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PARĀVARTYA – YOJAYET
‘Paravartya – Yojayet’ means 'transpose
and apply'
(i) Consider the division by divisors of more than one digit, and when the
divisors are slightly greater than powers of 10. Example 1 : Divide 1225 by 12.
Step 1 : (From left to right ) write the
Divisor leaving the first digit, write the other digit or digits using negative
(-) sign and place them below the divisor as shown.
Example 2 : Divide
1697 by 14.12 -2 ¯¯¯¯ Step 2 : Write down the dividend to the right. Set apart the last digit for the remainder. i.e.,, 12 122 5 - 2 Step 3 : Write the 1st digit below the horizontal line drawn under the dividend. Multiply the digit by –2, write the product below the 2nd digit and add. i.e.,, 12 122 5 -2 -2 ¯¯¯¯¯ ¯¯¯¯ 10 Since 1 x –2 = -2 and 2 + (-2) = 0 Step 4 : We get second digits’ sum as ‘0’. Multiply the second digits’ sum thus obtained by –2 and writes the product under 3rd digit and add. 12 122 5 - 2 -20 ¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯ 102 5 Step 5 : Continue the process to the last digit. i.e., 12 122 5 - 2 -20 -4 ¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯ 102 1 Step 6: The sum of the last digit is the Remainder and the result to its left is Quotient. Thus Q = 102 and R = 1 14 1 6 9 7 - 4 -4–8–4 ¯¯¯¯ ¯¯¯¯¯¯¯ 1 2 1 3 Q = 121, R = 3. Example 3 : Divide 2598 by 123. Note that the divisor has 3 digits. So we have to set up the last two digits of the dividend for the remainder. 1 2 3 25 98 Step ( 1 ) & Step ( 2 ) -2-3 ¯¯¯¯¯ ¯¯¯¯¯¯¯¯ Now proceed the sequence of steps write –2 and –3 as follows : 1 2 3 2 5 9 8 -2-3 -4 -6 ¯¯¯¯¯ -2–3 ¯¯¯¯¯¯¯¯¯¯ 2 1 1 5 Since 2 X (-2, -3)= -4 , -6; 5 – 4 = 1 and (1 X (-2,-3); 9 – 6 – 2 = 1; 8 – 3 = 5. Hence Q = 21 and R = 15. Example 4 : Divide 239479 by 11213. The divisor has 5 digits. So the last 4 digits of the dividend are to be set up for Remainder. 1 1 2 1 3 2 3 9 4 7 9 -1-2-1-3 -2 -4-2-6 with 2 ¯¯¯¯¯¯¯¯ -1-2-1-3 with 1 ¯¯¯¯¯¯¯¯¯¯¯¯¯ 2 1 4 0 0 6 Hence Q = 21, R = 4006. Example 5 : Divide 13456 by 1123 1 1 2 3 1 3 4 5 6 -1–2–3 -1-2-3 ¯¯¯¯¯¯¯ -2-4 –6 ¯¯¯¯¯¯¯¯¯¯¯¯¯ 1 2 0–2 0 Note that the remainder portion contains –20, i.e.,, a negative quantity. To over come this situation, take 1 over from the quotient column, i.e.,, 1123 over to the right side, subtract the remainder portion 20 to get the actual remainder. Thus Q = 12 – 1 = 11, and R = 1123 - 20 = 1103.
Find the Quotient and
Remainder for the problems using paravartya – yojayet method.
1) 1234 ÷ 112 2) 11329 ÷ 1132 3) 12349 ÷ 133 4) 239479 ÷ 1203 Example 1 : Divide 6x2 + 5x + 4 by x – 1 X - 1 6x2 + 5x + 4 ¯¯¯¯¯¯ 1 6 + 11 ¯¯¯¯¯¯¯¯¯¯¯¯ 6x + 11 + 15 Thus Q = 6x+11, R=15. Example 2 : Divide x3 – 3x2 + 10x – 4 by x - 5 X - 5 x3 – 3x2 + 10x – 4 ¯¯¯¯¯ 5 5 + 10 100 ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ x2 + 2x + 20, + 96 Thus Q= x2 + 2x + 20, R = 96. The procedure as a mental exercise comes as follows :
i) x3 / x gives x2 i.e.,, 1 the first
coefficient in the Quotient.
