Pair of Linear Equations in Two Variables | Revision Notes

Pair of Linear Equations in Two Variables | Revision Notes

 

Pair of Linear Equations in Two Variables

  • An equation which can be put in the form ax + by + c = 0, where a, b and c are real numbers and a and b are not both zero, is called a linear equation in two variables x and y.
  • Every solution of the equation ax + by + c = 0 is a point on the line representing it.
    Or each solution (x, y), of a linear equation in two variables ax + by + c = 0, corresponds to a point on the line representing the equation and vice-versa.
  • A linear equation in two variables has an infinite number of solutions.
  • If we consider two equations of the form
    a1x + b1y + c1 = 0,
    a2x + b2y + c2 = 0,
    a pair of such equations is called a system of linear equations.
  • We have three types of systems of two linear equations.
    • Independent System, which has a unique solution. Such system is termed as a consistent system with unique solution.
    • Inconsistent System, which has no solution.
    • Dependent System, which represents a pair of equivalent equations and has an infnite number of solutions. Such system is also termed as a consistent system with infnite solutions.
  • A pair of linear equations in two variables which has a common point, i.e., which has only one solution is called a consistent pair of linear equations.
  • A pair of linear equations in two variables which has no solution, i.e., the lines are parallel to each other is called an inconsistent pair of linear equations.
  • A pair of linear equations in two variables which are equivalent and has infinitely many solutions are called dependent pair of linear equations.
    Note that a dependent pair of linear equations is always consistent with infinite number of solutions.
  • If a pair of linear equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, represents
    • intersecting lines, then \frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}
       
    • parallel lines, then  \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}
       
    • coincident lines, then  \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}
      Converse of the above statement is also true.

 

  • Graphical method of Solving a Pair of Linear Equations
    • To solve a system of two linear equations graphically :
      • Draw graph of the frst equation.
      • On the same pair of axes, draw graph of the second equation.
    • After representing a pair of linear equations graphically, only one of the following three possibilities can happen :
      1. The two lines will intersect at a point.
        If the two lines intersect at a point, read the coordinates of the point of intersection to obtain the solution and verify your answer.
      2. The two lines will be parallel.
        If the two lines are parallel, i.e., there is no point of intersection, write the system as inconsistent. Hence, no solution.
      3. The two lines will be coincident.
        If the two lines have the same graph, then write the system as consistent with infnite number of solutions.
  • Algebraic methods of Solving a Pair of Linear Equations
    • Substitution method :
      Suppose we are given two linear equations in x and y. For solving these equations by the substitution method, we proceed according to the following steps :
      Step 1. Express y in terms of x in one of the given equations.
      Step 2. Substitute this value of y in terms of x in the other equation. This gives a linear equation in x.
      Step 3. Solve the linear equation in x obtained in step 2.
      Step 4. Substitute this value of x in the relation taken in step 1 to obtain a linear equation in y. 
      Step 5. Solve the above linear equation in y to get the value of y.
      Note : We may interchange the role of x and y in the above method.

      While solving a pair of linear equations, if we get statements with no variables, we conclude as below.
      (a) If the statement is true, we say that the equations have infnitely many solutions.
      (b) If the statement is false, we say that the equations have no solution.

      When the two given equations in x and y are such that the coeffcients of x and y in one equation are interchanged in the other, then we add and subtract the two equations to get a pair of very simple equations.
    • Elimination method :
      In this method, we eliminate one of the variables and proceed using the following steps.
      Step 1. Multiply the given equations by suitable numbers so as to make the coeffcients of one of the variables equal.
      Step 2. If the equal coeffcients are opposite in sign, then add the new equations Otherwise, subtract them.
      Step 3. The resulting equation is linear in one variable. Solve it to get the value of one of the unknown quantities.
      Step 4. Substitute this value in any of the given equations.
      Step 5. Solve it to get the value of the other variable.
    • Cross multiplication method :
      The system of two linear equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, where \frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}} has a unique solution, given by
      x=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}, y=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}
      We generally write it as
      \frac{x}{b_1c_2-b_2c_1}=\frac{y}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}

      Note:
      The system of equations
      a1x + b1y + c1 = 0,           ...(i)
      a2x + b2y + c2 = 0            ...(ii)
      (a) is consistent with unique solution, if  \frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}, i.e., lines represented by equations (i) and (ii) intersect at a point.
      (b) is inconsistent, if  \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} ,  i.e., lines represented by equations (i) and (ii) are parallel and non coincident.
      (c) is consistent with infinitely many solutions, if  \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}, i.e., lines represented by equations (i) and (ii) are coincident.

