Exercise. Question. Aftab tells his daughter, Seven ears ago, I was seven times as old as ou were then. Also, three ears from now, I shall be three times as old as ou will be. (Isn t this interesting?) Represent this situation algebraicall and graphicall. Solution Let the present age of father be x ears and the age of daughter be ears. Seven ears ago father s age ( x 7) r Seven ears ago daughter s age ( 7) r According to the problem, ( x 7) 7( 7) or x 7 4 After r father s age ( x + ) r After r daughter s age ( + ) r According to the condition given in the questions, x + ( + ) or x 6 To find the equivalent geometric representation, we find some points on the line representing each equation. These solutions are given below in the table. From Eq. (i), x 7 4 From Eq. (ii), x x 0 7 4 4 x + 4 7 6 7 8 Points A B C G 6 x 6 8 4 x 6 0 4 Points D E F G
Y X' 0 (0, 6) A O (7, 7) B (6, 0) (, ) E (4, 8) C F (8, 4) D 0 0 x 7 4 x 6 X (4,) G Y' Plot the points A( 06, ), B( 77, ), C( 48, ) and join them to get a straight line ABC. Similarl, plot the points D ( 6, 0), E(, ) and F ( 8, 4) and join them to get a straight line DEF. And these two lines intersect at point G ( 4, ). Question. The coach of a cricket team bus bats and 6 balls for ` 900. Later, she bus another bat and more balls of the same kind for ` 00. Represent this situation algebraicall and geometricall. Solution Let us denote the cost of one bat be ` x and one ball be `. Then, the algebraic representation is given b the following equations x + 6 900 and x + 00 To obtain the equivalent geometric representation, we find two points on the line representing each equation. That is, we find two solutions of each equations. These solutions are given below in the table For, x + 6 900 For, x + 00 x 0 00 60 0 x 00 00 400 600
Y X' O A (0, 60) D (00, 600) C (00, 400) x+ 6 900 x+ 00 B (00, 0) X Y' We plot the points A( 0, 60) and B( 00, 0) to obtain the geometric representation of x + 6 900 and C( 00, 400) and D( 00, 600) to obtain the geometric representation of x + 00. We observe that these line are coincident. Question. The cost of kg of apples and kg of grapes on a da was found to be ` 60. After a month, the cost of 4 kg of apples and kg of grapes is ` 00. Represent the situation algebraicall and geometricall. Solution Let us denote the cost of kg of apples be ` x and cost of kg of grapes be `. Then, the algebraic representation is given b the following equations x + 60 and 4x + 00 Then, the algebraic representation is given b the following equations x + 6 900 and x + 00 To obtain the equivalent geometric representation, we find two points on the line representing each equation. That is, we find two solutions of each equations. These solutions are given below in the table For, x + 6 900 For, x + 00 x 0 00 60 0 x 00 00 400 600
Y X' 00 90 80 70 60 0 40 0 0 0 x + 60 (40, 70) (4, 70) (0, 60) (0, 0) 4 x + 00 0 0 0 40 0 60 70 80 90 00 X Y' Geometric representation is shown in the above figure which is a pair of parallel lines.
Exercise. Question. From the pair of linear equations in the following problems and find their solution graphicall. (i) 0 students of class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of bos, find the number of bos and girls who took part in the quiz. (ii) pencils and 7 pens together cost ` 0, whereas 7 pencils and pens together cost ` 46. Find the cost of one pencil and that of one pen. Solution (i) Let the number of bos be x and the number of girls be. Then, the equations formed are x + 0 x + 4 Let us draw the graphs of Eqs. (i) and (ii) b finding two solutions for each of the equations. Table for line, x + 0 Table for line, X' x 0 8 0 x 0 7 Points A B E x + 4 x 0 x + 4 4 7 Points C D E Y 0 A (0, 0) 9 8 7 6 D 4 C(0, 4) (, ) x +4 E (, 7) x+ 0 B(8, ) 4 6 7 8 9 Plotting these points, we draw the lines AB and CE passing through them to represent the equations. The two lines AB and CE intersect at the point E ( 7, ). So, x and 7 is the required solution of the pair of linear equations. X
i.e., Number of bos, Number of girls 7 (ii) Let us denote the cost of one pencil be x and one pen be `. Then, the equations formed are x + 7 0 and 7x + 46 Table for line, x + 7 0 Table for line, 7x + 46 x 0 0 x 8 Y X' 6 4 0 A (, ) 7x+46 x + 7 0 4 6 7 8 9 C (8, ) B (0, 0) X 0 Plotting these points, we draw the lines AB and AC passing through them to represent the equations. The two lines AB and AC intersect at the point A (, ). So, x and is the required solution of the pair of linear equation, ie.., cost of one pencil, x ` and cost of one pen, `. Question. Y' On comparing the ratios a a b, and c, find out whether b c the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident. (i) x 4 + 8 0; 7x + 6 9 0 (ii) 9x + + 0; 8x + 6 + 4 0 (iii) 6x + 0 0; x + 9 0 Solution (i) The given pair of linear equations is x 4 + 8 0 and 7x + 6 9 0 On comparing with ax + b + c 0 We get, a, b 4, c 8; a 7, b 6, c 9
Here, 4 7 6 Q a a b b So, lines (i) and (ii) are intersecting lines. (ii) The given pair of linear equations is 9x + + 0 and 8x + 6 + 4 0 On comparing with ax + b + c 0 We get, a 9, b, c ; a 8, b 6, c 4 Here, 9 Q a b c 8 6 4 a b c So, lines (i) and (ii) are coincident lines. (iii) The given pair of linear equations is 6x + 0 0 and x + 9 0 On comparing with ax + b + c 0 We get, a 6, b, c 0; a, b, c 9 Here, 6 0 9 b c a b c 0 9 So, lines (i) and (ii) are parallel lines. Question. On comparing the ratios a b, and c, find out whether a b c the following pairs of linear equations are consistent or inconsistent. (i) x + ; x 7 (ii) x 8;4x 6 9 (iii) x + 7; 9x 0 4 (iv) x ; 0x + 6 (v) 4 x + 8; x + Solution (i) The given equation can be rewritten as x + 0, x 7 0 On comparing with ax + b + c 0 We get, a, b, c ; a, b, c 7 a Here, b, a b Thus,, i.e., a b a b Hence, the pair of linear equations is consistent.
