Question

A number is 5 times another number and their different is 32. find both numbers
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Answer 1

Step-by-step explanation:

let first no be x

and second no be y

ATQ =>

x= 5y

x-5y = 0..............(1)

x-y = 32...............(2)

subtract (2)&(1)=>

x-5y = 0

x-y = 32

- + -

____________

-4y = -32

y = 8

put y = 8 in (2)

x - y = 32

x -8 = 32

x = 40

Hope it helps uh

Answer 2

Answer:

One of them is 40 and another one is 8.

Step-by-step explanation:

Let 'x' be the first number and 'y' be the another

According to Qn,

x=5y

x-y=32

(**Note : x is larger number)

x-5y=0  .... (1)

x-y=32 .... (2)

(1)-(2) -> x-x-5y+y=0-32

             -4y=-32

                y=-32/-4

                 y=8

Put y=8 in eq. (1)

x-(5*8)=0

x-40=0

x=40