Example 3:ii) Multiply 1 by + 5,(obtained after reversing the sign of second term in the Quotient) and add to the next coefficient in the dividend. It gives 1 X( +5) = +5, adding to the next coefficient, i.e.,, –3 + 5 = 2. This is next coefficient in Quotient. iii) Continue the process : multiply 2 by +5, i.e.,, 2 X +5 =10, add to the next coefficient 10 + 10 = 20. This is next coefficient in Quotient. Thus Quotient is x2 + 2x + 20 iv) Now multiply 20 by + 5 i.e.,, 20 x 5 = 100. Add to the next (last) term, 100 + (-4) = 96, which becomes R, i.e.,, R =9. x4 – 3x3 + 7x2 + 5x + 7 ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ x + 4 Now thinking the method as in example ( 1 ), we proceed as follows. x + 4 x4 - 3x3 + 7x2 + 5x + 7 ¯¯¯¯¯ -4 - 4 + 28 - 140 + 540 ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ x3 - 7x2 + 35x - 135 547 Thus Q = x3 – 7x2 + 35x – 135 and R = 547. or we proceed orally as follows: x4 / x gives 1 as first coefficient.
i) -4 X 1 = - 4 : add to next coefficient – 4 + (-3) = - 7
which gives next coefficient in Q.
Thus Q = x3 – 7x2 + 35x – 135 , R =
547.ii) – 7 X - 4 = 28 : then 28 + 7 = 35, the next coefficient in Q. iii) 35 X - 4 = - 140 : then – 140 + 5 = - 135, the next coefficient in Q. iv) - 135 X - 4 = 540 : then 540 + 7 = 547 becomes R. Note :
1. We can follow the same procedure even the number of terms
is more.
Now consider the divisors of second degree or more as in the following
example.2. If any term is missing, we have to take the coefficient of the term as zero and proceed. Example :4 2x4 – 3x3 – 3x + 2 by x2 + 1. Here x2 term is missing in the dividend. Hence treat it as 0 . x2 or 0 . And the x term in divisor is also absent we treat it as 0 . x. Now x2 + 1 2x4 - 3x3 + 0 . x2 - 3x + 2 x2 + 0 . x + 1 0 - 2 ¯¯¯¯¯¯¯¯¯¯¯¯ 0 - 1 0 + 3 0 + 2 ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 2 - 3 - 2 0 4 Thus Q = 2x2 - 3x - 2 and R = 0 . x + 4 = 4. Example 5 : 2x5 – 5x4 + 3x2 – 4x + 7 by x3 – 2x2 + 3. We treat the dividend as 2x5 – 5x4 + 0. x3 + 3x2 – 4x + 7 and divisor as x3 – 2x2 + 0 . x + 3 and proceed as follows : x3 – 2x2 + 0 . x + 3 2x5 – 5x4 + 0.x3 + 3x2 – 4x + 7 ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 2 0 - 3 4 0 - 6 -2 0 + 3 - 4 0 + 6 ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 2 - 1 - 2 - 7 - 1 +13 Thus Q = 2x2 – x – 2, R = - 7 x2 – x + 13. You may observe a very close relation of the method paravartya in this aspect with regard to REMAINDER THEOREM and HORNER PROCESS of Synthetic division. And yet paravartya goes much farther and is capable of numerous applications in other directions also.
Apply paravartya – yojayet to find out the Quotient and
Remainder in each of the following problems.
