Question

❗❕A two-dogit number contains the smaller of the two digit in the units place the product of the digits is 40 and the difference between the digits is 3. find the number ❗❕

Plz those who don't know how to solve plz don't answer otherwise I'll report it to mods.... (╥﹏╥)

Answer 1

Hii

you are welcome in my ans

let smaller digit be having unit place x

let another digit having unit place be y

so number = 10y + x

now

according to question

10y *x = 40

x*y = 4

x = 4/y

10y - x = 3

10y - 4/y = 3

10y^2 - 3y -4 = 0

10y^2 - 8y + 5y - 4= 0

(5y - 4) ( 2y +1) = 0

y = 4/5 and -1/2

so

x = 5 and -8

there for

number = 10*4/5 + 5 = 8+5=13

or -10*1/2 - 8 = -13

__/\__☺️

Hope it may helps you

Answer 2

Answer:

10y ** = 40

10y ** = 40x*y = 4

10y ** = 40x*y = 4x = 4/y

10y ** = 40x*y = 4x = 4/y10y - X=3

10y ** = 40x*y = 4x = 4/y10y - X=310y - 4/y = 3

10y ** = 40x*y = 4x = 4/y10y - X=310y - 4/y = 310y^2 - 3y -4 = 0

10y ** = 40x*y = 4x = 4/y10y - X=310y - 4/y = 310y^2 - 3y -4 = 010y^2 - 8y + 5y - 4= 0

10y ** = 40x*y = 4x = 4/y10y - X=310y - 4/y = 310y^2 - 3y -4 = 010y^2 - 8y + 5y - 4= 0(5y - 4) (2y +1) = 0

10y ** = 40x*y = 4x = 4/y10y - X=310y - 4/y = 310y^2 - 3y -4 = 010y^2 - 8y + 5y - 4= 0(5y - 4) (2y +1) = 0y = 4/5 and -1/2

10y ** = 40x*y = 4x = 4/y10y - X=310y - 4/y = 310y^2 - 3y -4 = 010y^2 - 8y + 5y - 4= 0(5y - 4) (2y +1) = 0y = 4/5 and -1/2SO

10y ** = 40x*y = 4x = 4/y10y - X=310y - 4/y = 310y^2 - 3y -4 = 010y^2 - 8y + 5y - 4= 0(5y - 4) (2y +1) = 0y = 4/5 and -1/2SOx= 5 and -8

10y ** = 40x*y = 4x = 4/y10y - X=310y - 4/y = 310y^2 - 3y -4 = 010y^2 - 8y + 5y - 4= 0(5y - 4) (2y +1) = 0y = 4/5 and -1/2SOx= 5 and -8there for

10y ** = 40x*y = 4x = 4/y10y - X=310y - 4/y = 310y^2 - 3y -4 = 010y^2 - 8y + 5y - 4= 0(5y - 4) (2y +1) = 0y = 4/5 and -1/2SOx= 5 and -8there fornumber = 10*4/5 + 5 = 8+5=13

10y ** = 40x*y = 4x = 4/y10y - X=310y - 4/y = 310y^2 - 3y -4 = 010y^2 - 8y + 5y - 4= 0(5y - 4) (2y +1) = 0y = 4/5 and -1/2SOx= 5 and -8there fornumber = 10*4/5 + 5 = 8+5=13or -10*1/2 - 8 = -13