(ii) The given equation can be rewritten as x 8 0; 4x 6 9 0 On comparing with ax + b + c 0 We get, a, b, c 8; a 4, b 6, c 9 a Here, b c 8 8 a 4, b 6, c 9 9 8 Thus,, i.e., a b c 9 a b c Hence, the pair of linear equations is inconsistent. (iii) The given equation can be rewritten as x + 7 0; 9x 0 4 0 On comparing with ax + b + c 0 We get, a, b, c 7; a 9, b 0, c 4 a b Here,,, a 9 6 b 0 6 Thus,, i.e., a b 6 6 a b Hence, the pair of linear equations is consistent. (iv) The given equation can be rewritten as x 0 ; 0x + 6 + 0 On comparing with ax + b + c 0 We get, a, b, c ; a 0, b 6, c a Here, b,, a 0 b 6 c c Thus, a i.e., b c a b c Hence, the pair of linear equations is consistent (or dependent). (v) The given equation can be rewritten as 4 x + 8 0; x + 0 4 Here, a, b, c 8; a, b, c 4 a b c 8,, a b c Thus,, i.e., a b c a b c Hence, the pair of linear equations is consistent (or dependent).
Question 4. Which of the following pairs of linear equations are consistent or inconsistent? If consistent obtain the solution graphicall? (i) x +, x + 0 (ii) x 8, x 6 (iii) x + 6 0, (iv) x 0, 4x 4 0 4x 4 0 Solution (i) Given pair of lines are x + and x + 0 Here, a, b, c ; a, b, c 0 Here, a a, b and b c a b c 0 a b c So, the given pair of linear equations are coincident. Therefore, these lines have infinitel man common solutions. Hence, the given pair of linear equations is consistent. Now, x + x If x 0, If x, 0 x 0 x 0 Points A B and x + 0 0 x If x 0, If x, x 0 ( 0 x)/ 0 Points C D E Y A O C (0, ) x+ x + D (, ) x+ 0 B E (, 0) X
Plotting the points A ( 0, ) and B(, 0 ), we get the line AB. Again, plotting the points C( 0, ), D(, ) and E (, 0 ), we get the line CDE. We observe that the lines represented b Eqs. (i) and (ii) are coincident. (ii) Given pair of lines are x 8, x 6 Here, a, b, c 8; a, b, c 6 a Here,, b a b c and 8 c 6 a b c a b c Hence, the lines represented b the given equations are parallel. Therefore, it has no solution. Hence, the given pair of lines is inconsistent. (iii) Given pair of lines are x + 6 0and 4x 4 0 Here, a, b, c 6; a 4, b, c 4 a Here,, b a 4 b c and 6 c 4 a b c a b c So, the given pair of linear equations are intersecting, therefore these lines have unique solution. Hence, given pair of linear equation is consistent. We have, x + 6 0 6 x When x 0, 6 When x, 0 x 0 6 0 Point A B E and 4x 4 0 x When x 0, When x, 0 x 0 0 Point C D E
Y X' 6 A(0,6) 4 O x + 6 4 x 40 D(,0) E (,) B (,0) X C (0, ) Plotting the points A( 0, 6 ) and B(, 0 ), we get the straight line AB. Plotting the points C( 0, ) and D(, 0 ), we get the straight line CD. The lines AB and CE intersect at E (, ). (iv) Given, x 0and 4x 4 0 Here, a, b, c ; a 4, b 4, c Here, a a, b 4 b 4, c c Y' a b c a b c So, the given pairs of linear equations are parallel. Therefore, these lines have no solution. Hence, the given pair has inconsistent. Question. Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 6 m. Find the dimensions of the garden. Solution Let the length of the garden be x metre and its width be metre. Then, perimeter of rectangular garden (length + width) ( x + ) Therefore, half perimeter ( x + ) But it is given as 6. ( x + ) 6 Also, x + 4, i.e., x 4 For finding the solution of Eqs. (i) and (ii) graphicall, we form the following table For x + 6 For x 4 x 0 4 x 0 6 0 6 6 6
Y X' 4 0 8 6 4 0 8 6 4 () x + 6 (0, 6) (6, ) (4, ) x 4 4 6 8 0 4 6 8 0 4 Draw the graphs b joining points (0, 6) and (4, ) and points (0, 6) and (6, ). The two lines intersect at point (0, 6) as shown in the graph. Hence, length 0 m and width 6 m Question 6. Given the linear equation x + 8 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is (i) intersecting lines (ii) parallel lines (iii) coincident lines Solution Given, linear equation is x + 8 0 (i) For intersecting lines, a b a b An intersecting line ma be taken as x + 9 0. (ii) For parallel lines, a b c a b c An line parallel to Eq. (i) ma be taken as 6x + 9 + 7 0. (iii) For coincident lines, a b c a b c An line coincident to Eq. (i) ma be taken as 4x + 6 6 0. () (0, 6) () X ()
Question 7. Draw the graphs of the equations x + 0 and x + 0. Determine the coordinates of the vertices of the triangle formed b these lines and x-axis and shade the triangular region. Solution Pair of linear equations are x + 0 x 0 4 x + Points A B D x + 0 x 0 4 x 6 0 Points C D E Y 6 C (0, 6) B (4,) x +0 4 D (, ) (0, ) X' (, 0) F A O x + 0 E 4 (4, 0) X Y' Plot the points A( 0, ), B( 4, ) and join them to get a line AB. Similarl, plot the points C( 06, ), D( and, ) join them to form a line CD. Clearl, the two lines intersect each other at the point D(, ). Hence, x and is the solution of the given pair of equations. The line CD cuts the x-axis at the point E ( 4, 0 ) and the line AB cuts the x-axis at the point F ( 0., ) Hence, the coordinates of the vertices of the triangle are D(, ), E(, 40) and F( 0, ).
