \) \( 2(3 + t) + 3t = 36 \)

\( \implies \) \( 6 + 2t + 3t = 36 \)

\( \implies \) \( 5t = 30 \)

\( \implies \) \( t = 6 \)
Putting \( t = 6 \) in (1), we get
\( s - 6 = 3 \)

\( \implies \) \( s = 3 + 6 = 9 \)
Hence, the solution of the given system of equations is \( s = 9, t = 6 \).

(iii) The given system of equations is
\( 3x - y = 3 \) ...(1)
and \( 9x - 3y = 9 \) ...(2)
From (1), \( y = 3x - 3 \)
Substituting \( y = 3x - 3 \) in (2), we get
\( 9x - 3(3x - 3) = 9 \)

\( \implies \) \( 9x - 9x + 9 = 9 \)
or \( 9 = 9 \)
This statement is true for all value of \( x \). However, we do not get a specific value of \( x \) as a solution. So, we cannot obtain a specific value of \( y \). This situation has arisen because both the given equations are the same.
\( \therefore \) Equations (1) and (2) have infinitely many solutions.

(iv) The given system of equations is
\( 0.2x + 0.3y = 1.3 \)

\( \implies \) \( 2x + 3y = 13 \) ...(1)
and \( 0.4x + 0.5y = 2.3 \)

\( \implies \) \( 4x + 5y = 23 \) ...(2)
From (2), \( 5y = 23 - 4x \)

\( \implies \) \( y = \frac{23 - 4x}{5} \)
Substituting \( y = \frac{23 - 4x}{5} \) in (1), we get
\( 2x + 3 \left( \frac{23 - 4x}{5} \right) = 13 \)

\( \implies \) \( 10x + 69 - 12x = 65 \)

\( \implies \) \( -2x = -4 \)

\( \implies \) \( x = 2 \)
Putting \( x = 2 \) in (1), we get
\( 2 \times 2 + 3y = 13 \)

\( \implies \) \( 3y = 13 - 4 = 9 \)

\( \implies \) \( y = \frac{9}{3} = 3 \)
Hence, the solution of the given system of equations is \( x = 2, y = 3 \).

(v) The given system of equations is
\( \sqrt{2}x + \sqrt{3}y = 0 \) ...(1)
and \( \sqrt{3}x - \sqrt{8}y = 0 \) ...(2)
From (2), \( \sqrt{8}y = \sqrt{3}x \)

\( \implies \) \( y = \frac{\sqrt{3}x}{\sqrt{8}} \)
Substituting \( y = \frac{\sqrt{3}x}{\sqrt{8}} \) in (1), we get
\( \sqrt{2}x + \sqrt{3} \left( \frac{\sqrt{3}x}{\sqrt{8}} \right) = 0 \)

\( \implies \) \( \sqrt{2}x + \frac{3x}{\sqrt{8}} = 0 \)

\( \implies \) \( \frac{\sqrt{16}x + 3x}{\sqrt{8}} = 0 \)

\( \implies \) \( 4x + 3x = 0 \)

\( \implies \) \( 7x = 0 \)

\( \implies \) \( x = 0 \)
Putting \( x = 0 \) in (1), we get
\( 0 + \sqrt{3}y = 0 \)

\( \implies \) \( y = 0 \)
Hence, the solution of the given system of equations is \( x = 0, y = 0 \).

(vi) The given system of equations is
\( \frac{3x}{2} - \frac{5y}{3} = -2 \)

\( \implies \) \( 9x - 10y = -12 \) ...(1)
and \( \frac{x}{3} + \frac{y}{2} = \frac{13}{6} \)

\( \implies \) \( 2x + 3y = 13 \) ...(2)
From (1), \( 10y = 9x + 12 \)

\( \implies \) \( y = \frac{9x + 12}{10} \)
Substituting \( y = \frac{9x + 12}{10} \) in (2), we get
\( 2x + 3 \left( \frac{9x + 12}{10} \right) = 13 \)

\( \implies \) \( 20x + 27x + 36 = 130 \)

\( \implies \) \( 47x = 130 - 36 \)

\( \implies \) \( 47x = 94 \)

\( \implies \) \( x = \frac{94}{47} = 2 \)
Putting \( x = 2 \) in (1), we get
\( 9 \times 2 - 10y = -12 \)

\( \implies \) \( -10y = -12 - 18 \)

\( \implies \) \( -10y = -30 \)

\( \implies \) \( y = \frac{-30}{-10} = 3 \)
Hence, the solution of the given system of equations is \( x = 2, y = 3 \).

