Q: find the HCF of 81 and 237 and express it as a linear combination of 81 and 237 . Solution: 237=81*2+75..........(1) 81=75*1+6.........(2) 75=6*12+3..........(3) 6=3*2+0.........(4) so the HCF of 81 and 237 will be 3. now to expresss it as a linear combination of 81 and 237 . from(3)we have , 3=75-6*12 =>3=75-(81-75*1)*12 =>3=13*75-12*81 =>3=13*(237-81*2)-12*81 =>3=13*237-26*81-12*81 =>3=13*237-38*81 =>3=237x+81y, where x=130 and y=-38. please explain me the third step in the linear combination of 81 and 237 . i wnt to know from where 13 came ? also explain the last steps of linear combination .

See... 3 = 75 - 6 x 12 => 3 = 75 - (81 - 75 x 1) x 12...............(1) Now, apply multiplication and distribution rule i.e. a(b+c)=ab+bc ? (1) becomes => 3 = 75 - (81 x 12 - 75 x 12) Now open the bracket...so sign changes from -75 to +75 => 3 = 75 - 81 x 12 + 75 x 12 => 3 = 75 + 75 x 12 - 81 x 12 => 3 = 75 (1 + 12) - 81 x 12 => 3 = 75 x 13 - 81 x 12...............(2) 75 = 237 – 81 x 2 ? from your step 1..............(3) Substitue (3) in (2) => 3 = (237 - 81 x 2) x 13 - 81 x 12 Apply again multiplication and distribution rule, => 3 = 237 x 13 - 81 x 2 x 13 - 81 x 12 => 3 = 237 x 13 - 81 x 26 - 81 x 12 => 3 = 237 x 13 - 81 (26 + 12) => 3 = 237 x 13 - 81 x 38....................(4) HCF(237,81) = 3 = 237 x 13 + 81 x (-38) ? from (4)...........(5) Solve (4) and (5) the answer will be same i.e. 3 ? 3 = 237x + 81y where x = 13 and y = (-38) Hope u got it, Aditya!

This question is of r.d.sharma class 10 page 1.15 and 1.16 solved example 3 75-(81-75*1)*12 75-(81-75)*12 because 75*1=75 so now, 75-(81*12-75*12) because 12 is outside bracket so it will be multiplied by both nos. You must have heard that if minus sign is present before bracket then sign of no. Inside bracket changes. So, 75-81*12+75*12. Take this as equation 1. Now there are two 75 (1st no. in equation 1)->75 and then 75*12 75*12+75(the first no. In equation) Do 75 *12+75 the result will be the same as 75*13 so it is written 75*13 In the last step it is written 237*13-38*81 since 237 and 81 are nos. You are finding h.c.f. Of and 13 and -38 are nos. Which will be multiplied by both nos. So x=13 y=-38 please approve my answer by clicking approved if you liked it.

See... 3 = 75 - 6 x 12 => 3 = 75 - (81 - 75 x 1) x 12................(1) Now, apply the multiplication and distribution rule i.e. a(b+c)=ab+bc ? (1) becomes, => 3 = 75 - (81 x 12 - 75 x 12) Open up the bracket. So -75 changes to +75 => 3 = 75 - 81 x 12 + 75 x 12 => 3 = 75 + 75 x 12 - 81 x 12 Take 75 out coz its common => 3 = 75 (1 + 12) - 81 x 12 => 3 = 75 x 13 - 81 x 12..............................(2) From YOUR step 1, 75 can be written as 75 = 237 - 81 x 2.................(3) Substitute (3) in (2) => 3 = (237 - 81 x 2) x 13 - 81 x 12 Again use multiplication and distribution rule => 3 = 237 x 13 - 81 x 2 x 13 - 81 x 12 => 3 = 237 x 13 - 81 x 26 - 81 x 12 => 3 = 237 x 13 - 81 (26 + 12) {81 common} => 3 = 237 x 13 - 81 x 38...............(4) => 3 = 237 x 13 + 81 x (-38)................(5) Solve (4) & (5) u get the same answer i.e. 3 ? HCF (237,81) = 3 = 237 x 13 + 81 x (-38)?from (5) ? 3 = 237x + 81y where x = 13 & y = (-38) Hope u got it Aditya.....