Exercise. Question. Solve the following pair of linear equations b the substitution method. (i) x + 4, x 4 (ii) s t, s + t 6 (iii) x, 9x 9 (iv) 0.x + 0.., 0.4x + 0.. (v) x + 0, x 8 0 (vi) x, x + 6 Solution (i) We have, x + 4 and x 4 From Eq. (ii), x 4 (iii) Substituting from Eq. (iii) in Eq. (i), we get x + x 4 4 x 8 x 9 On substituting x 9 in Eq. (iii), we get 9 4 x 9, (ii) We have, s t and s t + 6 From Eq. (i), s t + (iii) On substituting s from Eq. (iii) in Eq. (ii), we get t + t + 6 ( t + ) + t 6 6 ( t + ) + t 6 t + 6 6 t 0 t 6 From Eq. (iii), s 6+ 9 Hence, s 9and t 6 (iii) We have, x and 9x 9
From Eq. (i), x (iii) On substituting from Eq. (iii) in Eq. (ii), we get 9x ( x ) 9 9 9 It is a true statement. Hence, ever solution of Eq. (i) is a solution of Eq. (ii) and vice-versa. On putting x k in Eq. (i), we get k k x k, k is a solution for ever real k. Hence, infinitel man solutions exist. (iv) We have, 0.x + 0.. and 0.4x + 0.. 4 x + 0 0 0 From Eq. (i), x + 0 0 0 x + x x (iii) On substituting from Eq. (iii) in Eq. (ii), we get 4 ( x) x + 0 0 0 4 x + ( x) x + 6 0x 69 x 69 6 4 x On substituting x in Eq. (iii), we get 9 Hence, x and (v) We have, x + 0 and x 8 0 From Eq. (ii), 8 x x 8 (iii) On substituting from Eq. (iii) in Eq. (i), we get x x + 0 8 x x + 0 8 8x + x 0 6x + x 0 4x + x 0 7x 0 x 0 Putting x 0 in Eq. (iii), 0
(vi) We have, x and x + 6 From Eq. (ii), x x, 6 6 ( x) 6 ( x) (iii) On substituting from Eq. (iii) in Eq. (i), we get x ( x) x ( 9 x) 8 x 8 8 ( 9 x) (Multipl b 8) 7x 0( x) 6 7x 0 + 0x 6 7x + 0x 0 6 47x 94 x 94 47 x On substituting x in Eq. (iii), we get 9 i.e., Question. Solve x + and x 4 4 and hence find the value of m for which mx +. Solution We have, x + and x 4 4 From Eq. (ii), 4 x + 4 or x + (iii) On substituting from Eq. (iii) in Eq. (i) x x + + 4x + ( x + ) 4x + x + 6 7x 6 7x 4 x +
On substituting x in Eq. (iii), + 0 On substituting x and in the equation mx +, we get m ( ) + m + m m Question. Form the pair of linear equations for the following problems and find their solution b substitution method. (i) The difference between two numbers is 6 and one number is three times the other. Find them. (ii) The larger of two supplementar angles exceeds the smaller b 8 degrees. Find them. (iii) The coach of a cricket team bus 7 bats and 6 balls for ` 800. Later, she bus bats and balls for ` 70. Find the cost of each bat and each ball. (iv) The taxi charges in a cit consist of a fixed charge together with the charge for the distance covered. For a distance of 0 km, the charge paid is ` 0 and for a journe of km, the charge paid is `. What are the fixed charges and the charge per km? How much does a person have to pa for travelling a distance of km? (v) A fraction becomes 9, if is added to both the numerator and the denominator. If, is added to both the numerator and the denominator it becomes. Find the fraction. 6 (vi) Five ears hence, the age of Jacob will be three times that of his son. Five ears ago, Jacob s age was seven times that of his son. What are their present ages? Solution (i) Let the two numbers be x and ( x > ). We are given that, x 6 and x The linear equations are x 6 and x On substituting x from Eq. (ii) in Eq. (i), we get 6 6 From Eq. (ii), x 9 x 9 Hence, the two numbers are 9 and. (ii) Let the supplementar angles be x and ( x > ) Now, we are given that x + 80 and x 8 From Eq. (ii), x 8 (iii) On substituting from Eq. (iii) in Eq. (i), x + x 8 80 x 98 x 99
98 99 Then, from Eq. (iii), 99 8 8 8 Hence, the angles are 99 and 8. (iii) Let cost of one bat ` x Cost of one ball ` We are given that, Cost of 7 bats and 6 balls ` 800, i.e., 7x + 6 800 Cost of bats and balls ` 70, i.e., x + 70 70 x From Eq. (ii), 70 x (iii) On substituting from Eq. (iii) in Eq. (i), we get ( 70 x) 7x + 6 800 On multipling b, we get x + 6 ( 70 x) 800 x + 000 8x 9000 x 8x 9000 000 7x 800 x 00 From Eq. (iii), 70 00 70 00 0 0 Hence, cost of one bat ` 00 Cost of one ball ` 0 (iv) Let fixed charge be ` x and charge per km be `. Then, b given conditions, x + 0 0 and x + From Eq. (i), x ( 0 0 ) (iii) On substituting x from Eq. (iii) in Eq. (ii), we get 0 0 + 0 0 0 From Eq. (iii), x 0 0 0 0 00 x Hence, fixed charges ` Rate per kilometre ` 0 Amount to be paid for travelling km x + + 0 + 0 ` x