Question. Solve for \( x \) and \( y \):
\( \frac{x + 1}{2} + \frac{y - 1}{3} = 8 \)
\( \frac{x - 1}{3} + \frac{y + 1}{2} = 9 \)

Answer: We have,
\( \frac{x + 1}{2} + \frac{y - 1}{3} = 8 \)

\( \implies \) \( 3x + 3 + 2y - 2 = 48 \)

\( \implies \) \( 3x + 2y = 47 \) ...(1)
And \( \frac{x - 1}{3} + \frac{y + 1}{2} = 9 \)

\( \implies \) \( 2x - 2 + 3y + 3 = 54 \)

\( \implies \) \( 2x + 3y = 53 \) ...(2)
From equation (1), \( y = \frac{47 - 3x}{2} \) ...(3)
Substituting the value of \( y \) from (3) in (2), we get
\( 2x + 3 \left( \frac{47 - 3x}{2} \right) = 53 \)

\( \implies \) \( 4x + 141 - 9x = 106 \)

\( \implies \) \( 5x = 35 \)

\( \implies \) \( x = 7 \)
Putting \( x = 7 \) in (3), we get
\( y = \frac{47 - 21}{2} = \frac{26}{2} = 13 \)
Hence, the solution of the given system is \( x = 7 \) and \( y = 13 \).

Question. Form the pair of linear equations for the following problems and find their solutions 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 charges per kilometer? How much does a person have to pay for travelling a distance of 25 km?
(v) A fraction becomes \( \frac{9}{11} \), if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator it becomes \( \frac{5}{6} \). 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.

Answer: (i) Let the numbers be \( x \) and \( y \). Difference of two numbers is 26.
i.e., \( x - y = 26 \) ...(1)
One number is three times the other.
i.e., \( x = 3y \) ...(2)
Putting \( x = 3y \) in (1), we get
\( 3y - y = 26 \)

\( \implies \) \( 2y = 26 \)

\( \implies \) \( y = 13 \)
Putting \( y = 13 \) in (2), we get \( x = 3 \times 13 = 39 \)
Hence, the numbers are \( x = 39 \) and \( y = 13 \).

(ii) Let the angles be \( x \) and \( y \). Then according to the question.
\( x + y = 180 \) ...(1)
and \( x = y + 18 \) ...(2)
Putting \( x = y + 18 \) in (1), we get
\( y + 18 + y = 180 \)

\( \implies \) \( 2y = 180 - 18 \)

\( \implies \) \( 2y = 162 \)

\( \implies \) \( y = 81 \)
Putting \( y = 81 \) in (2), we get \( x = 81 + 18 = 99 \)
Hence, angles are \( x = 99^\circ \) and \( y = 81^\circ \).

(iii) Let the cost of one bat and one ball be Rs. \( x \) and Rs. \( y \) respectively. Then,
\( 7x + 6y = 3800 \) ...(1)
\( 3x + 5y = 1750 \) ...(2)
From (2), \( 5y = 1750 - 3x \) or \( y = \frac{1750 - 3x}{5} \)
Substituting \( y = \frac{1750 - 3x}{5} \) in (1), we get
\( 7x + 6 \left( \frac{1750 - 3x}{5} \right) = 3800 \)

\( \implies \) \( 35x + 10500 - 18x = 19000 \)

\( \implies \) \( 17x = 19000 - 10500 \)

\( \implies \) \( 17x = 8500 \) or \( x = 500 \)
Putting \( x = 500 \) in (2), we get
\( 3(500) + 5y = 1750 \)

\( \implies \) \( 5y = 1750 - 1500 \)

\( \implies \) \( 5y = 250 \)

\( \implies \) \( y = \frac{250}{5} = 50 \)
Hence, the cost of one bat is Rs. 500 and the cost of one ball is Rs. 50.

(iv) Let fixed charges of taxi = Rs. \( x \)
And running charges of taxi = Rs. \( y \) per km.
According to the question,
Expenses of travelling 10 km = Rs. 105
\( \therefore x + 10y = 105 \) ...(1)
Again expenses of travelling 15 km = Rs. 155
\( \therefore x + 15y = 155 \) ...(2)

\( \implies \) \( x = 155 - 15y \)
Putting \( x = 155 - 15y \) in (1), we get
\( 155 - 15y + 10y = 105 \)

\( \implies \) \( 155 - 5y = 105 \)

\( \implies \) \( -5y = 105 - 155 \)

\( \implies \) \( -5y = -50 \)

\( \implies \) \( y = 10 \)
Putting \( y = 10 \) in (2), we get
\( x + 15 \times 10 = 155 \)

\( \implies \) \( x + 150 = 155 \)

\( \implies \) \( x = 155 - 150 = 5 \)
Hence, fixed charges of taxi is Rs. 5 and running charges per km is Rs. 10.
A person should pay for travelling 25 km \( = 5 + 25 \times 10 = 5 + 250 = \) Rs. 255.

