Question

One of the two numbers is 2 less than twice the other number. If the

product of these numbers equals to 40, find the numbers. Only algebraic

solutions will be accepted.

Answer 1

hey

let s note a and b the two numbers

we can write

(1) a = 2b - 2

(2) ab = 40

from (2) we can say for b different from 0 that

a = 40/b

replace it in (1)

40/b=2b-2

so

2(b-1)b = 40

b(b-1) = 20

$b^2-b-20 = 0$

5 is a trivial solution and as the product of the two solutions is -20 then the second solution is -4 therefore

$b^2-b-20 = 0$

<=> (b-5)(b+4) = 0

<=> b = 5 or b = -4

so we have

a= 40/b = 8 and b = 5

or

a = 40/b = -10 and b = -4

(8,5) and (-10.-4) are solutions

do not hesitate if you have any question

thanks