+ (v) Let x be the fraction where x and are positive integers. Then, b given conditions, x + + 9 and x + + 6 ( x + ) ( + ) + + ( x + ) 9 ( + ) and 6 x + 9 + 8 6x 8 x 9 + 4 0 and 6x + 0 From Eq. (ii), 6x + or 6x + (iii) On substituting from Eq. (iii) in Eq. (i), 6x x 9 + + 4 0 x 9 ( 6x + ) + 0 0 x 4x 7 + 0 0 x 7 6 7 + 4 + 4 From Eq. (iii), 9 Hence, the fraction is 7 9. (vi) Let x (in ear) be the present age of Jacob s son and (in ear) be the present age of Jacob. Five ears hence, it has relation b given condition, ( + ) ( x + ) + x + x + 0 0 Five ears ago, it has relation ( ) 7( x ) 7x 0 0 From Eq. (i), x + 0 (iii) On substituting from Eq. (iii) in Eq. (ii), 7x ( x + 0) 0 0 4x 40 0 4x 40 x 0 On substituting x 0 in Eq. (iii), we get 40 0 + 0 Hence, the present age of Jacob is 40 r and present age of his son is 0 r.
Exercise.4 Question. Solve the following pair of linear equations b the elimination method and the substitution method. (i) x + and x 4 (ii) x + 4 0 and x (iii) x 4 0and 9x + 7 (iv) x + and x Solution (i) Solution b elimination method We have, x + and x 4 On multipling Eq. (i) b and Eq. (ii) b and adding, we get ( x + ) + ( x ) + 4 x + + x + 4 x 9 x 9 From Eq. (i), substituting x 9, we get 9 + 9 9 9 6 Hence, x and Solution b substitution method We have, x + and x 4 From Eq. (i), x (iii) On substituting from Eq. (iii) in Eq. (ii), x ( x) 4 x + x 4 x 9 x 9 Then, from Eq. (iii), substituting x 9, 9 9 9 6 Hence, x and 6 6
(ii) Solution b elimination method We have, x + 4 0 and x On multipling Eq. (i) b and Eq. (ii) b and adding, we get ( x + 4) + ( x ) 0+ x + 4 + 4x 4 0+ 4 7x 4 x From Eq. (i), substituting x in Eq. (i), we get + 4 0 4 4 Hence, x and Solution b substitution method We have, x + 4 0 and x From Eq. (ii), x x (Divide b ) (iii) On substituting from Eq. (iii) in Eq. (i), x + 4( x ) 0 x + 4x 4 0 7x 4 x From Eq. (iii), Hence, x and (iii) Solution b elimination method We have, x 4 and 9x + 7 On multipling Eq. (i) b and Eq. (ii) b, we get 9x (iii) and 9x 7 (iv) On subtracting Eq. (iv) from Eq. (iii), ( 9x ) ( 9x ) 7 + From Eq. (i), substituting, we get x 4 9x + 9x 7 x 9 9 Hence, x and
Solution b substitution method We have, x 4 9x + 7 9x 7 From Eq. (ii) 9x 7or (iii) Substituting from Eq. (iii) in Eq. (i), we get 9x 7 x 4 6x 4x + 8 9x 7 x 9 From Eq. (iii), substituting x 9, we get 9 9 7 8 9 0 6 9 Hence, x and (iv) Solution b elimination method We have, x + and x On multipling Eq. (i) b and Eq. (ii) b, we get x + (iii) and x 6 (iv) On adding Eq. (iii) and Eq. (iv), we get x + x + 6 x x Now, from Eq. (ii) substituting x, we get Hence, x and Solution b substitution method We have, x + and x From Eq. (ii), x ( x ) (iii)
On substituting ( x )in Eq. (i), we get x + ( x ) x + ( x ) x + 4( x ) x x 0 x Now, from Eq. (iii) substituting x, we get ( ) Hence, x and Question. Form the pair of linear equations in the following problems and find their solutions (if the exist) b the elimination method (i) If we add to the numerator and subtract from the denominator, a fraction reduces to. It becomes, if we onl add to the denominator. What is the fraction? (ii) Five ears ago, Nuri was thrice as old as Sonu. Ten ears 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 b reversing the order of the digits. Find the number. (iv) Meena went to a bank to withdraw ` 000. She asked the cashier to give her ` 0 and ` 00 notes onl. Meena got notes in all. Find how man notes of ` 0 and ` 00 she received. (v) A lending librar has a fixed charge for the first three das and an additional charge for each da thereafter. Saritha paid ` 7 for a book kept for seven das, while Sus paid ` for the book she kept for five das. Find the fixed charge and the charge for each extra da. Solution (i) Let the fraction be x. According to the given conditions, x + ; x + x + ; x + x and x On subtracting Eq. (i) from Eq. (ii), ( x ) ( x ) + x On substituting x in Eq. (i), Hence, the fraction is.
