Hi,

The 'clue' to the problem is the numbers must end with something that

results in a zero in the hundredths place when multiplied: 

x + y = 2.7



 x^2 - 2.7x + 1.8 = 0   |

 

 x =  1.5  0r  x = 1.2

 



 

Answer by Maths68(1474)   (Show Source): You can put this solution on YOUR website!
 Let x and y be the two numbers

x+y=2.7

x+y=27/10

10(x+y)=27

10x+10y=27.............(1)





xy=1.8

xy=18/10

x=18/10y...............(2)





Put the value of x from (2) to (1)

10(18/10y)+10y=27

18/y+10y=27

(18+10Y^2)/y=27

18+10Y^2=27y

10y^2-27y+18=0



 
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation  (in our case ) has the following solutons:

  

  

  

  For these solutions to exist, the discriminant  should not be a negative number.

  

  First, we need to compute the discriminant : .

  

  Discriminant d=9 is greater than zero. That means that there are two solutions: .

  

    

    

    

    Quadratic expression  can be factored:

  

  Again, the answer is: 1.5, 1.2.
Here's your graph:





So

y=1.5 or 1.2





Put the values of y in (1)

10x+10y=27.............(1)

Put y=1.5

10x+10(1.5)=27

10x+15=27

10x=27-15

10x=12

x=12/10

x=1.2





Put the values of y in (1)

10x+10y=27.............(1)

Put y=1.2

10x+10(1.2)=27

10x+12=27

10x=27-12

10x=15

x=15/10

x=1.5



So numbers are 1.2 and 1.5







Check

=====

xy=1.8

(1.2)(1.5)=1.8

1.8=1.8





x+y=2.7

1.2+1.5=2.7

2.7=2.7 

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