y= (1, 4) times x + (15, 4)
Let x and y be the 2 parts of 15 ==> x + y=15 (given)
Reciprocal of x and y ==> 1/x +1/y ==> 1/x + 1/y = 3/10 (given)
Let's solve 1/x + 1/y = 3/10 . Common denominator = 10.x.y (reduce to same denominator)
==> (10y+10x)/10xy = 3xy/10xy ==> 10x+10y =3xy
But x+y = 15 , then 10x+10y =150 ==> 150=3xy and xy = 50
Now we have the sum S of the 2 parts that is S = 15 and
their Product = xy =50
Let's use the quadratic equation for S and P==> X² -SX +P =0
Or X² - 15X + 50=0, Solve for X & you will find:
The 1st part of 15 is 10 & the 2nd part is 5
Answer:
48 ways
Step-by-step explanation:
Let me take a guess
S₁_₁₅ = (1+15)*7 + 8 = 120
There are 48 combinations of distinct digits from 1 to 15 to make 20
120-20=100
So every 20 has a corresponding 100
I wish I got it right, otherwise report it.