1) (4x2 + 3x + 5) ÷ (x+1) 2) (x3 – 4x2 + 7x + 6) ÷ (x – 2) 3) (x4 – x3 + x2 + 2x + 4) ÷ (x2 - x – 1) 4) (2x5 + x3 – 3x + 7) ÷ (x3 + 2x – 3) 5) (7x6 + 6x5 – 5x4 + 4x3 – 3x2 + 2x – 1) ÷ (x-1) Recall that 'paravartya yojayet' means 'transpose and apply'. The rule relating to transposition enjoins invariable change of sign with every change of side. i.e., + becomes - and conversely ; and X becomes ÷ and conversely. Further it can be extended to the transposition of terms from left to right and conversely and from numerator to denominator and conversely in the concerned problems. Type ( i ) : Consider the problem 7x – 5 = 5x + 1 7x – 5x = 1 + 5 i.e.,, 2x = 6 x = 6 ÷ 2 = 3. Observe that the problem is of the type ax + b = cx + d from which we get by ‘transpose’ (d – b), (a – c) and d - b. x = ¯¯¯¯¯¯¯¯ a - c In this example a = 7, b = - 5, c = 5, d = 1 Hence 1 – (- 5) 1+5 6 x = _______ = ____ = __ = 3 7 – 5 7-5 2 Example 2: Solve for x, 3x + 4 = 2x + 6 d - b 6 - 4 2 x = _____ = _____ = __ = 2 a - c 3 - 2 1 Type ( ii ) : Consider problems of the type (x + a) (x+b) = (x+c) (x+d). By paravartya, we get cd - ab x = ______________ (a + b) – (c + d) It is trivial form the following steps (x + a) (x + b) = (x + c) (x + d) x2 + bx + ax + ab = x2 + dx + cx + cd bx + ax – dx – cx = cd – ab x( a + b – c – d) = cd – ab cd – ab cd - ab x = ____________ x = _________________ a + b – c – d ( a + b ) – (c + d.) Example 1 : (x – 3) (x – 2 ) = (x + 1 ) (x + 2 ). By paravartya cd – ab 1 (2) – (-3) (-2) x = __________ = ______________ a + b – c –d - 3 – 2 – 1 – 2 2 - 6 - 4 1 = _______ = ___ = __ - 8 - 8 2 Example 2 : (x + 7) (x – 6) = (x +3) (x – 4). Now cd - ab (3) (-4) – (7) (-6) x = ___________ = ________________ a + b – c – d 7 + (-6) – 3 - (-4) - 12 + 42 30 = ____________ = ___ = 15 7 – 6 – 3 + 4 2 Note that if cd - ab = 0 i.e.,, cd = ab, i.e.,, if the product of the absolute terms be the same on both sides, the numerator becomes zero giving x = 0. For the problem (x + 4) (x + 3) = (x – 2 ) ( x – 6 ) Solution is x = 0 since 4 X 3 = - 2 X - 6. = 12 Type ( iii) : Consider the problems of the type ax + b m ______ = __ cx + d n By cross – multiplication, n ( ax + b) = m (cx + d) nax + nb = mcx + md nax - mcx = md – nb x( na – mc ) = md – nb md - nb x = ________ na - mc. Now look at the problem once again ax + b m _____ = __ cx + d n paravartya gives md - nb, na - mc and md - nb x = _______ na - mc Example 1: 3x + 1 13 _______ = ___ 4x + 3 19 md - nb 13 (3) - 19(1) 39 - 19 20 x = ______ = ____________ = _______ = __ na - mc 19 (3) - 13(4) 57 - 52 5 = 4 Example 2: 4x + 5 7 ________ = __ 3x + 13/2 8 (7) (13/2) - (8)(5) x = _______________ (8) (4) - (7)(3) (91/2) - 40 (91 - 80)/2 11 1 = __________ = _________ = ______ = __ 32 – 21 32 – 21 2 X 11 2 Type (iv) : Consider the problems of the type m n _____ + ____ = 0 x + a x + b Take L.C.M and proceed. m(x+b) + n (x+a) ______________ = 0 (x + a) (x +b) mx + mb + nx + na ________________ = 0 (x + a)(x + b) (m + n)x + mb + na = 0  (m + n)x = - mb - na -mb - na x = ________ (m + n) Thus the problem m n ____ + ____ = 0, by paravartya process x + a x + b gives directly -mb - na x = ________ (m + n) Example 1 : 3 4 ____ + ____ = 0 x + 4 x – 6 gives -mb - na x = ________ Note that m = 3, n = 4, a = 4, b = - 6 (m + n) -(3)(-6) – (4) (4) 18 - 16 2 = _______________ = ______ = __ ( 3 + 4) 7 7 Example 2 : 5 6 ____ + _____ = 0 x + 1 x – 21 gives -(5) (-21) - (6) (1) 105 - 6 99 x = ________________ = ______ = __ = 9 5 + 6 11 11
I . Solve the following problems using the sutra
Paravartya – yojayet.
