Question

x/8+y/5=31/40
in the equation above, if x and y are positive integers, what is the value of x+y?

Answer 1

Answer:

y = 80

x = 46

Step-by-step explanation:

$\frac{x}{8} + \frac{y}{5} = \frac{31}{40}$

$\frac{5x + 8y}{40} = \frac{31}{40}$

2ox + 32y = 1240

10x + 16y = 620

5x + 8y = 310

5x = 310 - 8y

$x = \frac{310 - 8y}{5}$

y = 80

$x = \frac{310 - 8(10)}{5} = \frac{310 - 80}{5} = \frac{230}{5} = 46$

Answer 2

Not something to solve per se. Solving implies one unknown but you have two. However another restriction is that x and y are positive integers.

Simply “fit” x and y to make 31 and only one of the combinations “fit” as such.

using x = 1, 2, 3 etc

so you get

5 + 26

10 + 21

15 + 16

20 + 11

25 + 6

Only 15 + 16 = 31 actually fits 5x + 8y = 31 if x = 3 and y = 2. Therefore x+y = 5