(v) Let the numerator be \( x \) and denominator be \( y \). Then, according to the question,
Case I.
\( \frac{x + 2}{y + 2} = \frac{9}{11} \)

\( \implies \) \( 11(x + 2) = 9(y + 2) \)

\( \implies \) \( 11x + 22 = 9y + 18 \)

\( \implies \) \( 11x - 9y = -4 \) ...(1)
Case II.
\( \frac{x + 3}{y + 3} = \frac{5}{6} \)

\( \implies \) \( 6(x + 3) = 5(y + 3) \)

\( \implies \) \( 6x + 18 = 5y + 15 \)

\( \implies \) \( 6x - 5y = -3 \) ...(2)

\( \implies \) \( x = \frac{5y - 3}{6} \)
Putting \( x = \frac{5y - 3}{6} \) in (1), we get
\( 11 \left( \frac{5y - 3}{6} \right) - 9y = -4 \)

\( \implies \) \( 11(5y - 3) - 54y = -24 \)

\( \implies \) \( 55y - 33 - 54y = -24 \)

\( \implies \) \( y = 33 - 24 = 9 \)
Putting \( y = 9 \) in (1), we get
\( 11x - 9 \times 9 = -4 \)

\( \implies \) \( 11x = -4 + 81 = 77 \)

\( \implies \) \( x = 7 \)
Hence, the required fraction is \( \frac{7}{9} \).

(vi) Let the present age of Jacob and his son be \( x \) and \( y \) respectively.
Case I. After five years age of Jacob \( = (x + 5) \),
After five years the age of his son \( = (y + 5) \).
According to question,
\( x + 5 = 3(y + 5) \)

\( \implies \) \( x - 3y = 10 \) ...(1)
Case II. Five years ago Jacob's age \( = x - 5 \), and his son's age \( = y - 5 \). Then, according to question,
\( x - 5 = 7(y - 5) \)

\( \implies \) \( x = 7y - 30 \) ...(2)
Putting \( x = 7y - 30 \) from (2) in (1), we get
\( 7y - 30 - 3y = 10 \)

\( \implies \) \( 4y = 40 \)

\( \implies \) \( y = 10 \)
Putting \( y = 10 \) in (1), we get
\( x - 3 \times 10 = 10 \)

\( \implies \) \( x = 10 + 30 \)

\( \implies \) \( x = 40 \)
Hence, age of Jacob is 40 years, and age of his son is 10 years.

Question. Solve the following system of equations by the method of elimination (substitution).
\( (a + b)x + (a - b)y = a^2 + b^2 \)
\( (a - b)x + (a + b)y = a^2 + b^2 \)

Answer: The given system of equations is
\( (a + b)x + (a - b)y = a^2 + b^2 \) ...(1)
\( (a - b)x + (a + b)y = a^2 + b^2 \) ...(2)
From (2), we get \( (a + b)y = a^2 + b^2 - (a - b)x \)

\( \implies \) \( y = \frac{a^2 + b^2}{a + b} - \frac{a - b}{a + b}x \) ...(3)
Substituting \( y = \frac{a^2 + b^2}{a + b} - \frac{a - b}{a + b}x \) in (1), we get
\( (a + b)x + (a - b) \left[ \frac{a^2 + b^2}{a + b} - \frac{a - b}{a + b}x \right] = a^2 + b^2 \)

\( \implies \) \( (a + b)x + \frac{(a - b)(a^2 + b^2)}{a + b} - \frac{(a - b)^2}{a + b}x = a^2 + b^2 \)

\( \implies \) \( (a + b)x - \frac{(a - b)^2}{a + b}x = a^2 + b^2 - \frac{(a - b)(a^2 + b^2)}{a + b} \)

\( \implies \) \( (a + b)x - \left[ \frac{a^2 - 2ab + b^2}{a + b} \right]x = (a^2 + b^2) \left[ 1 - \frac{a - b}{a + b} \right] \)

\( \implies \) \( \frac{(a^2 + 2ab + b^2)x - (a^2 - 2ab + b^2)x}{a + b} = (a^2 + b^2) \left[ \frac{a + b - a + b}{a + b} \right] \)

\( \implies \) \( \frac{4ab}{a + b}x = \frac{(a^2 + b^2)2b}{a + b} \)

\( \implies \) \( 4abx = 2b(a^2 + b^2) \)