(ii) Let present age of Nuri x (in ears). Present age of Sonu (in ears) According to the given conditions, Five ears ago, x ( ) x 0 Ten ears later, x + 0 ( + 0) x 0 On subtracting Eq. (i) from Eq. (ii), ( x ) ( x ) 0 + 0 + 0 0 From Eq. (ii), substituting 0, we get x + 0 0 + 0 x 0 Therefore, present age of Nuri 0 r and present age of Sonu 0 r (iii) Let x be the digit at unit s place and be the digit at ten s place of the value of the number x + 9 Value of the number x + 0 When we reverse the digits, the value of the new number + 0x We are given that 9 ( x + 0) ( + 0x) 9x + 90 + 0x 88 x x 8 On substituting x from Eq. (ii) in Eq. (i), 8 + 9 9 9 Then, from Eq. (ii), x 8 x 8 Hence, the number is x + 0 8 + 0 8 + 0 8 (iv) Let number of ` 0 notes x Number of ` 00 notes B given condition, x + Also, 0 x + 00 000 or x + 40 (Divide b 0) Subtracting Eq. (i) from Eq. (ii) ( x + ) ( x + ) 40 On substituting in Eq. (i), we get Number of ` 0 notes x 0 Number of ` 00 notes x + x 0
(v) Let the fixed charges for the first three das be ` x. Let the additional charge per da be `. Saritha paid for 7 das ` 7 B given condition, x + 4 7 (Q` 4 are to be paid for 4 extra das) x + 4 7 and in the case of Sus, On subtracting Eq. (ii) from Eq. (i), we get x + 7 From Eq. (i), substituting, we get x + 4 7 x Fixed charge for first three das ` Addition charge per extra da `
Exercise. Question. Which of the following pairs of linear equations has unique solution, no solution, or infinitel man solutions. In case there is a unique solution, find it b using cross multiplication method. (i) x 0and x 9 0 (ii) x and x 8 (iii) x 0 and 6x 0 40 (iv) x 7 0 and x 0 Solution (i) We have, x 0 and x 9 0 Here, a, b, c ; a, b 9, c a b c a b 9 c a b c a b c Hence, no solution exists. (ii) We have, x and x 8 Here, a, b, c ; a, b, c 8 a b, a b a b a b Hence, we have a unique solution. x 0 x 8 0 B cross multiplication, we get x 8 8 x {( ) ( 8) ( ) ( )} {( )( ) ( 8) ( )} {( ) ( ) ( ) ( ) } x ( 8 0) ( 6) ( 4 ) x
x and x and (iii) We have, x 0 and 6x 0 40 Here, a, b, c 0 ; a 6, b 0, c 40 a a 6, c 0 0 c 40 a a There are infinitel man solutions. b c Each b c (iv) We have, x 7 0 and x 0 Here, a, b, c 7; a, b, c a b a b Here, a b a b Hence, unique solution exists. x 7 0 x 0 B cross multiplication, we have x 7 7 x {( ) ( ) ( ) ( 7 )} {( 7 ) ( ) ( )( )} {( ) ( ) ( ) ( )} x ( 4 ) ( ) ( 9) x 4 6 6 x and 4 6 6 6 x 4 6 and 6 6 x 4 and Question. (i) For which values of a and b does the following pair of linear equations have an infinite number of solutions? x 7 ( a b) x ( a b) a b
(ii) For which value of k will the following pair of linear equations have no solution? x + ( k ) x + ( k ) k + Solution (i) We have, x + 7 0 and ( a b) x + ( a + b) ( a + b ) 0 Here, a, b, c 7; a a bb, a+ bc, a+ b For infinite number of solutions, we have 7 a b a + b ( a + b ) a b a + b a + b 7 From first and second, we have a b a + b a b a + b Q a a b c b c a b From second and third, we have a + b a + b 7 7a + 7b 9a + b 6 4b a 6 b a (Divide b ) From Eqs. (i) and (ii), eliminating a, Substituting b in Eq. (i), we get b b b a a (ii) We have, x + 0 and ( k ) x + ( k ) ( k + ) 0 Here, a, b, c ; a k, b k, c ( k + ) a a k, b b k, c c ( k + ) For no solution a a b ie.., b k k + k Now, if k, then we have k k ( k ) k
a a 4, b b Thus, for k, we have a b c a b c Hence, it has no solution when k., c c k ( ) 4 Question. Solve the following pair of linear equations b the substitution and cross multiplication methods 8x 9 x 4 Solution We have, 8x 9 and x 4 or 8x 9 0 and x 4 0 B substitution method From Eq. (ii), 4 x (iii) Substitute the value of x, from Eq. (iii) in Eq. (i), we get 8 4 9 0 6 7 0 0 From Eq. (iii) substitute, we get 4 4 0 x 6 B cross multiplication method We have, 8x 9 0 and x 4 0 x 9 9 8 8 4 4 x Then, we have, {( ) ( 4) ( ) ( 9)} {( 9) ( ) ( 4) ( 8)} {() 8 () () ()} x ( 0 8) ( 7 ) ( 6 ) x x and
Question 4. Form the pair of linear equations in the following problems and find their solution (if the exist) b an algebraic method (i) A part of monthl hostel charges is fixed and the remaining depends on the number of das one has taken food in the mess. When a student A takes food for 0 das she has to pa ` 000 as hostel charges whereas a student B, who takes food for 6 das, pas ` 80 as hostel charges. Find the fixed charges and the cost of food per da. (ii) A fraction becomes when is subtracted from the numerator and it becomes 4 when 8 is added to its denominator. Find the fraction. (iii) Yash scored 40 marks in a test, getting marks for each right answer and losing mark for each wrong answer. Had 4 marks been awarded for each correct answer and marks been deducted for each incorrect answer, then Yash would have scored 0 marks. How man questions were there in the test? (iv) Places A and B are 00 km apart on a highwa. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, the meet in hours. If the travel towards each other, the meet in hour. What are the speeds of the two cars? (v) The area of a rectangle gets reduced b 9 sq units, if its length is reduced b units and breadth is increased b units. If we increase the length b units and the breadth b units, the area increases b 67 sq units. Find the dimensions of the rectangle. Solution. (i) Let fixed hostel charge (monthl) ` c and cost of food for one da ` x In the case of A student, 0x + c 000 In the case of B student, 6x + c 80 On subtracting Eq. (i) from Eq. (ii), we get 6x 0x 80 000 6x 80 x 0 From Eq. (i), c 000 0x 000 0 0 000 600 400 Hence, monthl fixed charges ` 400 Cost of food per da ` 0