 

Important Questions

  1. Solve for x and y : (a – b)x + (a + b)y = a2 – 2ab – b2 and (a + b)(x + y) = a2 + b2.
  2. Solve the following equations:
    1. \frac { 2 x - 4 } { 4 } = \frac { 3 y - 2 } { 2 } ; \frac { 4 x - 3 } { 2 } - \frac { 6 - y } { 3 } = 2 \frac { 5 } { 6 }
       
    2. \frac { 2 x y } { x + y } + \frac { 3 } { 2 } , \frac { x y } { 2 x - y } = \frac { - 3 } { 10 } ,\text{where } x + y \neq 0,2 x - y \neq 0
       
    3. \frac { 1 } { 2 x } - \frac { 1 } { y } = - 1 , \frac { 1 } { x } + \frac { 1 } { 2 y } = 8 , x , y \neq 0
  3. Solve for x and y:
    \frac { x + y } { x y } = 2 , \frac { x - y } { x y } = 6
  4. Solve:
    \frac { 2 } { 3 x + 2 y } + \frac { 3 } { 3 x - 2 y } = \frac { 17 } { 5 } ; \frac { 5 } { 3 x + 2 y } + \frac { 1 } { 3 x - 2 y } = 2
  5. Solve:
    \frac { x } { a } + \frac { y } { b } = a + b , \frac { x } { a ^ { 2 } } + \frac { y } { b ^ { 2 } } = 2 , \text{where }a , b \neq 0
  6. Solve:
    a x + b y = 1 , b x + a y = \frac { 2 a b } { a ^ { 2 } + b ^ { 2 } }
  7. Solve:
    a x + b y = a - b ; b x - a y = a + b
  8. Solve:
    \frac { b } { a } x + \frac { a } { b } y = a ^ { 2 } + b ^ { 2 } ; x + y = 2 a b
  9. Solve:
    \frac { 5 } { x - 1 } + \frac { 1 } { y - 2 } = 2 ; \frac { 6 } { x - 1 } + \frac { 3 } { y - 2 } = 1
  10. Draw the graphs of 2x + y = 6 and 2x – y + 2 = 0. Shade the region bounded by these lines and x-axis. Find the area of the shaded region.
  11. Solve graphically 4x – y = 4 and 4x + y = 12. Shade the triangular region formed by these lines and x-axis. Also, write the coordinate of the vertices of the triangle formed by these lines and x-axis.
  12. Six years hence a man’s age will be three times his son’s age and three years ago, he was nine times as old as his son. Find their present ages.
  13. Ages of two friends A and B differ by 3 years. A’s father D is twice as old as A, and B is twice as old as his sister C. Ages of C and D differ by 30 years. Find the ages of A and B.
  14. A two digit number is obtained by either multiplying the sum of the digits by 8 and adding 1, or by multiplying the difference of the digits by 13 and adding 2. Find the number. How many such numbers are there?
  15. A man wished to give Rs 12 to each person and found that he fell short of Rs 6 when he wanted to give to all persons. He therefore, distributed Rs 9 to each person and found that Rs 9 were left over. How much money did he have and how many persons were there?
  16. In a rectangle, if the length is increased and breadth reduced each by 2 meters, the area is reduced by 28 sq. m. If the length is reduced by 1 m and the breadth is increased by 2 m, the area increases by 33 sq. m. Find the length and the breadth of the rectangle.
  17. 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?
  18. Students of a class are made to stand in rows. If four students are extra in each row, there would be two rows less. If 4 students are less in each row, there would be 4 more rows. Find the number of students in the class.
  19. A person sells two articles together for Rs. 46, making a profit of 10% on one and 20% on the other. If he had sold each article at 15% profit, the result would have been the same. At what price does he sell each article ?
  20. The auto fare for the first kilometer is fxed and is different from the rate per km for the remaining distance. A man pays Rs. 57 for the distance of 16 km and Rs. 92 for a distance of 26 km. Find the auto fare for the first kilometer and for each successive kilometer.
  21. A railway half ticket costs half the full fare, but the reservation charges are the same on a half ticket as on a full ticket. One reserved frst class ticket from station A to B cots Rs 2530. Also, one reserved first class ticket and one reserved frst class half ticket from A to B cost Rs 3810. Find the full first class fare from station A to B and also the reservation charges for a ticket.
  22. There are two classrooms A and B containing students. If 5 students are shifted from room A to room B, the resulting number of students in the two rooms become equal. If 5 students are shifted from room B to room A, the resulting number of students in room A becomes double the number of students left in room B. Find the original number of students in the two rooms.
  23. A person invested some amount @ 12% simple interest and some other amount @ 10% simple interest. He received an yearly interest of Rs. 13000. But if he had interchanged the invested amounts, he would have received Rs. 400 more as interest. How much amount did he invest at different rates?
  24. It takes 12 hours to fll a swimming pool using two pipes. If the pipe of larger diameter is used for 4 hours and the pipe of smaller diameter for 9 hours, only half the pool can be flled. How long would it take for each pipe to fll the pool separately?
  25. A sailor goes 8 km downstream in 40 minutes and returns in 1 hour. Find the speed of the sailor in still water and the speed of current.
  26. 8 men and 12 boys can fnish a piece of work in 10 days while 6 men and 8 boys can fnish it in 14 days. Find the time taken by one man alone and that by one boy alone to fnish the work.
  27. A boat goes 24 km upstream and 28 km downstream in 6 hours. It goes 30 km upstream and 21 km downstream in 6 hours 30 minutes. Find the speed of boat in still water and also the speed of the stream.
  28. A person travels 600 km partly by train and partly by car. If he covers 400 km by train and the rest by car, it takes 6 hours 30 minutes. But if he travels 200 km by train and the rest by car, he takes half an hour longer. Find the speed of the car and that of the train.