\( \implies \) \( x = \frac{a^2 + b^2}{2a} \)
Putting the value of \( x \) in (3), we get
\( y = \frac{a^2 + b^2}{a + b} - \left( \frac{a - b}{a + b} \right) \times \frac{a^2 + b^2}{2a} \)
\( = \frac{a^2 + b^2}{a + b} \left[ 1 - \frac{a - b}{2a} \right] \)
\( = \frac{a^2 + b^2}{a + b} \left[ \frac{2a - a + b}{2a} \right] \)
\( = \frac{a^2 + b^2}{a + b} \left( \frac{a + b}{2a} \right) = \frac{a^2 + b^2}{2a} \)
Hence, \( x = \frac{a^2 + b^2}{2a} \) and \( y = \frac{a^2 + b^2}{2a} \).

SOLVED EXAMPLES

Question. Solve the following system of linear equations by using the method of elimination by equating the coefficients.
\( \frac{x}{7} + \frac{y}{3} = 5 \); \( \frac{x}{2} - \frac{y}{9} = 6 \)

Answer: The given system of equations is
\( \frac{x}{7} + \frac{y}{3} = 5 \) ...(1)
\( \frac{x}{2} - \frac{y}{9} = 6 \) ...(2)
Multiplying equation (1) by 21 and equation (2) by 18, we get
\( 3x + 7y = 105 \) ...(3)
\( 9x - 2y = 108 \) ...(4)
Let us eliminate \( y \). The coefficients of \( y \) are 7 and \( -2 \). The L.C.M. of 7 and 2 is 14. So, we make the coefficients of \( y \) as 14 and \( -14 \). Multiplying equation (3) by 2 and equation (4) by 7, we get
\( 6x + 14y = 210 \) ...(5)
\( 63x - 14y = 756 \) ...(6)
Adding equation (5) & equation (6), we get
\( 69x = 966 \)

\( \implies \) \( x = 14 \)
Putting \( x = 14 \) in equation (3), we get
\( 3 \times 14 + 7y = 105 \)

\( \implies \) \( 7y = 105 - 42 = 63 \)

\( \implies \) \( y = 9 \)
Hence, the solution is \( x = 14, y = 9 \).
Verification. Substituting \( x = 14 \) and \( y = 9 \) in the given equations, we find that both the equations are satisfied. Hence, the solution is correct.

Question. Solve the following system of linear equations by using the method of elimination by equating the coefficients.
\( \sqrt{3}x - \sqrt{2}y = \sqrt{3} \)
\( \sqrt{5}x + \sqrt{3}y = \sqrt{2} \)

Answer: The given equations are
\( \sqrt{3}x - \sqrt{2}y = \sqrt{3} \) ...(1)
\( \sqrt{5}x + \sqrt{3}y = \sqrt{2} \) ...(2)
Let us eliminate \( y \). To make the coefficients of \( y \) equal, we multiply the equation (1) by \( \sqrt{3} \) and equation (2) by \( \sqrt{2} \) to get
\( 3x - \sqrt{6}y = 3 \) ...(3)
\( \sqrt{10}x + \sqrt{6}y = 2 \) ...(4)
Adding equation (3) and equation (4), we get
\( 3x + \sqrt{10}x = 5 \)

\( \implies \) \( (3 + \sqrt{10})x = 5 \)

\( \implies \) \( x = \frac{5}{3 + \sqrt{10}} = \left( \frac{5}{\sqrt{10} + 3} \right) \left( \frac{\sqrt{10} - 3}{\sqrt{10} - 3} \right) \)
\( = \frac{5(\sqrt{10} - 3)}{10 - 9} = 5(\sqrt{10} - 3) \)
Putting \( x = 5(\sqrt{10} - 3) \) in (1), we get
\( \sqrt{3} \times 5(\sqrt{10} - 3) - \sqrt{2}y = \sqrt{3} \)

\( \implies \) \( 5\sqrt{30} - 15\sqrt{3} - \sqrt{2}y = \sqrt{3} \)

\( \implies \) \( \sqrt{2}y = 5\sqrt{30} - 15\sqrt{3} - \sqrt{3} \)

\( \implies \) \( \sqrt{2}y = 5\sqrt{30} - 16\sqrt{3} \)

\( \implies \) \( y = \frac{5\sqrt{30}}{\sqrt{2}} - \frac{16\sqrt{3}}{\sqrt{2}} = 5\sqrt{15} - 8\sqrt{6} \)
Hence, the solution is \( x = 5(\sqrt{10} - 3) \) and \( y = 5\sqrt{15} - 8\sqrt{6} \).
Verification. After verifying, we find that solution is correct.