(ii) Let the fraction be x. According to the given condition, x and x + 8 x and 4x 8 On subtracting Eq. (i) from Eq. (ii), we get ( 4x ) ( x ) 8 x On putting x in Eq. (i), we get Hence, the fraction is. (iii) Let number of right answers is x and number of wrong answers is. Total number of questions x + In first case, marks awarded for x right answer x Marks lost for wrong answer Then, x 40 (B given condition) In second case, marks awarded for x right answer 4x Marks lost for wrong answer 4 (QLosing marks we take sign) Then, 4x 0 (B given condition) From Eq. (i), x 40 (iii) From Eq. (ii) and Eq. (iii) eliminating, 4x ( x 40) 0 4x 6x + 80 0 x 0 x On putting x in Eq. (iii), 40 4 40 Total number of questions x + + 0 (iv) Let speed of car I x kmh and speed of car II kmh Car I starts from point A and car II starts from point B. First case Car I Car II A B 00 km Two cars meet at point C after h. AC distance travelled b car I in h x km? C
BC distance travelled b car II in h km Here, AC BC AB x 00 (QAB 00 km) x 0 (Divide b ) Second case Two cars meet at C after one hour. ie.., x + 00 On adding Eq. (i) and Eq. (ii), we get x 0 x 60kmh On putting x 60 in Eq. (ii), we get 60 + 00 40 km h Hence, the speed of the two cars are respectivel 60 kmh and 40 kmh. (v) Let x and be the length and breadth of a rectangle. In first case, area is reduced b 9 sq units. When length x units and breadth + Q units Area of rectangle Length Breadth x According to the given condition, x ( x ) ( + ) 9 x ( x + x ) 9 x + + 9 x 6 In second case, area increase b 67 sq units. When length x + and breadth +. ( x + ) ( + ) x 67 x + x + + 6 x 67 x + 6 On multipling Eq. (i) b and Eq. (ii) b, we get On adding Eq. (iii) and Eq. (iv), A 9x 8 (iii) 0x + 0 (iv) 9x x 7 On substituting x 7 in Eq. (ii), we get 4 + 6 7 9 Hence, length 7 units and breadth 9 units x C B
Exercise.6 Question. Solve the following pairs of equations b reducing them to a pair of linear equations (i) + and + x x 6 (ii) + and 4 9 x x (iii) 4 + 4 and 4 x x 6 (iv) + and ( x ) ( ) ( x ) ( ) (v) 7 x and 8 x + 7 x x (vi) 6x + 6x and x + 4 x 0 (vii) + 4 and ( x + ) ( x ) ( x + ) ( x ) (viii) + and ( x + ) ( x ) 4 ( x + ) ( x ) 8 Solution (i) We have, + and + x x On putting x u and v, we get u + v and u + v On multipling b 6 on both sides, in both equations, we get u + v u + v On multipling Eq. (i) b and Eq. (ii) b, then subtracting later from first, we get ( u + v) ( u + v) 9u 4u 6 6 u 0 u Then, substituting u in Eq. (i), we get 6 + v v 6 v Now, u and v x and 6 6
x and 4 9 (ii) We have, + and x x On putting u and v, we get x u + v and 4u 9v On multipling Eq. (i) b and Eq. (ii) b and then adding, we get ( u + v) + ( 4u 9v) x 0u u On putting u in Eq. (i), + v v v Now, u and v and x x and x 4 and 9 (iii) We have, 4 + 4 and 4 x x On putting X, we get x 4X + 4 and X 4 On multipling Eq. (i) b 4 b Eq. (ii) and and then adding, we get 6X + 9X 4 4 + x 6 + 69 X X Then, x x From Eq. (i), substituting X, we get 4 + 4 4 0 6 Hence, x and (iv) We have, 6 + and ( x ) ( ) ( x ) ( )
On putting u x and v, we get u + v and 6u v On multipling Eq. (i) b and Eq. (ii) b, then adding, we get u + 6u 6 + u 7 u From Eq. (i), + v v v Now, u and v x and x and x + and + x 4 and (v) We have, 7 x and 8 x + 7 x x 7x and 8 x 7 + x x x x 7 8 7 and + x x On putting x u and v, we get 7v u and 8v + 7u On multipling Eq. (i) b 7 and Eq. (ii) b, then adding, we get 49v + 6v + 0 6v 6 v On substituting v in Eq. (ii), we get 8 + 7u 7u 8 7u 7 u Now, u and v and x x and (vi) We have, 6x + 6x and x + 4 x Dividing b x, we get
6 + 6 and + 4 x x or 6 + 6 and 4 + x x On putting x u and v, we get u + 6v 6 and 4u + v On multipling Eq. (i) b 4 and Eq. (ii) b, subtracting later from first, we get 4( u + 6v) ( 4u + v) 4 6 u + 4v u 6v 4 8v 9 v On substituting v in Eq. (ii), 4u + 4u 4 u Now, u and v and x x and 0 (vii) We have, + 4 and ( x + ) ( x ) ( x + ) ( x ) On putting u and v x + x 0u + v 4 and u v On multipling Eq. (i) b and Eq. (ii) b and then adding, we get ( 0u + v) + ( u v) 4 + ( ) 0u + 0v + 0u 0v 0 4 80u 6 u From Eq. (i), substituting u, we get 0 + v 4 v v Now, u and v and x + x x + (iii) and x (iv) On adding Eqs. (iii) and (iv), we get
x 6x On putting x in Eq. (iii), + Hence, x and (viii) We have, + and ( x + ) ( x ) 4 ( x + ) ( x ) 8 On putting u and v, we get ( x + ) ( x ) u + v 4 and u 8 v u v 4 On adding Eqs. (i) and (ii), we get u 4 4 u 4 From Eq. (i), + v 4 4 v 4 Now, u and v 4 and x + 4 x x + 4 (iii) and x (iv) On adding Eqs. (iii) and (iv), we get 6x 6 x Substituting x in Eq. (iii), we get + 4 Hence, x and Question. Formulate the following problems as a pair of equations and hence find their solutions. (i) Ritu can row downstream 0 km in hours and upstream 4 km in hours. Find her speed of rowing in still water and the speed of the current. (ii) women and men can together finish an embroider work in 4 das, while women and 6 men can finish it in das. Find the time taken b woman alone to finish the work and also that taken b man alone.