1) 3x + 5 = 5x – 3 6) (x + 1) ( x + 2) = ( x – 3) (x – 4) 2) (2x/3) + 1 = x - 1 7) (x – 7) (x – 9) = (x – 3) (x – 22) 3) 7x + 2 5 8) (x + 7) (x + 9) = (x + 3 ) (x + 21) ______ = __ 3x - 5 8 4) x + 1 / 3 _______ = 1 3x - 1 5) 5 2 ____ + ____ = 0 x + 3 x – 4 1. Show that for the type of equations m n p ____ + ____ + ____ = 0, the solution is x + a x + b x + c - mbc – nca – pab x = ________________________ , if m + n + p =0. m(b + c) + n(c+a) + p(a + b) 2. Apply the above formula to set the solution for the problem Problem 3 2 5 ____ + ____ - ____ = 0 x + 4 x + 6 x + 5 m n m + n ____ + ____ = _____ x + a x + b x + c Now this can be written as, m n m n ____ + ____ = _____ + _____ x + a x + b x + c x + c m m n n ____ - ____ = _____ - _____ x + a x + c x + c x + b m(x +c) – m(x + a) n(x + b) – n(x + c) ________________ = ________________ (x + a) (x + c) (x + c) (x + b) mx + mc – mx – ma nx + nb – nx – nc ________________ = _______________ (x + a) (x + c) (x +c ) (x + b) m (c – a) n (b –c) ____________ = ___________ x +a x + b m (c - a).x + m (c - a).b = n (b - c). x + n(b - c).a x [ m(c - a) - n(b - c) ] = na(b - c) – mb (c - a) or x [ m(c - a) + n(c - b) ] = na(b - c) + mb (a - c) Thus mb(a - c) + na (b - c) x = ___________________ m(c-a) + n(c-b). By paravartya rule we can easily remember the formula. Example 1 : solve 3 4 7 ____ + _____ = ____ x + 1 x + 2 x + 3 In the usual procedure, we proceed as follows. 3 4 7 ____ + ____ = ____ x + 1 x + 2 x + 3 3(x + 2) + 4(x + 1) 7 ________________ = _____ (x + 1) (x + 2) x + 3 3x + 6 + 4x + 4 7 _____________ = ____ x2 + 2x + x + 2 x + 3 7x + 10 7 _________ = ____ x2 + 3x + 2 x + 3 (7x + 10) (x + 3) = 7(x2 + 3x + 2) 7x2 + 21x + 10x + 30 = 7x2 + 21x + 14. 31x + 30 = 21x + 14 31 x – 21 x = 14 – 30 i.e.,, 10x = - 16 x = - 16 / 10 = - 8 / 5 Now by Paravartya process 3 4 7 ____ + ____ = ____ ( ... N1 + N2 = 3+4 = 7 = N3) x + 1 x + 2 x + 3 mb( a – c ) + na ( b – c ) x = _____________________ here N1 = m = 3 , N2 = n = 4 ; m ( c – a ) + n ( c – b ) a = 1, b = 2, c = 3 3 . 2 ( 1 – 3 ) + 4 . 1 . ( 2 – 3) = __________________________ 3 ( 3 – 1 ) + 4 ( 3 – 2 ) 6 ( -2)+ 4 (-1) - 12 – 4 - 16 - 8 = _____________ = _______ = ____ = ___ 3 (2) + 4(1) 6 + 4 10 5 Example 2 : 3 5 8 ____ + ____ = _____ Here N1 + N2 = 3 + 5 = 8. x - 2 x – 6 x + 3 mb ( a – c ) + na ( b – c) x = _____________________ m ( c – a ) + n ( c – b ) 3 . ( -6 ) ( - 2 - 3 ) + 5 .( -2 ) ( -6 – 3 ) = __________________________________ 3 ( 3 – ( -2 ) ) + 5 ( 3 – ( - 6 ) ) 3 ( - 6 ) ( - 5 ) + 5 ( - 2 ) ( - 9 ) = ____________________________ 3( 3 + 2 ) + 5 ( 3 + 6 ) 90 + 90 = _______ = 180 / 60 = 3. 