(iii) Roohi travels 00 km to her home partl b train and partl b bus. She takes 4 hours, if she travels 60 km b train and the remaining b bus. If she travels 00 km b train and the remaining b bus, she takes 0 min longer. Find the speed of the train and the bus separatel. Solution (i) Let the speed of Ritu in still water x kmh and speed of current kmh Then, speed downstream ( x + ) kmh Speed upstream ( x ) kmh Time Speed Distance 0 and 4 (B condition) x + x x + 0 and x On adding Eqs. (i) and (ii), we get x x 6 On putting x 6 in Eq. (i), 6 + 0 4 Speed of Ritu in still water 6 kmh Speed of current 4 kmh (ii) Let woman finish the work in x das and let man finish the work in das. Work of woman in da x Work of man in da x + Work of women and men in one da + x x x The number of das required for complete work x + x We are given that, 4 x + Similarl, in second case x (Given) 6x + x + Then, and 6 x + x 4 x 0 8 + and 8 + 9 x x On putting x u and v
0v + 8u and 8v + 9u On multipling Eq. (i) b 9 and Eq. (ii) b 8, then subtracting later from first, we get 80v 44v 9 8 6v v 6 On substituting v in Eq. (ii), we get 6 8 + 9u 6 9u u 8 Now, u and v 8 6 and x 8 6 x 8 and 6 Hence, single woman finishes the work in 8 das and single man finishes the work in 6 das. (iii) Let the speed of the train x kmh and the speed of the bus kmh In first case, Roohi travels 60 km b train and 40 km b bus in 4 h. 60 40 Thus, + 4 x 60 + (B given condition) x Similarl, in second case, she travels 00 km b train and 00 km b bus in 4 6 h. 00 00 + 4 + Q0 min h x 6 6 00 00 + x 6 4 48 + (Multipl b 6 x ) On putting x u and v, we get u + 60v (iii) and 4u + 48v (iv)
On multipling Eq. (i) b 4 and Eq. (iv) b, then subtracting from Eq. (iv) 4 ( u + 60v) ( 4u + 48v) 4 440 v 70 v 9 70v 9 9 v 70 80 Substituting v in Eq. (iii), we get 80 u + 60 80 u 4 4 u 60 Now, u and v 60 80 x 60 and 80 x 60 and 80 Hence, the speed of the train 60 kmh and the speed of the bus 80 kmh
Exercise.7 (Optional)* Question. The ages of two friends Ani and Biju differ b ears. Ani s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cath. The ages of Cath and Dharam differ b 0 ears. Find the ages of Ani and Biju. Solution Let the ages of Ani and Biju be x ear and ear respectivel. According to the given condition, x ± Also, age of Ani s father Dharam x ear and age of Biju s sister ear According to the given condition, x 0 4x 60 Case I When x (iii) On subtracting Eq. (iii) from Eq. (i), we get x 7 x 9 r On putting x 9 in Eq. (iii), we get 9 6 r x 9 r and 6 r Case II When x (iv) On subtracting Eq. (iv) from Eq. (ii), we get x 60 + x 6 x On putting x in Eq. (iv), we get 4 r Hence, age of Ani is 9 r and age of Biju is 6 r or age of Ani is r and age of Biju is 4 r. Question. One sas, Give me a hundred, friend! I shall then become twice as rich as ou. The other replies, If ou give me ten, I shall be six times as rich as ou. Tell me what is the amount of their (respective) capital? [From the Bijaganita of Bhaskara II] [Hint x + 00 ( 00), + 0 6 ( x 0)].
Solution Let the amount of their respective capital s be `x and `. According to the given condition, x + 00 ( 00) x 00 and 6 ( x 0) + 0 6x 70 On multipling Eq. (i) b and subtracting from Eq. (i) x x 00 40 x 440 x ` 40 On putting x 40 in Eq. (i), we get 40 00 40 ` 70 Hence, the amount of their respective capitals are ` 40 and ` 70. Question. A train covered a certain distance at a uniform speed. If the train would have been 0 kmh faster, it would have taken hours less than the scheduled time. And, if the train were slower b 0 kmh ; it would have taken hours more than the scheduled time. Find the distance covered b the train. Solution Let the actual speed of the train be x kmh and actual time taken b r. Q Distance Speed Times ( x) km According to the given condition, x ( x + 0)( ) x x x + 0 0 x 0 + 0 0 x + 0 0 (Divide b ) and x ( x 0)( + ) x x + x 0 0 x 0 0 0 On multipling Eq. (i) b and subtracting Eq. (ii) from Eq. (i), ( x + 0) ( x 0 0) 0 60 On putting in Eq. (i), we get x + 0 0 x 60 + 0 0 x 0 Hence, the distance covered b the train 0 600 km
Question 4. The students of a class are made to stand in rows. If students are extra in a row, there would be row less. If students are less in a row, there would be rows more. Find the number of students in the class. Solution Let the number of students in the class be x and the number of rows be. The number of rows in each row x According to the given condition, x x + ( ) x x x + x + 0 and x x ( + ) x x x + 6 x 6 0 Putting x u in Eqs. (i) and (ii), u + 0 (iii) and u 6 0 (iv) On subtracting Eq. (iii) from Eq. (iv), u 9 0u 9 On substituting u 9 in Eq. (iii), we get 9 + 0 4 (v) Now, u 9 x 9 x 9 [From Eq. (v), 4] 4 x 6 The number of students in the class x 6 4 44 Question. In a ABC, C, B ( A + B). Find the three angles. Solution We have, C B ( A + B)
QThe sum of the three angles of a triangle is 80. A + B + C 80 A + B + ( A + B) 80 [From Eq. (i)] A + B 80 A + B 60 (Divide b ) Again A + B + C 80 A + B + B 80 [From Eq. (i)] A + 4 B 80 (iii) On subtracting Eq. (ii) from Eq. (iii), B 0 B 40 On putting B 40 in Eq. (ii), we get A + 40 60 A 0 On putting B 40 in Eq. (i), we get C 40 0 Hence, the angles are A 0, B 40 and C 0. Question 6. Draw the graphs of the equations x and x. Determine the coordinates of the vertices of the triangle formed b these lines and the Y-axis. Solution The given equations are x and x Table for x Table for x x 0 0 Points A B x 0 0 Points C D Now, we plot the points A (, 0 ) and B (, 0 ) on a graph paper and join these points to form a line AB. Also we plot the points C (, 0 ) and D (, 0 ) on the same graph paper and join these points form the line CD.
Y X' D (0, ) (, 0) x x C A (,0) X B (0, ) Required triangle is of ABD and whose vertices are A (, 0 ), B (, 0 ) and D (, 0 ). Question 7. Y' Solve the following pair of linear equations (i) px + q p q (ii) ax + b c qx p p + q bx + a + c (iii) x 0 (iv) ( a b) x + ( a + b) a ab b a b ax + b a + b ( a + b)( x + ) a + b (v) x 78 74 78x + 604 Solution (i) Given pair of linear equations is px + q p q and qx p p + q On multipling Eq. (i) b p and Eq. (ii) b q and then adding, we get p x + q x ( p pq) + ( pq + q ) x ( p + q ) p + q x On putting x in Eq. (i), we get p + q p q q q Hence, solution is x and. (ii) Given pair of linear equations is ax + b c and bx + a + c On multipling Eq. (i) b b and Eq. (ii) b a and subtracting Eq. (ii) from Eq. (i), b a bc ( a + ac) bc a ac ( b a ) b a a + ac bc a b
On putting the value of in Eq. (i), we get a + ac bc ax + b c a b ab + abc b c ax c a b a c b c ab abc + b c ax a b a ( ac b bc) ax a b ac b bc x a b ac b bc a + ac bc Hence, solution is x and. a b a b (iii) Given pair of linear equations is x 0 a b and ax + b a + b bx a 0 and ax + b a + b On multipling Eq. (i) b b and Eq. (ii) b a and adding, we get bx+ ax aa ( + b) aa ( + b) x a ( b + a ) On putting x ain Eq. (i), we get ba a 0 b Hence, solution is x aand b. (iv) Given pair of linear equations is ( a b) x + ( a + b) a ab b and ( a + b)( x + ) a + b x ( a + b) + ( a + b) a + b On subtracting Eq. (ii) from Eq. (i), we get ( a b) x ( a + b) x ab b bx b ( a + b) x ( a + b) On putting x ( a + b)in Eq. (i), we get ( a b) ( a + b) + ( a + b) a ab b a b + ( a + b) a ab b ab a + b Hence, the solution is x a + b and ab. a + b
7 x + (v) Given pair of linear equations is x 78 74 and 78x + 604 On adding Eqs. (i) and (ii), we get 6x 6 678 x + (Divide b 6) (iii) On subtracting Eq. (ii) from Eq. (i), we get 0x 0 0 x (Divide b 0) (iv) On adding Eqs. (iii) and (iv), we get x 4x On putting x in Eq. (iii), we get + Hence, the solution is x and. Question 8. ABCD is a cclic quadrilateral. Find the angles of the cclic quadrilateral. + 4x C D 4 + 0 A Solution Since, in cclic quadrilateral, the sum of two opposite angles is 80. B + D 80 and A + C 80 7x + 80 and 4 + 0 4x 80 7x 80 and 4 4x 60 7x 80 and x 40 (Divide b 4) On multipling Eq. (ii) b 7 and subtracting from Eq. (i), we get 4 80 80 4 00 On putting in Eq. (ii), we get x 40 x Hence, solution is x and. D