15 + 45 1) 2 3 5 ____ + ____ = ____ x + 2 x + 3 x + 5 2) 4 6 10 ____ + ____ = ____ x + 1 x + 3 x + 4 3) 5 2 3 ____ + ___ = ____ x - 2 3 - x x – 4 4) 4 9 15 _____ + _____ = _____ 2x + 1 3x + 2 3x + 1 But 3 (4) 2 (9) 2(15) ________ + ________ = _______ gives 3(2x + 1) 2( 3x + 2) 2(3x + 1) 12 18 30 _____ + _____ = _____ Now proceed. 6x + 3 6x + 4 6x + 2 Simultaneous simple equations: By applying Paravartya sutra we can derive the values of x and y which are given by two simultaneous equations. The values of x and y are given by ration form. The method to find out the numerator and denominator of the ratio is given below. Example 1: 2x + 3y = 13, 4x + 5y = 23.
i) To get x, start with y coefficients and the independent
terms and cross-multiply forward, i.e.,, right ward. Start from the upper row
and multiply across by the lower one, and conversely, the connecting link
between the two cross-products being a minus. This becomes numerator.
Example 2: 5x – 3y = 11i.e.,, 2x + 3y = 13 4x + 5y = 23 Numerator of the x – value is 3 x 23 – 5 x 13 = 69 – 65 = 4 ii) Go from the upper row across to the lower one, i.e.,, the x- coefficient but backward, i.e.,, leftward. Denominator of the x – value is 3 x 4 – 2 x 5 = 12 – 10 = 2 Hence value of x = 4 ÷ 2 = 2. iii) To get y, follow the cyclic system, i.e.,, start with the independent term on the upper row towards the x–coefficient on the lower row. So numerator of the y–value is 13 x 4 – 23 x 2 = 52 – 46 = 6. iv) The denominator is the same as obtained in Step(ii) i.e.,, 2. Hence value of y is 6÷2=3. Thus the solution to the given equation is x = 2 and y = 3. 6x – 5y = 09 Now Nr. of x is (-3) (9) – (5) (11) = - 27 + 55 = 28 Dr. of x is (-3) (6) – (5) (-5) = - 18 + 25 = 07 x = Nr ÷ Dr = 28 ÷ 7 = 4 and for y, Nr is (11) (6) – (9)(5) = 66 – 45 = 21 Dr is 7 Hence y = 21 ÷ 7 = 3. Example 3: solve 3x + y = 5 4x – y = 9 Now we can straight away write the values as follows: (1)(9) – (-1)(5) 9 + 5 14 x = _____________ = _____ = ___ = 2 (1)(4) – (3)(-1) 4 + 3 7 (5)(4) – (9)(3) 20 – 27 -7 y = ____________ = _______ = ___ = -1 (1)(4) – (3)(-1) 4 + 3 7 Hence x = 2 and y = -1 is the solution. Algebraic Proof: ax + by = m ……… ( i ) cx + dy = n ………. ( ii ) Multiply ( i ) by d and ( ii ) by b, then subtract adx + bdy = m.d cbx + dby = n.b ____________________ ( ad – cb ) .x = md – nb md - nb bn - md x = ______ = ______ ad - cb bc - ad Multiply ( i ) by c and ( ii ) by a, then subtract acx + bcy = m.c cax + day = n.a _____________________ ( bc – ad ) . y = mc - na mc - na y = ______ bc - ad You feel comfort in the Paravartya process because it avoids the confusion in multiplication, change of sign and such other processes. 1. 2x + y = 5 2. 3x – 4y = 7 3x – 4y = 2 5x + y = 4 3. 4x + 3y = 8 4. x + 3y = 7 6x - y = 1 2x + 5y = 11 |
